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Normal Moore Spaces, PMEA, and Consistency Strength

1 · Prerequisites

2 · Summary

This page develops the normal Moore space problem from its two topological ingredients, Bing's metrization theory and the Q-set construction of separable counterexamples, into the measure-theoretic and inner-model interfaces that decide its consistency strength. It is built on the choice-strength page that supplies the DMC/DC landscape, and on the measure-theory and large-cardinal pages listed in its prerequisites.

The topological spine is as follows. A Moore space is a regular $T_{1}$ space carrying a development, and developments may always be taken decreasing; a metrizable space is collectionwise normal, every Moore space is subparacompact, and a collectionwise normal Moore space is screenable. Any space with a $σ$ -cellular base is metrizable, by an explicit level metric, and this converts the screenability of a normal Moore space into metrizability: a normal screenable Moore space is metrizable, and hence so is every collectionwise normal Moore space. That last equivalence is Bing's classical theorem, and it makes the normal Moore space conjecture equivalent to the question whether every normal Moore space is collectionwise normal.

The counterexample side begins with Q-sets: an uncountable set of reals all of whose subsets are relatively $G_{δ}$ . Under Martin's axiom and the failure of the continuum hypothesis, every set of reals of cardinality $ω_{1}$ is a Q-set, and Bing's tangent-disk construction turns any uncountable Q-set into a separable normal nonmetrizable Moore space, whose axis part is closed discrete and therefore obstructs metrizability. So $M A + \neg C H$ refutes the normal Moore space conjecture.

The measure-theoretic side is the product measure extension axiom PMEA and its countably additive weakening PMEA- $σ$ : every fair-coin product measure on $2^{λ}$ extends to a full power-set measure with the stated additivity. The three-quarter separation estimate turns a normal space with a discrete family and small local bases into a collectionwise normal one, so PMEA makes every normal space of character below the continuum collectionwise normal, and PMEA- $σ$ already does so for first countable spaces. Since Moore spaces are first countable, PMEA- $σ$ alone proves the normal Moore space conjecture, and PMEA is consistent relative to a strongly compact cardinal.

The inner-model side records the covering interface that closes the circle: HYP is the combinatorial axiom combining a singular strong limit cardinal of countable cofinality, the $κ$ -continuum hypothesis, and a nonreflecting stationary set, and the ladders of that set can be separated level by level. The Dodd-Jensen covering and square package produces HYP data from the nonexistence of inner models with measurable cardinals. Fleissner's construction is then carried out in full: from the level data --- the cofinal sequence of cardinals, $2^{κ} = κ^{+}$ , the stationary set of cofinality- $ω$ ordinals and the ladder separation --- the page builds the space $F \cup Q$ with its basic sets $B (σ) = [σ] \cup C (σ)$ , proves the development and the uniform base, proves normality through the club and the two separation cases, and proves non-metrizability through the stafull extraction, the Erdős–Rado/Ramsey colouring and the closing chain. The CH instance is the case $κ = ω$ , $κ_{n} = n$ , with the ladder separation proved directly; $V = L$ therefore refutes the normal Moore space conjecture, the failure of the conjecture yields an inner model with a measurable cardinal, the metatheoretic consistency lower bound follows, and with the strongly compact upper bound this is the consistency-strength sandwich for NMSC.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-22Open item page →

Moore spaces and developments

Definition

Let $(X, T)$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Stars. For a family $U$ of subsets of $X$ and a set $A \subseteq X$ put $St (A, U) : = ⋃ {U \in U : U \cap A \neq \emptyset},$ the star of $A$ with respect to $U$ (Refinements, locally finite families, point-finite families, and star refinements); write $St (x, U)$ for $St ({x}, U)$ . If $U$ covers $X$ then $x \in St (x, U)$ , and $St (x, U)$ is exactly the union of the members of $U$ containing $x$ ; it is open as soon as the members of $U$ are open.

Development. A development for $X$ is a sequence $(G_{n})_{n \in N}$ of open covers of $X$ such that for every $x \in X$ and every open $D$ with $x \in D$ there is $n \in N$ with $St (x, G_{n}) \subseteq D .$ Equivalently: the family ${St (x, G_{n}) : n \in N}$ is a neighbourhood base at $x$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open). A development is decreasing when $G_{n + 1}$ refines $G_{n}$ for every $n$ (Refinements, locally finite families, point-finite families, and star refinements).

$X$ is developable when it admits a development, and a Moore space is a regular $T_{1}$ space (Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly) that is developable.

Every development can be made decreasing, and then it still is one. Let $(G_{n})$ be a development and let $G_{n}^{'}$ be the family of all intersections $G_{0} \cap \dots \cap G_{n}$ with $G_{i} \in G_{i}$ for $0 \leq i \leq n$ . Each $G_{n}^{'}$ is an open cover of $X$ ; a member of $G_{m}^{'}$ is contained in a member of $G_{i}$ whenever $i \leq m$ , so $G_{m}^{'}$ refines $G_{i}$ and $St (x, G_{m}^{'}) \subseteq St (x, G_{i}) (i \leq m);$ in particular $G_{m}^{'}$ refines $G_{i}^{'}$ for $i \leq m$ . Given $x \in D$ open, choose $n$ with $St (x, G_{n}) \subseteq D$ ; then $St (x, G_{n}^{'}) \subseteq D$ by the displayed inclusion. So $(G_{n}^{'})$ is a decreasing development, and a space is developable if and only if it has a decreasing development.

Developments give first countability. If $(G_{n})$ is a development, then for each $x$ the sets $St (x, G_{n})$ are open neighbourhoods of $x$ , and every open set containing $x$ contains one of them; hence ${St (x, G_{n}) : n \in N}$ is a countable local base at $x$ and $X$ is first countable (First countable space: a countable neighbourhood base at every point). In particular every Moore space is first countable.

Remarks

Conventions. Regular and normal name separation conditions alone in this library, with $T_{1}$ written separately; Moore space is defined as regular $T_{1}$ plus developable.
Why a development and not a metric. Both structures define the topology by countably many "approximations"; a development survives in spaces that carry no compatible metric, and the whole point of this page is that a later ZFC theorem Collectionwise normal Moore spaces are metrizable concerns collectionwise normal Moore spaces, including the regular $T_{1}$ hypotheses. It does not assert metrization of arbitrary developable spaces. That later theorem is separate from the ZF definitions and finite normalization argument here.
The normalization uses no choice. The family $G_{n}^{'}$ is described by a formula from the given $G_{0}, \dots, G_{n}$ , and per point one selects one member of each of finitely many covers, which is finite choice and hence available in $Z F$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Normalized families and collectionwise normality

Definition

Let $(X, T)$ be a topological space and let $A = {A_{i} : i \in I}$ be a family of subsets of $X$ that is pairwise disjoint: $A_{i} \cap A_{j} = \emptyset$ for all distinct $i, j$ .

$A$ is separated when it has a pairwise disjoint open expansion: there are open sets $U_{i} \supseteq A_{i}$ with $U_{i} \cap U_{j} = \emptyset$ for all distinct $i, j$ . For $I = \emptyset$ this is vacuous.
$A$ is normalized when every subunion can be separated from its complementary subunion: for every $J \subseteq I$ there are disjoint open $U, V \subseteq X$ with $⋃_{i \in J} A_{i} \subseteq U, ⋃_{i \in I ∖ J} A_{i} \subseteq V .$ The two halves $J$ and $I ∖ J$ play symmetric roles, so it is enough to test one representative of each complementary pair.
$X$ is collectionwise normal (cwn) when every discrete family of closed subsets of $X$ (Discrete families and $σ$ -locally-finite and $σ$ -discrete bases) is separated.

Remarks

Discrete closed families are normalized in a normal space. Let $F$ be a discrete family of closed sets and $J$ a set of indices. A discrete family is locally finite (Every discrete family is locally finite, so every $σ$ -discrete basis is $σ$ -locally finite), and a locally finite union of closed sets is closed (Locally finite families remain locally finite after taking closures, closure commutes with their union, and a locally finite union of closed sets is closed); hence $⋃_{i \in J} F_{i}$ and $⋃_{i \notin J} F_{i}$ are disjoint closed sets. If $X$ is normal (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly) they can be separated by disjoint open sets, so $F$ is normalized. Normality is thus the two-set case of the normalization of discrete closed families.
Collectionwise normality implies normality. For disjoint closed $A, B \subseteq X$ the two-member family ${A, B}$ is discrete: a point of $A$ has a neighbourhood missing $B$ , one of $B$ has a neighbourhood missing $A$ , and a point outside $A \cup B$ has a neighbourhood missing both, since $A$ and $B$ are closed. Separating that discrete family yields disjoint open sets containing $A$ and $B$ , so every cwn space is normal.
Separation implies normalization. If $U_{i} \supseteq A_{i}$ are pairwise disjoint open sets and $J \subseteq I$ , then $U : = ⋃_{i \in J} U_{i}$ and $V : = ⋃_{i \notin J} U_{i}$ are disjoint open sets containing the two subunions. Hence every separated family is normalized, and the two-member cases of the two conditions coincide.
No choice is hidden. The definitions distinguish between "there exist open sets $U_{i}$ " and "there is a family $i \mapsto U_{i}$ " only in the usual way: a separation of a family is a family of open sets, so it is a single function together with its verification, not an application of choice.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Metrizable spaces are collectionwise normal

Statement

In $Z F$ , every metrizable space (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not) is collectionwise normal (Normalized families and collectionwise normality).

Facts & Assumptions

Given: A metrizable space $(X, T)$ together with one metric $d$ with $T = T_{d}$ (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), and a discrete family $F = {F_{i} : i \in I}$ of closed subsets of $X$ (Discrete families and $σ$ -locally-finite and $σ$ -discrete bases).

[F1]

A discrete family is locally finite, and a locally finite union of closed sets is closed; hence every subunion $⋃_{i \in J} F_{i}$ is closed (Every discrete family is locally finite, so every $σ$ -discrete basis is $σ$ -locally finite, Locally finite families remain locally finite after taking closures, closure commutes with their union, and a locally finite union of closed sets is closed).

[F2]

Collectionwise normality asks that a discrete family of closed sets be separated, that is, have a pairwise disjoint open expansion (Normalized families and collectionwise normality).

[L1]

For nonempty $A \subseteq X$ and $u \in X$ the distance $d (u, A) = \inf {d (u, a) : a \in A}$ exists, is $\geq 0$ , and equals $0$ when $u \in A$ ; the map $u \mapsto d (u, A)$ differs by at most $d (u, v)$ at two points $u, v$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz).

[L2]

If $A \subseteq B$ are nonempty then $d (u, B) \leq d (u, A)$ , because every lower bound of the set of distances to $B$ is one for $A$ and the infimum is the greatest lower bound (Greatest lower bound (infimum)). If $A$ is closed and $u \notin A$ then $d (u, A) > 0$ : $d (u, A) = 0$ would let balls of every radius about $u$ meet $A$ , so $u$ would lie in the closure of $A$ and hence in $A$ (Open ball, closed ball and sphere in a metric space, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Proof

technique · direct

1.1

Fix $d$ and $F$ as in the Given. For each $i \in I$ put $G_{i} : = ⋃_{j \neq i} F_{j}$ , a possibly empty closed set by [F1].

givenF1

2.1

For each $i$ , define $U_{i}$ by cases: $U_{i} : = \emptyset$ if $F_{i} = \emptyset$ ; $U_{i} : = X$ if $F_{i} \neq \emptyset$ and $G_{i} = \emptyset$ ; and $U_{i} : = {x \in X : d (x, F_{i}) < \frac{1}{3} d (x, G_{i})}$ if both $F_{i}$ and $G_{i}$ are nonempty. This definition is a formula in $F_{i}$ and $G_{i}$ , so the assignment $i \mapsto U_{i}$ is a single definable function and no selection is used.

step 1.1L1

3.1

Each $U_{i}$ is open. The first two cases are clear. In the third, let $x \in U_{i}$ and put $c : = d (x, F_{i}) \geq 0$ and $h : = d (x, G_{i}) > 0$ , so $3 c < h$ by definition of $U_{i}$ ; set $r : = (h - 3 c) / 8 > 0$ . For $y$ with $d (x, y) < r$ we get $d (y, F_{i}) \leq c + r$ and $d (y, G_{i}) \geq h - r$ , and $3 c + 3 r = 3 c + \frac{3}{8} (h - 3 c) < \frac{1}{2} (h + 3 c) < h - r$ , where the last inequality is $3 c < h$ ; hence $3 d (y, F_{i}) < d (y, G_{i})$ and $y \in U_{i}$ . So every point of $U_{i}$ has a ball around it inside $U_{i}$ , and $U_{i}$ is open in the metric topology (Open ball, closed ball and sphere in a metric space, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

step 2.1L1

3.2

Each $F_{i} \subseteq U_{i}$ . If $F_{i} = \emptyset$ this is clear; if $G_{i} = \emptyset$ then $U_{i} = X$ ; otherwise $y \in F_{i}$ gives $d (y, F_{i}) = 0$ and $d (y, G_{i}) > 0$ because $G_{i}$ is closed and $y \notin G_{i}$ , so $0 < \frac{1}{3} d (y, G_{i})$ and $y \in U_{i}$ .

step 2.1L1L2

3.3

The sets $U_{i}$ are pairwise disjoint. Let $x \in U_{i} \cap U_{k}$ with $i \neq k$ . If either of the first two cases produced $U_{i}$ or $U_{k}$ , then one of them is empty and the other is $X$ only when every $F_{j}$ with $j \neq i$ is empty, in which case $U_{k} = \emptyset$ for $k \neq i$ ; so both sets are given by the third case. Then $F_{k} \subseteq G_{i}$ and $F_{i} \subseteq G_{k}$ give by [L2] that $d (x, F_{i}) < \frac{1}{3} d (x, G_{i}) \leq \frac{1}{3} d (x, F_{k})$ and $d (x, F_{k}) < \frac{1}{3} d (x, G_{k}) \leq \frac{1}{3} d (x, F_{i})$ , hence $d (x, F_{i}) < \frac{1}{9} d (x, F_{i})$ , so that $d (x, F_{i}) = 0$ and then $d (x, F_{k}) < 0$ , contradicting $d (x, F_{k}) \geq 0$ .

step 2.1L1L2

4.1

By steps 3.1, 3.2 and 3.3 the family $(U_{i})_{i \in I}$ is a pairwise disjoint open expansion of $F$ , so $F$ is separated and $X$ is collectionwise normal by [F2].

step 3.1step 3.2step 3.3F2∎

Remarks

The empty cases are real cases. If $F_{i} = \emptyset$ then $G_{i}$ may be everything, and the formula $\frac{1}{3} d (x, G_{i})$ with $G_{i} = X$ would give $d (x, X) = 0$ and force $U_{i} = \emptyset$ ; the first case records that directly. If all other $F_{j}$ are empty, $G_{i} = \emptyset$ and the distance $d (x, G_{i})$ is undefined, which is why the second case is separated out. Both are decided by the given data, so no choice enters.
No choice anywhere. One metric is fixed by the hypothesis, the sets $U_{i}$ are defined by a formula, and the three cases are decided by definable conditions; the argument therefore runs in $Z F$ .

TheoremStatement: Literature-sourcedProof: Literature-sourcedaudited 2026-09-22Open item page →

Moore spaces are subparacompact

Statement

In $Z F C$ every Moore space is subparacompact: every open cover of the space has a refinement that is a countable union of discrete families of closed sets and covers the space (Moore spaces and developments, Discrete families and $σ$ -locally-finite and $σ$ -discrete bases).

The single use of choice is the well-ordering of the given open cover, which is why the statement is formulated in $Z F C$ rather than $Z F$ (The Axiom of Choice).

Facts & Assumptions

Given: A Moore space $X$ , a decreasing development $(G_{n})_{n \in N}$ of $X$ , and an open cover $U$ of $X$ together with a well-ordering $<_{W}$ of the set $U$ itself.

[F1]

A Moore space is regular $T_{1}$ and developable, and a development may be assumed decreasing: every member of $G_{m}$ is contained in a member of $G_{n}$ when $n \leq m$ (Moore spaces and developments).

[F2]

A development's stars form a local base: for $x$ and open $D ∋ x$ there is $n$ with $St (x, G_{n}) \subseteq D$ ; each $St (x, G_{n})$ is open and contains $x$ (Moore spaces and developments, Refinements, locally finite families, point-finite families, and star refinements).

[F3]

A family is discrete when every point has a neighbourhood meeting at most one member (Discrete families and $σ$ -locally-finite and $σ$ -discrete bases), and a countable union of discrete families is what the conclusion asks for.

[L1]

Well-ordering principle: since $W$ well-orders $U$ , every nonempty subfamily of $U$ has a $W$ -least element, and "the $W$ -least $U$ with a property" is a definable description (The Axiom of Choice).

[L2]

Point $z$ lies in $\overline{F}$ exactly when every neighbourhood of $z$ meets $F$ ; consequently an open set disjoint from $F$ is disjoint from $\overline{F}$ (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set).

Proof

technique · direct

1.1

Fix $(G_{n})$ , $U$ and $<_{W}$ . For $n \in N$ and $U \in U$ put $F (U, n) : = {x \in X : x \in U, x \notin V for every V <_{W} U, St (x, G_{n}) \subseteq U}$ .

givenF1L1

2.1

Every $F (U, n)$ is contained in $U$ , and $⋃_{U \in U, n} F (U, n) = X$ : given $x$ , let $U$ be the $<_{W}$ -least cover member containing $x$ , which exists by [L1], and choose $n$ with $St (x, G_{n}) \subseteq U$ by [F2]; then $x \in F (U, n)$ .

step 1.1F2L1

2.2

Every $F (U, n)$ is closed. Let $z \in \overline{F (U, n)}$ and let $G \in G_{n}$ contain $z$ . By [L2] and openness of $G$ there is $y \in G \cap F (U, n)$ ; then $G \subseteq St (y, G_{n}) \subseteq U$ . Thus every member of $G_{n}$ containing $z$ lies in $U$ , so $St (z, G_{n}) \subseteq U$ and in particular $z \in U$ . If $V <_{W} U$ , then $V \cap F (U, n) = \emptyset$ by definition, and openness of $V$ with [L2] gives $z \notin V$ . Hence $z \in F (U, n)$ by step 1.1.

step 1.1F2L2

2.3

For fixed $n$ the family ${F (U, n) : U \in U}$ is discrete. Let $x \in X$ , let $V$ be the $<_{W}$ -least cover member containing $x$ , and choose $k \geq n$ with $St (x, G_{k}) \subseteq V$ . Suppose $y \in St (x, G_{k}) \cap F (U, n)$ . Some $G \in G_{k}$ contains $x, y$ ; as $G_{k}$ refines $G_{n}$ , some $H \in G_{n}$ contains $G$ . Hence $x \in St (y, G_{n}) \subseteq U$ , so minimality gives either $V = U$ or $V <_{W} U$ . But $y \in V$ , and the second alternative contradicts the defining exclusion in $F (U, n)$ . Therefore $U = V$ , and the open neighbourhood $St (x, G_{k})$ meets at most the one family member $F (V, n)$ .

step 1.1F1F2F3L1

3.1

The family $⋃_{n} {F (U, n) : U \in U}$ is a countable union of discrete families of closed sets (steps 2.2 and 2.3), covers $X$ (step 2.1), and refines $U$ because $F (U, n) \subseteq U$ . Hence $X$ is subparacompact.

step 2.1step 2.2step 2.3∎

Remarks

Where the choice is spent. The development is a single given sequence and the sets $F (U, n)$ are defined by a formula, but the well-ordering $W$ of the cover is an application of the well-ordering principle and is used in step 2.1 to select the least cover member containing a point. Without it the same construction is not available, which is why the item is stated over $Z F C$ .
Discreteness, not just local finiteness. The argument produces, for each $n$ , one open neighbourhood of each point meeting at most one member, which is discreteness and not merely local finiteness; no local-finiteness closure lemma is needed, because closedness of each $F (U, n)$ is proved directly in step 2.2.

LemmaStatement: Literature-sourcedProof: Literature-sourcedaudited 2026-09-22Open item page →

Collectionwise normal Moore spaces are screenable

Statement

In $Z F C$ every collectionwise normal Moore space is screenable: every open cover of the space has a refinement that is a countable union of pairwise disjoint families of open sets and covers the space (Moore spaces and developments, Normalized families and collectionwise normality, The Axiom of Choice).

Facts & Assumptions

Given: A collectionwise normal Moore space $X$ with a decreasing development $(G_{n})_{n \in N}$ (Moore spaces and developments) and an open cover $H = {H_{α} : α \in A}$ well-ordered by $W$ .

[F1]

Each $St (x, G_{n})$ is open and contains $x$ , and for open $D ∋ x$ there is $n$ with $St (x, G_{n}) \subseteq D$ ; members of $G_{m}$ are contained in members of $G_{n}$ when $n \leq m$ (Moore spaces and developments, Refinements, locally finite families, point-finite families, and star refinements).

[F2]

Collectionwise normality: every discrete family of closed sets has a pairwise disjoint open expansion (Normalized families and collectionwise normality, Discrete families and $σ$ -locally-finite and $σ$ -discrete bases).

[L1]

$z \in \overline{F}$ exactly when every neighbourhood of $z$ meets $F$ ; hence an open set disjoint from $F$ is disjoint from $\overline{F}$ (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set).

[L2]

The well-ordering $W$ provides least elements, so "the $W$ -least $H$ with a property" is a definable description (The Axiom of Choice).

Proof

technique · direct

1.1

Fix $(G_{n})$ , $H$ and $W$ . For $n \in N$ and $α \in A$ put $X (α, n) : = {x \in X : x \in H_{α}, x \notin H_{β} for all β < α, every G \in G_{n} with x \in G satisfies G \subseteq H_{α}}$ .

givenF1L2

2.1

$⋃_{α, n} X (α, n) = X$ , and $X (α, n) \subseteq H_{α}$ for all $α, n$ : given $x$ , let $α$ be the $W$ -least index with $x \in H_{α}$ and choose $n$ with $St (x, G_{n}) \subseteq H_{α}$ ; every $G \in G_{n}$ containing $x$ is then contained in $H_{α}$ , so $x \in X (α, n)$ .

step 1.1F1L2

2.2

Each $X (α, n)$ is closed. Let $z \in \overline{X (α, n)}$ and $G \in G_{n}$ with $z \in G$ . By [L1] there is $y \in G \cap X (α, n)$ ; then $z, y \in G \in G_{n}$ , and since every member of $G_{n}$ containing $y$ lies in $H_{α}$ , we get $G \subseteq H_{α}$ ; hence every member of $G_{n}$ containing $z$ lies in $H_{α}$ . Also $z \in St (z, G_{n}) \subseteq H_{α}$ , and $z \notin H_{β}$ for $β < α$ because $H_{β} \cap X (α, n) = \emptyset$ with $H_{β}$ open and [L1]. So $z \in X (α, n)$ .

step 1.1F1L1

2.3

For each fixed $n$ the family ${X (α, n) : α \in A}$ is discrete. Let $x \in X$ , let $β$ be $W$ -least with $x \in H_{β}$ and choose $k \geq n$ with $St (x, G_{k}) \subseteq H_{β}$ . If $y \in St (x, G_{k}) \cap X (α, n)$ , pick $G \in G_{k}$ with $x, y \in G$ ; since $G$ lies in some $H \in G_{n}$ , we have $x \in St (y, G_{n}) \subseteq H_{α}$ , so $β \leq α$ . Also $y \in St (x, G_{k}) \subseteq H_{β}$ , so if $β < α$ then $y \in H_{β}$ contradicts $y \in X (α, n)$ ; hence $α = β$ and the open neighbourhood $St (x, G_{k})$ of $x$ meets at most one member.

step 1.1F1

3.1

For each $n$ , apply [F2] to the discrete family ${X (α, n) : α \in A}$ of closed sets, obtaining pairwise disjoint open sets $W (α, n) \supseteq X (α, n)$ , and put $Y (α, n) : = W (α, n) \cap H_{α}$ . Then each $Y (α, n)$ is open, contains $X (α, n)$ , lies in $H_{α}$ , and the family ${Y (α, n) : α \in A}$ is pairwise disjoint.

step 2.2step 2.3F2

4.1

Since $⋃_{α, n} X (α, n) = X$ by step 2.1, the family $⋃_{n} {Y (α, n) : α \in A}$ covers $X$ , refines $H$ by step 3.1, and is a countable union of pairwise disjoint families of open sets. Hence the arbitrary open cover $H$ has such a refinement and $X$ is screenable.

step 2.1step 2.2∎

Remarks

Bing's Theorem 9 is steps 1.1-4.1. The sets $X (α, n)$ are Bing's $x (h, i)$ , each $X_{i}$ is his discrete family of closed sets, and the well-order of the cover is exactly where choice enters; the proof of closedness follows his displayed argument, with the closure criterion used at the two places where an open set disjoint from a member must remain disjoint from its closure.
Collectionwise normality is used once, in step 3.1, and it is applied to a family of closed sets; the expansion is then intersected with the corresponding cover member so that the refinement property survives. A merely normal space would not suffice at this step.

LemmaStatement: AI-adaptedProof: AI-generatedaudited 2026-09-22‡ rests on unproved material (inherited)Open item page →

‡ Rests on 2 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Under choice, every metric space has a σ-discrete basis and ‡ Under choice, a regular T₁ space with a σ-locally-finite basis has a compatible normal sequence. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

A sigma-cellular base metrizes a normal Moore space

Statement

Assume $Z F C$ . Let $X$ be a normal Moore space carrying a base of the form $⋃_{n \in N} B_{n}$ (Basis and subbasis for a topology, and the topology generated by a family of sets) in which every $B_{n}$ is a pairwise disjoint family of open sets. Then $X$ is metrizable.

The normality and development hypotheses are essential to this conclusion: a $T_{1}$ space with a sigma-disjoint base need not be metrizable.

Facts & Assumptions

Given: A normal Moore space $X$ , a development $(G_{i})_{i \in N}$ , and a base $⋃_{n} B_{n}$ whose levels are pairwise disjoint open families.

[F1]

A Moore space is regular and $T_{1}$ and has a development: every $G_{i}$ covers $X$ , and for every open $O ∋ x$ some $i$ satisfies $St (x, G_{i}) \subseteq O$ (Moore spaces and developments).

[F2]

For every open $O ∋ x$ , some member of the displayed base contains $x$ and is contained in $O$ (Basis and subbasis for a topology, and the topology generated by a family of sets).

[F3]

If $A$ is closed, $U$ is open, $A \subseteq U$ , and $X$ is normal, then there is open $D$ with $A \subseteq D \subseteq \overline{D} \subseteq U$ (A space is normal if and only if every closed $A$ inside an open $U$ admits an open $V$ with $A \subseteq V \subseteq \overline{V} \subseteq U$ , Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly).

[F4]

A family is discrete when every point has a neighbourhood meeting at most one member; a sigma-discrete open basis of a regular $T_{1}$ space yields a compatible metric in $Z F C$ (Discrete families and $σ$ -locally-finite and $σ$ -discrete bases, Under choice, a space is metrizable if and only if it is regular, $T_{1}$ , and has a $σ$ -discrete basis, The Axiom of Choice).

Proof

technique · direct

1.1

For $n, i \in N$ , put $B_{n}^{*} = ⋃ B_{n}$ , $W_{n} = X ∖ B_{n}^{*}$ , and $X_{n, i} = X ∖ ⋃ {G \in G_{i} : G \cap W_{n} \neq \emptyset}$ . The set $W_{n}$ is closed and $X_{n, i}$ is closed. Also $X_{n, i} \subseteq B_{n}^{*}$ : a member of the cover $G_{i}$ containing a point of $X_{n, i}$ is disjoint from $W_{n}$ .

givenF1

2.1

Every point of $B_{n}^{*}$ belongs to some $X_{n, i}$ . Indeed, if $x \in B \in B_{n}$ , choose $i$ with $St (x, G_{i}) \subseteq B$ by [F1]. Every member of $G_{i}$ containing $x$ is then disjoint from $W_{n}$ , which is precisely $x \in X_{n, i}$ . Empty levels cause no exception: then $B_{n}^{*} = X_{n, i} = \emptyset$ .

F1step 1.1

2.2

Apply [F3] to the closed set $X_{n, i}$ inside the open set $B_{n}^{*}$ and obtain open $D_{n, i}$ with $X_{n, i} \subseteq D_{n, i} \subseteq \overline{D_{n, i}} \subseteq B_{n}^{*}$ .

F3step 1.1

3.1

The family $H_{n, i} = {D_{n, i} \cap B : B \in B_{n}}$ is a discrete family of open sets. A point outside $\overline{D_{n, i}}$ has an open neighbourhood missing every member. A point of $\overline{D_{n, i}}$ lies in $B_{n}^{*}$ by step 2.2 and hence in a unique $B_{0} \in B_{n}$ ; the open neighbourhood $B_{0}$ meets no $D_{n, i} \cap B$ with $B \neq B_{0}$ .

givenF4step 2.2

4.1

The countable union $⋃_{n, i} H_{n, i}$ is an open sigma-discrete basis. To verify the basis property, let $O$ be open and $x \in O$ . By [F2] choose $n$ and $B \in B_{n}$ with $x \in B \subseteq O$ . Step 2.1 gives $i$ with $x \in X_{n, i} \subseteq D_{n, i}$ , so $x \in D_{n, i} \cap B \subseteq O$ and this set belongs to $H_{n, i}$ .

F2F4step 2.1step 2.2step 3.1

5.1

By [F1], $X$ is regular and $T_{1}$ . The sigma-discrete basis of step 4.1 therefore satisfies the reverse direction of Bing's metrization theorem [F4], so $X$ admits a compatible metric.

F1F4step 4.1∎

Remarks

Why the naive block metric fails. Although $B_{n}^{*}$ is open, its complement $W_{n}$ need not be open. Thus $B_{n} \cup {W_{n}}$ need not be an open partition, and agreement on those blocks does not directly define the original topology. Normality and the development are exactly what replace each cellular level by the countable family of discrete open families in step 3.1.
Source route. Step 3.1 is Bing's normal-development conversion from screenable to strongly screenable (Theorem 8). Step 5.1 uses the sigma-discrete-basis form of his metrization theorem (Theorem 3).

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22‡ rests on unproved material (inherited)Open item page →

Normal screenable Moore spaces are metrizable

Statement

In $Z F C$ every normal (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly) screenable Moore space is metrizable. Here a space is screenable when every open cover has a refinement that is a countable union of pairwise disjoint families of open sets and covers the space (Refinements, locally finite families, point-finite families, and star refinements).

Facts & Assumptions

Given: A normal Moore space $X$ with a decreasing development $(G_{n})_{n \in N}$ (Moore spaces and developments), and for every open cover $U$ of $X$ a refinement $⋃_{i \in N} H_{i}$ of $U$ by pairwise disjoint families of open sets covering $X$ .

[F1]

A Moore space is regular $T_{1}$ and developable, and a development may be taken decreasing (Moore spaces and developments). Together with the normality in the Given clause, this supplies the normal-Moore-space hypothesis of A sigma-cellular base metrizes a normal Moore space.

[F2]

A base of a space is a family of open sets such that every open set is a union of members; equivalently every point of an open set has a base member between it and the set (Basis and subbasis for a topology, and the topology generated by a family of sets).

[F3]

Screenability applied to an open cover yields a covering refinement that is a countable union of pairwise disjoint open families; a refinement of a cover is a family each of whose members lies in a member of the cover (Refinements, locally finite families, point-finite families, and star refinements, the definition of screenable in the Statement).

[L1]

Countably many choices are available: selecting one screening of $G_{n}$ for each $n \in N$ uses countable choice, which is a theorem of $Z F C$ (The Axiom of Choice).

Proof

technique · direct

1.1

Fix $(G_{n})$ and, for each $n$ , a screening of the open cover $G_{n}$ , say $⋃_{i \in N} H_{n, i}$ , where each $H_{n, i}$ is a pairwise disjoint family of open sets and $⋃_{i} H_{n, i}$ covers $X$ and refines $G_{n}$ .

givenF3L1

2.1

The family $⋃_{(n, i) \in N \times N} H_{n, i}$ is a base for $X$ : let $D$ be open and $x \in D$ . Choose $n$ with $St (x, G_{n}) \subseteq D$ and then $i$ and $H \in H_{n, i}$ with $x \in H$ . Since $H_{n, i}$ refines $G_{n}$ there is $G \in G_{n}$ with $H \subseteq G$ , so $x \in G$ and therefore $G \subseteq St (x, G_{n}) \subseteq D$ ; hence $x \in H \subseteq D$ and $H$ is a member of the displayed family.

step 1.1F1F2F3

3.1

Each $H_{n, i}$ is a pairwise disjoint family of open sets, and the index set $N \times N$ is countable, so the base of step 2.1 is $σ$ -cellular in the sense of A sigma-cellular base metrizes a normal Moore space.

step 2.1

4.1

The Given clause and [F1] say that $X$ is a normal Moore space, and step 3.1 supplies its $σ$ -cellular base. Therefore A sigma-cellular base metrizes a normal Moore space supplies a metric whose metric topology is the topology of $X$ , so $X$ is metrizable.

givenstep 3.1F1∎

Remarks

Relation to Bing's route. Bing proves this theorem by showing that a normal screenable developable space is strongly screenable (Theorem 8), that strongly screenable developable spaces are perfectly screenable (Theorem 6), and that perfectly screenable regular spaces are metrizable (Theorems 3 and 7). Steps 1.1-2.2 above are the first two of those reductions in the equivalent language of a $σ$ -cellular base, and step 3.1 replaces Bing's displayed weighted metric by the explicit level metric of A sigma-cellular base metrizes a normal Moore space; the conclusion is the same.
Where normality enters. Screenability produces the $σ$ -cellular base in steps 1.1-3.1. Normality is then an essential hypothesis of A sigma-cellular base metrizes a normal Moore space, whose proof uses normal shrinking to turn the cellular levels into a $σ$ -discrete base. Thus normality is used at step 4.1 rather than in the screening construction itself.
Choice. The only choice is the countable selection of one screening per development level in step 1.1; the metric is then defined by a formula.

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22‡ rests on unproved material (inherited)Open item page →

Collectionwise normal Moore spaces are metrizable

Statement

In $Z F C$ , every collectionwise normal Moore space is metrizable (Normalized families and collectionwise normality, Moore spaces and developments, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

Facts & Assumptions

Given: A collectionwise normal Moore space $X$ .

[F1]

Collectionwise normality implies normality: for disjoint closed $A, B$ the two-member family ${A, B}$ is discrete, and separating it gives disjoint open sets containing $A$ and $B$ (Normalized families and collectionwise normality, Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly, Discrete families and $σ$ -locally-finite and $σ$ -discrete bases).

[F2]

Every collectionwise normal Moore space is screenable (Collectionwise normal Moore spaces are screenable).

[F3]

Every normal screenable Moore space is metrizable (Normal screenable Moore spaces are metrizable).

Proof

technique · direct

1.1

The space $X$ is normal by [F1] and screenable by [F2]; it is a Moore space by hypothesis.

givenF1F2

2.1

By [F3] applied to the normal screenable Moore space $X$ , the space is metrizable.

step 1.1F3∎

Remarks

This is Bing's Theorem 10 through Theorem 8. The separate screenability lemma is the combinatorial half of Bing's Theorem 10, and the metrization theorem is his Theorem 8 with the metrization criterion of Theorem 3; the two items are kept apart because the first carries the well-ordering argument and the second carries the metric construction.
No recorded result is used. Both suppliers are items of this page, proved before this one; the argument is short only because the work sits in those two items.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-22Open item page →

Q-sets and Bing's tangent-disk Moore-space interface

Definition

Work in $Z F C$ . All Euclidean distances are the usual ones on $R^{2}$ .

Q-sets. A subset $E \subseteq R$ is a Q-set when $E$ is uncountable and every subset $A \subseteq E$ is relatively $G_{δ}$ : there are open sets $V_{n} \subseteq R$ , $n \in N$ , with $A = E \cap ⋂_{n \in N} V_{n} .$ (Equivalently, replacing $V_{n}$ by $⋂_{k \leq n} V_{k}$ , the sets may be assumed decreasing.) Nothing here asserts that a Q-set exists; existence is proved from Martin's axiom together with $\neg C H$ in the later item Martin's axiom produces an uncountable Q-set.

The tangent-disk space. Let $E \subseteq R$ and put $Z (E) : = (R \times (0, \infty)) \cup (E \times {0}),$ with the subspace convention that $E \times {0}$ is the axis part and $P : = R \times (0, \infty)$ the open part. For $n \in N$ with $n \geq 1$ and $p \in Z (E)$ define $U (p, n) : = {\begin{cases} {z \in P : dist (p, z) < 1 / n} & p \in P, \\ {p} \cup {z \in P : dist ((p_{1}, 1 / n), z) < 1 / n} & p = (p_{1}, 0) \in E \times {0} . \end{cases}$ The tangent-disk topology (the Moore plane topology restricted to $Z (E)$ ) is the topology generated by the basis $B : = {U (p, n) : p \in Z (E), n \geq 1}$ (Basis and subbasis for a topology, and the topology generated by a family of sets, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). This is a basis: every point belongs to its own sets $U (p, n)$ . At an axis point in an intersection, both sets are tangent disks based at that same point, and the disk of radius $1 / \max (n, m)$ lies in both. At a point of $P$ , their intersection is Euclidean open in $P$ and contains a sufficiently small $U (p, k)$ . Therefore unions of these sets form a topology, since intersections are unions of such smaller members. In this topology a point $p = (a, 0)$ of the axis part has the tangent disk $U (p, n)$ as basic neighbourhoods, an open disk whose boundary is internally tangent to the axis at $(a, 0)$ , together with that point; the axis part is closed in $Z (E)$ and each $U (p, n)$ meets it in ${p}$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

The pair $(E, Z (E))$ is the interface recorded here: development, separability, normality and nonmetrizability of $Z (E)$ for uncountable Q-sets $E$ are proved in Bing's Q-set space is a normal nonmetrizable Moore space.

Remarks

Why the tangent disks. The closure of the Euclidean disk $B ((a, 1 / n), 1 / n)$ touches the axis exactly at $(a, 0)$ , so $U ((a, 0), n) \cap (R \times {0}) = {(a, 0)}$ ; the axis points are therefore isolated within the axis part, and the axis part is a closed discrete subspace of $Z (E)$ . That discreteness is what turns an uncountable Q-set into a nonmetrizable Moore space, and the Q-set property is exactly what restores normality.
The open part is Euclidean. $P$ carries its usual metric topology (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Open ball, closed ball and sphere in a metric space), and it is dense in $Z (E)$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

Martin's axiom extends families almost disjoint from a subfamily

Statement

Assume $M A$ (Martin's Axiom at a cardinal and Martin's Axiom) and let $B \subseteq P (ω)$ be a family with $∣ a \cap b ∣ < ω for all distinct a, b \in B,$ and $∣ B ∣ < c$ . Then for every $A \subseteq B$ all of whose members are infinite there is $d \subseteq ω$ with $∣ a \cap d ∣ = ω (a \in A), ∣ b \cap d ∣ < ω (b \in B ∖ A) .$

Facts & Assumptions

Given: An almost disjoint family $B$ of subsets of $ω$ with $∣ B ∣ < c$ , a subfamily $A \subseteq B$ , and Martin's axiom.

[F1]

$M A$ is the scheme $M A (κ)$ for every infinite $κ < c$ , where $M A (κ)$ says that every nonempty ccc partial order and every family of at most $κ$ dense subsets has a filter meeting all of them (Martin's Axiom at a cardinal and Martin's Axiom).

[F2]

$∣ B ∣ < c$ and $ω < c$ give $∣ B ∣ \cdot ω < c$ , and there is a bijection $ω \times ω \to ω$ ; cardinal arithmetic is in $Z F C$ (Cardinal (initial ordinal) and cardinality, The natural numbers $N$ (von Neumann), The Axiom of Choice).

[L1]

Finite subsets of $ω$ and finite subsets of $B$ form sets, and a condition is a pair $(s, F)$ of such sets; a subset of $ω$ is a function-like set of natural numbers (A function is a relation $f$ with $(a, b) \in f$ and $(a, c) \in f$ implying $b = c$ ; $f : A \to B$ , the value $f (a)$ , domain and codomain, The cardinality $∣ A ∣$ of a finite set).

Proof

technique · direct

1.1

Let $P$ be the set of pairs $(s, F)$ with $s \subseteq ω$ finite and $F \subseteq B ∖ A$ finite, ordered by $(s, F) \leq (s^{'}, F^{'})$ if and only if $s \supseteq s^{'}$ , $F \supseteq F^{'}$ and $(s ∖ s^{'}) \cap ⋃ F^{'} = \emptyset$ . This is a partial order with greatest (weakest) element $(\emptyset, \emptyset)$ ; every condition is below it in the stronger-smaller convention.

givenL1

2.1

$P$ is ccc, indeed a countable union of centered sets: conditions with the same first coordinate $s$ are pairwise compatible, since for $(s, F_{1})$ and $(s, F_{2})$ the pair $(s, F_{1} \cup F_{2})$ is a common extension; and there are only countably many finite $s \subseteq ω$ .

step 1.1L1

2.2

For $a \in A$ and $n \in ω$ the set $D_{a, n} : = {(s, F) \in P : ∣ s \cap a ∣ \geq n}$ is dense. Given $(s^{'}, F^{'})$ , the set $a ∖ ⋃ F^{'}$ is infinite: $a$ is infinite by hypothesis, the finite family $F^{'}$ is disjoint from $A$ by step 1.1, and each $b \in F^{'}$ is distinct from $a$ and hence meets $a$ in a finite set; choose $s : = s^{'} \cup s^{''}$ where $s^{''}$ is a set of $n$ elements of $a ∖ (s^{'} \cup ⋃ F^{'})$ , and put $F : = F^{'}$ . Then $(s, F) \leq (s^{'}, F^{'})$ because $s ∖ s^{'}$ avoids $⋃ F^{'}$ , and $∣ s \cap a ∣ \geq n$ .

step 1.1L1

2.3

For $b \in B ∖ A$ the set $E_{b} : = {(s, F) \in P : b \in F}$ is dense. Given $(s^{'}, F^{'})$ , the pair $(s^{'}, F^{'} \cup {b})$ lies in $P$ by step 1.1 and extends $(s^{'}, F^{'})$ : indeed $s^{'} \supseteq s^{'}$ , $F^{'} \cup {b} \supseteq F^{'}$ and $(s^{'} ∖ s^{'}) \cap ⋃ F^{'} = \emptyset$ . Once $b$ belongs to the second coordinate of a member of the filter, no later first coordinate adds an element of $b$ , so the intersection of $d$ with $b$ is computed from a single finite $s$ .

step 1.1L1

3.1

The family of dense sets ${D_{a, n} : a \in A, n \in ω} \cup {E_{b} : b \in B ∖ A}$ has cardinality at most $∣ B ∣ \cdot ω < c$ by [F2], and $P$ is ccc by step 2.1, so $M A$ gives a filter $G \subseteq P$ meeting all of them. Put $d : = ⋃ {s : (s, F) \in G}$ .

step 2.1step 2.2step 2.3F1F2

4.1

For $b \in B ∖ A$ we have $∣ d \cap b ∣ < ω$ : by step 2.3 and step 3.1 some $(s, F) \in G$ has $b \in F$ ; for any $(s^{'}, F^{'}) \in G$ , compatibility gives $(s^{''}, F^{''}) \in G$ extending both, and then $s^{'} \cap b \subseteq s^{''} \cap b = (s \cap b) \cup ((s^{''} ∖ s) \cap b) = s \cap b$ because $(s^{''} ∖ s) \cap b = \emptyset$ . As $(s^{'}, F^{'}) \in G$ was arbitrary, $d \cap b \subseteq s \cap b$ , so $d \cap b = s \cap b$ is a finite set.

step 2.3step 3.1

4.2

For $a \in A$ we have $∣ d \cap a ∣ = ω$ : for every $n$ step 2.2 and step 3.1 give $(s, F) \in G$ with $∣ s \cap a ∣ \geq n$ , and $s \subseteq d$ .

step 2.2step 3.1

5.1

Steps 4.1 and 4.2 exhibit $d \subseteq ω$ with the two required properties for the given $A \subseteq B$ .

step 4.1step 4.2∎

Remarks

Where the hypotheses are used. Only in step 2.2, and there twice over: adding new elements of $a$ is possible because $a$ is infinite and meets the finitely many members of $F^{'}$ — which lie outside $A$ — in finite sets. The restriction of the second coordinate to $B ∖ A$ is what keeps $D_{a, n}$ extendable, since a condition carrying $a$ in its second coordinate could never add an element of $a$ again. Without infiniteness of the members of $A$ the statement is false, as $A = {\emptyset}$ shows, and the consumer Martin's axiom produces an uncountable Q-set therefore uses a base of intervals in which every real lies in infinitely many members.
The filter is supplied by MA, not by choosing conditions. The extension conditions in steps 2.2 and 2.3 are explicit constructions, so no choice beyond the given filter is used; the filter itself comes from $M A$ .

LemmaStatement: Literature-sourcedProof: Literature-sourcedaudited 2026-09-22Open item page →

Martin's axiom produces an uncountable Q-set

Statement

In $Z F C + M A + \neg C H$ every subset $E \subseteq R$ of cardinality $ω_{1}$ is a Q-set (Q-sets and Bing's tangent-disk Moore-space interface); in particular an uncountable Q-set exists.

Facts & Assumptions

Given: Work in ZFC. Assume $M A + \neg C H$ and a subset $E \subseteq R$ with $∣ E ∣ = ω_{1}$ .

[F1]

$\neg C H$ says that there is a set $A$ with $N ≺ A ≺ P (N)$ (The continuum hypothesis, and what this page does not prove); under choice every uncountable cardinal is at least $ω_{1}$ , so $ω_{1} \leq ∣ A ∣ < 2^{ℵ_{0}}$ and hence $ω_{1} < 2^{ℵ_{0}}$ (The successor cardinal $κ^{+}$ , the alephs $ℵ_{α}$ , the beths $ℶ_{α}$ , successor and limit cardinals, and the identifications $ℵ_{0} = ω$ and $ℵ_{1} = ω_{1}$ , Cardinal (initial ordinal) and cardinality, The Axiom of Choice). Moreover $P (N)$ injects into $R$ : send a subset to its characteristic binary sequence, then use the stated bijection from binary sequences onto the Cantor subset of $R$ (The Cantor set is exactly the set of $\sum_{k \geq 1} a_{k} 3^{- k}$ with every $a_{k} \in {0, 2}$ , and this gives a bijection with ${0, 1}^{N}$ ).

[F2]

$M A$ is the scheme $M A (κ)$ for every infinite $κ < c$ (Martin's Axiom at a cardinal and Martin's Axiom).

[F3]

Solovay's almost-disjoint extension lemma: if $B \subseteq P (ω)$ is almost disjoint and $∣ B ∣ < c$ , then every $A \subseteq B$ all of whose members are infinite is extended by a $d \subseteq ω$ that is infinite on $A$ and finite on $B ∖ A$ (Martin's axiom extends families almost disjoint from a subfamily).

[L1]

Put $W (k, n) = (k / 2^{n} - 2^{- n - 1}, k / 2^{n} + 2^{- n - 1})$ , $k \in Z$ , $n \in N$ . Enumerate pairs without repetition: encode $k \geq 0$ by $2 k$ and $k < 0$ by $- 2 k - 1$ , and use the bijection of $N \times N \approx N$ . For each $x, n$ , the floor of $2^{n} x$ supplied by Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ shows that either $x$ lies in an interval of scale $n$ , or $x = (2 k + 1) / 2^{n + 1}$ is exactly a midpoint. In the latter case it is a grid center at every scale $m \geq n + 1$ and therefore belongs to an interval at every such scale. If no midpoint case occurs, it belongs at every scale. Thus every $x$ belongs to infinitely many distinct indexed intervals. These intervals form a base: a containing interval of sufficiently fine scale lies in any prescribed neighbourhood of $x$ , since its diameter is $2^{- n}$ and these diameters tend to zero. Indeed $2^{n} \geq n + 1$ by induction and For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ supplies arbitrarily small reciprocal bounds. Relative $G_{δ}$ and Q-set have the meaning of Q-sets and Bing's tangent-disk Moore-space interface.

[L2]

If $x \neq y$ both belong to $W (k, n)$ , then $∣ x - y ∣ < 2^{- n}$ . At each fixed scale the intervals are pairwise disjoint, so at most one contains both points. By the decay proved in [L1], only finitely many scales can satisfy $2^{- n} > ∣ x - y ∣$ . Because the pair enumeration has no repetitions, ${i : x, y \in W_{i}}$ is finite. This uses the triangle inequality on the real line and the explicit interval endpoints.

Proof

technique · direct

1.1

Fix $E$ with $∣ E ∣ = ω_{1}$ and the dyadic base $(W_{i})_{i \in ω}$ of [L1]; for $x \in E$ put $s (x) : = {i \in ω : x \in W_{i}}$ .

givenL1

2.1

$B : = {s (x) : x \in E}$ satisfies $∣ B ∣ \leq ω_{1} < 2^{ℵ_{0}}$ by [F1], and $∣ s (x) \cap s (y) ∣ < ω$ for distinct $x, y \in E$ by [L2]. Each $s (x)$ is infinite by [L1], so distinct points have distinct codes: equality would make their intersection infinite. Thus $x \mapsto s (x)$ is injective.

step 1.1F1L1L2

3.1

Let $X \subseteq E$ . Each $s (x)$ is infinite by [L1], so [F3] applies to $A : = {s (x) : x \in X} \subseteq B$ and gives $d \subseteq ω$ with $∣ s (x) \cap d ∣ = ω$ for $x \in X$ and $∣ s (z) \cap d ∣ < ω$ for $z \in E ∖ X$ .

step 2.1F3F2L1

4.1

If $d$ is infinite, enumerate it increasingly as $d = {p (0) < p (1) < \dots}$ with domain $ω$ ; if $d$ is finite then $X = \emptyset$ by step 3.1, and $X = E \cap \emptyset$ is relatively $G_{δ}$ trivially. In the infinite case, $X = E \cap ⋂_{n \in ω} ⋃_{k \geq n} W_{p (k)}$ : for $x \in X$ the set ${k : x \in W_{p (k)}}$ is infinite by step 3.1, so $x$ lies in every tail union; conversely, if $x \in E$ lies in every tail union, then ${k : x \in W_{p (k)}}$ is infinite, so $∣ s (x) \cap d ∣ = ω$ , and step 3.1 excludes $x \in E ∖ X$ .

step 3.1L1

5.1

The sets $⋃_{k \geq n} W_{p (k)}$ are open in $R$ , so step 4.1 exhibits every subset $X \subseteq E$ as a relative $G_{δ}$ set in $E$ ; hence $E$ is a Q-set, and since $∣ E ∣ = ω_{1}$ it is uncountable, so it is uncountable. Such an $E$ exists: [F1] gives an injection of $ω_{1}$ into $R$ , and its image has cardinality $ω_{1}$ . Hence an uncountable Q-set exists.

step 4.1L1F1∎

Remarks

Why $ω_{1}$ and not $c$ . The almost-disjoint lemma needs $∣ B ∣ < c$ ; a set of reals of cardinality $c$ has $2^{c}$ subsets but only $c$ $G_{δ}$ sets, so it cannot be a Q-set. Under $M A + \neg C H$ the cardinal $ω_{1}$ is below $c$ , which is exactly the range in which the lemma applies.
Consistency. Under $C H$ there are no Q-sets: a Q-set would give a separable normal nonmetrizable Moore space (Bing's Q-set space is a normal nonmetrizable Moore space), while Jones' argument refutes that when $2^{ℵ_{0}} < 2^{ℵ_{1}}$ , which $C H$ gives. Nothing in this item asserts a Q-set under $C H$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

Bing's Q-set space is a normal nonmetrizable Moore space

Statement

Work in $Z F C$ (The Axiom of Choice), as on this page. For every uncountable Q-set $E \subseteq R$ (Q-sets and Bing's tangent-disk Moore-space interface) the tangent-disk space $Z (E)$ is a separable normal nonmetrizable Moore space (Moore spaces and developments, Separability: the existence of an at most countable dense subset, Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly).

Facts & Assumptions

Given: An uncountable Q-set $E \subseteq R$ , the space $Z : = Z (E)$ with open part $P = R \times (0, \infty)$ and axis part $E_{0} = E \times {0}$ (Q-sets and Bing's tangent-disk Moore-space interface), and for $n \geq 1$ the open covers $H_{n} : = {V (z, n) : z \in Z}, V (z, n) : = {\begin{cases} B (z, \min (1 / n, z_{2} / 3)) & z \in P, \\ U (z, n) & z \in E_{0}, \end{cases}$ where $z_{2}$ is the second coordinate and $U (z, n)$ is the tangent disk. Put $G_{m} : = H_{m + 1}$ for every $m \in N$ , including $m = 0$ .

[F1]

The tangent disks $U ((a, 0), n)$ are basic open sets, $U ((a, 0), n) \cap E_{0} = {(a, 0)}$ , and they form a local base at $(a, 0)$ ; the balls $B (z, r) \cap P$ are basic open sets of $P$ and form a local base at $z \in P$ (Q-sets and Bing's tangent-disk Moore-space interface, Open ball, closed ball and sphere in a metric space, Basis and subbasis for a topology, and the topology generated by a family of sets, Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open).

[F2]

On $P$ , the topology is Euclidean: every point of an open set lies in an ordinary ball contained in that set. Distances satisfy the triangle inequality (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[F3]

$E_{0}$ is closed in $Z$ and is a Q-set-indexed axis; every subset of $E$ is relatively $G_{δ}$ in $E$ , and for $A = E \cap ⋂_{n} S_{n}$ with $S_{n}$ open, decreasing, and $A \subseteq S_{n}$ one has $A = E \cap ⋂_{n} S_{n}$ (Q-sets and Bing's tangent-disk Moore-space interface, Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[F4]

Normality means that every two disjoint closed sets have disjoint open neighbourhoods, including empty closed sets (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly).

[L1]

A development's stars are open sets containing the point, and closeness of a point to a closed set is tested by neighbourhoods; closures in $Z$ of subsets of $P$ are computed with $E_{0}$ closed (Moore spaces and developments, Q-sets and Bing's tangent-disk Moore-space interface, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[L2]

A countable dense set $D$ in a metric space gives a countable base of balls $B (d, 1 / k)$ , $d \in D$ , $k \geq 1$ : given $x \in B (x, δ)$ , choose $k$ with $2 / k < δ$ and then $d \in D \cap B (x, 1 / k)$ ; the ball $B (d, 1 / k)$ contains $x$ and lies in $B (x, δ)$ by the triangle inequality. Enumerate the pairs using the fixed countable enumeration of $D$ (Separability: the existence of an at most countable dense subset, Open ball, closed ball and sphere in a metric space).

Proof

technique · direct

1.1

Fix $E$ , $Z$ , $P$ , $E_{0}$ and the covers $H_{n}$ .

givenF1

2.1

Each $H_{n}$ is an open cover by open sets, because $z \in V (z, n)$ and each $V (z, n)$ is a basic open set by [F1].

step 1.1F1

2.2

The zero-based sequence $(G_{m})_{m \in N}$ is a development, with positive scale $n = m + 1$ . At $z = (a, 0) \in E_{0}$ , if $q \in Z$ satisfies $z \in V (q, n)$ and $q \neq z$ , then $q \in P$ and $dist (q, z) \geq q_{2}$ , while $z \in V (q, n)$ forces $dist (q, z) < q_{2} / 3$ , a contradiction; hence $St (z, H_{n}) = U (z, n)$ , and the tangent disks form a local base at $z$ by [F1], so every neighbourhood of $z$ contains a star. At $z \in P$ , every member $V (q, n)$ containing $z$ satisfies $V (q, n) \subseteq B (z, 2 / n)$ when $q \in P$ , and $V (q, n) \subseteq B (z, 4 / n)$ when $q \in E_{0}$ ; hence $St (z, H_{n}) \subseteq B (z, 4 / n)$ , which is contained in any prescribed Euclidean neighbourhood for $n$ large.

step 1.1F1L1

2.3

$Z$ is separable: the set of points of $P$ with both coordinates rational is countable and dense: every ordinary ball in $P$ contains a rational point, and the open disk part of $U ((a, 0), n)$ contains $(a, 1 / (2 n))$ , hence also a rational point of $P$ sufficiently close to it (Separability: the existence of an at most countable dense subset, Open ball, closed ball and sphere in a metric space).

step 1.1F1

2.4

$E_{0}$ is closed in $Z$ and discrete: it is the trace of the closed set $R \times {0}$ , and each of its points $z$ has the neighbourhood $U (z, 1)$ with $U (z, 1) \cap E_{0} = {z}$ by [F1].

step 1.1F1F3

2.5

(Axis separation.) Identify the axis with $E$ via $a \mapsto (a, 0)$ for this step. Let $A \subseteq E$ and put $B : = E ∖ A$ ; symbols for their neighbourhoods refer to the corresponding axis points. Since every subset of the Q-set $E$ is relatively $G_{δ}$ there are decreasing sequences $(S_{n})$ and $(T_{n})$ of open subsets of $R$ with $A = E \cap ⋂_{n} S_{n}$ and $B = E \cap ⋂_{n} T_{n}$ , so $A \subseteq S_{n}$ and $B \subseteq T_{n}$ for every $n$ ; for $n \in N$ put $B_{n} : = B ∖ S_{n}$ and $A_{n} : = A ∖ T_{n}$ . Fix $n$ . For $a \in A$ the set $S_{n}$ is open and contains $a$ , so let $l_{n} (a)$ be the least positive integer with $(a - 1 / l_{n} (a), a + 1 / l_{n} (a)) \subseteq S_{n}$ , and put $ε_{n} (a) = 1 / l_{n} (a)$ . This is a positive real even when $S_{n} = R$ . Every $b \in B_{n} = E ∖ S_{n}$ has $∣ a - b ∣ \geq ε_{n} (a)$ . Let $k_{n} (a)$ be the least positive integer with $2 / \sqrt{k_{n} (a)} \leq ε_{n} (a)$ . Then $U ((a, 0), k_{n} (a)) \cap U ((b, 0), 1) = \emptyset$ for every $a \in A$ and $b \in B_{n}$ : tangent disks of radii $1 / k$ and $1$ at axis points at distance $ρ$ are disjoint whenever $ρ \geq 2 / \sqrt{k}$ , because their centre distance $\sqrt{ρ^{2} + (1 - 1 / k)^{2}}$ is then at least $1 + 1 / k$ . Hence the open sets $V_{n} : = ⋃_{b \in B_{n}} U ((b, 0), 1)$ and $W_{n} : = ⋃_{a \in A_{n}} U ((a, 0), k_{n} (a))$ satisfy $B_{n} \subseteq V_{n}$ , $A_{n} \subseteq W_{n}$ , $U ((a, 0), k_{n} (a)) \cap V_{n} = \emptyset$ for $a \in A$ , $V_{n} \cap E_{0} = B_{n}$ , $W_{n} \cap E_{0} = A_{n}$ , and $V_{n} \cap W_{n} = \emptyset$ : a tangent disk meets the axis only at its own tangency point, and every disk occurring in $W_{n}$ was chosen disjoint from $V_{n}$ . Put $V : = ⋃_{n} (V_{n} ∖ ⋃_{k < n} cl (W_{k}))$ and $W : = ⋃_{n} (W_{n} ∖ ⋃_{k < n} cl (V_{k}))$ . Both are open and they are disjoint. If $x \in V_{n} ∖ ⋃_{k < n} cl (W_{k})$ and $x \in W_{m} ∖ ⋃_{k < m} cl (V_{k})$ with $n < m$ , then $x \in V_{n} \subseteq ⋃_{k < m} cl (V_{k})$ , contradicting the choice of $x \in W_{m}$ ; symmetrically for $m < n$ ; and $n = m$ is excluded by $V_{n} \cap W_{n} = \emptyset$ . Also $B \subseteq V$ and $A \subseteq W$ : for $b \in B$ the fact that $b \notin A = E \cap ⋂_{n} S_{n}$ gives an $m$ with $b \notin S_{m}$ , that is $b \in B_{m} \subseteq V_{m}$ , while $b \notin ⋃_{k < m} cl (W_{k})$ because $cl (W_{k}) \cap E_{0} = A_{k} \subseteq A$ is disjoint from $B$ ; symmetrically for $a \in A$ . The closure identities $cl (W_{k}) \cap E_{0} = A_{k}$ and $cl (V_{k}) \cap E_{0} = B_{k}$ hold because $A_{k} = E ∖ T_{k}$ and $B_{k} = E ∖ S_{k}$ are closed in the Euclidean subspace $E$ , not merely in the discrete axis: if a sequence of points of $W_{k}$ converges to $(c, 0) \in E_{0}$ , then its tangency points converge to $c$ , since a point $(x, y) \in U ((a, 0), r)$ with $r \leq 1$ satisfies $∣ x - a ∣^{2} < 2 r y - y^{2}$ , so $∣ x - a ∣ \to 0$ as $y \to 0$ ; hence $c$ lies in the Euclidean relatively closed set $A_{k}$ , and dually for $B_{k}$ .

step 1.1F1F3

3.1

Reduction to arbitrary closed sets. Let $C, D$ be disjoint closed subsets of $Z$ . By step 2.5 choose disjoint open $O, Q$ containing $C \cap E_{0}$ and $E_{0} ∖ C$ , respectively. For each $a \in C \cap E_{0}$ choose the least positive $n (a)$ with $U (a, n (a)) \subseteq O ∖ D$ , and put $R_{C} = ⋃_{a \in C \cap E_{0}} U (a, 2 n (a))$ . Its closure misses $D \cap E_{0}$ , since $R_{C} \subseteq O$ and $D \cap E_{0} \subseteq Q$ . Its closure also misses $D \cap P$ . Indeed, if a point $q = (x, y)$ lies in the disk of radius $r / 2$ , tangent at $a$ , where $r = 1 / n (a) \leq 1$ , then $(x - a_{1})^{2} + y^{2} < r y$ , so its distance to the complement of the disk of radius $r$ is at least $y / 2$ : the distance to the outer centre has square less than $r^{2} - r y$ , and $r - \sqrt{r^{2} - r y} = r y / (r + \sqrt{r^{2} - r y}) \geq y / 2$ . The outer disk misses $D$ . A sequence in $R_{C}$ converging to $d \in D \cap P$ would eventually have height at least $d_{2} / 2$ , and therefore distance at least $d_{2} / 4$ from $D$ , impossible. A closure point in $P$ supplies such a sequence by its ordinary ball base (AC is available). Thus $\overline{R_{C}} \cap D = \emptyset$ . Reversing the roles, using $Q$ around $D \cap E_{0}$ and $O$ around $C \cap E_{0}$ , gives open $R_{D}$ with $D \cap E_{0} \subseteq R_{D}$ and $\overline{R_{D}} \cap C = \emptyset$ .

step 2.5F1F2F4L1

3.2

$Z$ is $T_{1}$ and regular: distinct points are separated by small balls or tangent disks, using that $E_{0}$ is closed and $P$ is open; for $z = (a, 0) \in E_{0}$ and a closed $C ∌ z$ , choose $n$ with $U (z, n) \cap C = \emptyset$ by [F1], let $D$ be the closed disk of radius $1 / n$ about $(a, 1 / n)$ and put $U : = U (z, 2 n)$ and $V : = Z ∖ cl (U)$ ; every point of $cl (U)$ other than $z$ lies in $U (z, n) \subseteq Z ∖ C$ and $z \notin C$ , so $C \subseteq V$ , while $U \cap V = \emptyset$ . Together with step 2.2 the space $Z$ is a Moore space.

step 2.2F1F2L1

3.3

$Z$ is not metrizable. Suppose $d$ induces its topology. By step 2.3 the metric space is separable, so by [L2] it has a countable base, say $(W_{i})_{i \in ω}$ . For $z \in E_{0}$ the set $U (z, 1)$ is open and contains $z$ , so by [F1] some $i$ has $z \in W_{i} \subseteq U (z, 1)$ , and then $W_{i} \cap E_{0} = {z}$ by step 2.4; the map $z \mapsto \min {i : z \in W_{i} \subseteq U (z, 1)}$ is therefore a definable injection $E_{0} \to ω$ , contradicting the uncountability of $E$ and hence of $E_{0}$ .

step 2.3step 2.4F1L2

4.1

Every point of $C \cap P$ lies in an ordinary ball with rational centre and rational positive radius whose Euclidean closed ball is contained in $P ∖ D$ : first take a sufficiently small ball inside that open set, then a rational centre sufficiently close to the point and a rational radius between the required bounds. Such a closed ball has positive distance from the axis and is closed in $Z$ . The family of all such rational balls is at most countable and covers $C \cap P$ ; list it with empty sets as padding, and prepend $R_{C}$ . This gives $(I_{n})_{n \in N}$ covering $C$ with $\overline{I_{n}} \cap D = \emptyset$ . Similarly obtain $(J_{n})$ covering $D$ with $\overline{J_{n}} \cap C = \emptyset$ , starting with $R_{D}$ . The sets $I = ⋃_{n} (I_{n} ∖ ⋃_{k \leq n} \overline{J_{k}})$ and $J = ⋃_{n} (J_{n} ∖ ⋃_{k < n} \overline{I_{k}})$ are open, contain $C, D$ , and are disjoint: for a point in the terms indexed by $n, m$ , if $m \leq n$ the first term excludes $J_{m}$ , and if $n < m$ the second excludes $I_{n}$ . This includes empty traces and proves normality by [F4].

step 3.1F1F2F4

5.1

Steps 2.2, 3.2, 2.3, 3.3 and 4.1 show that $Z$ is a Moore space, separable, nonmetrizable and normal, as claimed.

step 2.2step 3.2step 2.3step 3.3step 4.1∎

Remarks

Disjointness of the tangent disks in step 2.5. Tangent disks of radii $1 / k$ and $1$ with centres $(a_{1}, 1 / k)$ and $(b_{1}, 1)$ are disjoint exactly when their centre distance $\sqrt{ρ^{2} + (1 - 1 / k)^{2}}$ is at least $1 + 1 / k$ , where $ρ = ∣ a_{1} - b_{1} ∣$ ; squaring, this is equivalent to $ρ \geq 2 / \sqrt{k}$ , which is the estimate used in the definition of $k_{n} (a)$ . Since a tangent disk meets the axis only at its tangency point, this also gives $V_{n} \cap E_{0} = B_{n}$ and $W_{n} \cap E_{0} = A_{n}$ .
Why normality needs the Q-set property. The relative- $G_{δ}$ presentations in step 2.5 provide the axis separation; steps 3.1 and 4.1 then establish full normality. No assertion that every arbitrary uncountable $E$ gives a nonnormal space is made.

TheoremStatement: Literature-sourcedProof: Literature-sourcedaudited 2026-09-22Open item page →

MA plus not-CH yields a normal nonmetrizable Moore space

Statement

$Z F C + M A + \neg C H$ proves that there is a separable normal nonmetrizable Moore space.

Facts & Assumptions

Given: $M A + \neg C H$ and a subset $E_{0} \subseteq R$ with $∣ E_{0} ∣ = ω_{1}$ .

[F1]

Every set of reals of cardinality $ω_{1}$ is a Q-set and an uncountable Q-set exists (Martin's axiom produces an uncountable Q-set, Q-sets and Bing's tangent-disk Moore-space interface).

[F2]

For every uncountable Q-set $E$ the tangent-disk space $Z (E)$ is a separable normal nonmetrizable Moore space (Bing's Q-set space is a normal nonmetrizable Moore space, Q-sets and Bing's tangent-disk Moore-space interface, Moore spaces and developments).

Proof

technique · direct

1.1

By [F1] the set $E_{0}$ , of cardinality $ω_{1}$ , is an uncountable Q-set.

givenF1

2.1

Applying [F2] to $E_{0}$ produces a separable normal nonmetrizable Moore space, namely the tangent-disk space $Z (E_{0})$ .

step 1.1F2∎

Remarks

This is the second of the two standard refutations of the normal Moore space conjecture, and the only one available at $ω_{1}$ . It uses no large cardinal, only $M A$ and the failure of $C H$ ; the construction is separable, in contrast with the $C H$ construction recorded elsewhere on this page.
The two ingredients are independent. The Q-set comes from Martin's axiom through almost-disjoint forcing (Martin's axiom extends families almost disjoint from a subfamily), and the space comes from Bing's tangent-disk construction; the theorem spends both.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

PMEA and PMEA-sigma

Definition

Work in $Z F C$ , and write $c = 2^{ℵ_{0}}$ (The Axiom of Choice, Cardinal (initial ordinal) and cardinality). For a cardinal $λ$ , put $2^{λ} = {x : x : λ \to {0, 1}}$ .

For finite $F \subseteq λ$ and $s : F \to {0, 1}$ , let $C (F, s) = {x \in 2^{λ} : x ∣_{F} = s}$ . The finite-cylinder sigma-algebra $C_{λ}$ is generated by these sets. It equals the cylinder sigma-algebra of Coordinate maps, finite-coordinate cylinders, and the cylinder $σ$ -algebra, since any subset of a finite binary coordinate space is a finite union of singleton assignments. In particular $C (\emptyset, \emptyset) = 2^{λ}$ .

Give ${0, 1}$ its full power set and the probability assigning each point mass $1 / 2$ . This is standard Borel: the discrete metric is complete and the finite space itself is a countable dense set, and its Borel sets are all its subsets (Standard Borel spaces). Thus Arbitrary products of standard Borel probability spaces, with index set $λ$ and these factors, supplies a unique probability $μ_{λ}$ on $C_{λ}$ with $μ_{λ} (C (F, s)) = 2^{- ∣ F ∣}$ . Indeed its finite marginals are the finite coin products; conversely these cylinder masses determine every such marginal by finite disjoint unions. This is the fair-coin product measure. For finite $λ$ it is normalized counting measure on $2^{λ}$ ; for $λ = 0$ it is the unit mass on the singleton empty function.

A full extension is a probability measure $ν : P (2^{λ}) \to [0, 1]$ restricting to $μ_{λ}$ on $C_{λ}$ . Here a probability measure has total mass $1$ and is countably additive in the sense of Measures on sigma-algebras. Its full domain makes it complete (Complete measure spaces), but completeness alone does not mean that every subset of the underlying space is measurable.

A full extension is $c$ -additive if for every pairwise disjoint family $(A_{j})_{j \in J}$ with $∣ J ∣ < c$ , $ν (⋃_{j \in J} A_{j}) = \sum_{j \in J} ν (A_{j}),$ where the sum denotes the supremum of the finite subsums. Equivalently its null ideal is closed under unions of fewer than $c$ sets. To verify this equivalence, well-order a null family and disjointify it by subtracting earlier members; the resulting sets remain null, so the displayed identity makes their union null. Conversely, for a disjoint family in a probability space, at most $n$ members have mass at least $1 / n$ for each integer $n \geq 1$ . Thus only countably many members have positive mass. Countable additivity sums those members, and the remaining fewer than $c$ null members have null union by the assumed null-ideal closure. This proves the identity. The cardinal bound is strictly less than $c$ ; no condition on increasing families of length $c$ is intended.

PMEA (the product measure extension axiom) asserts that for every cardinal $λ$ , $μ_{λ}$ has a $c$ -additive full extension.

PMEA- $σ$ asserts that for every cardinal $λ$ , $μ_{λ}$ has a countably additive full extension.

Remarks

PMEA implies PMEA- $σ$ . The additional null-ideal closure in PMEA is the clause available to later consumers dealing with fewer than continuum many null sets. The product-measure corollary constructs $μ_{λ}$ on $C_{λ}$ ; it does not construct either full extension asserted by these axioms. Neither axiom asks for a two-valued measure on the index set $λ$ .

LemmaStatement: Literature-sourcedProof: Literature-sourcedaudited 2026-09-22Open item page →

The PMEA three-quarter separation estimate

Statement

Assume PMEA (PMEA and PMEA-sigma). Let $X$ be a normal space (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly), let ${F_{i} : i \in I}$ be a discrete family of subsets of $X$ (Discrete families and $σ$ -locally-finite and $σ$ -discrete bases), and for each $x \in X$ let $U_{x}$ be a downwards-directed family of neighbourhoods of $x$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open) such that $∣ U_{x} ∣ < c$ and: whenever $G \subseteq X$ is open, $i \in I$ and $x \in F_{i} \subseteq G$ , there is $U \in U_{x}$ with $U \subseteq G$ . Then there is a function $x \mapsto U^{x}$ with $U^{x} \in U_{x}$ and $U^{x} \cap U^{y} = \emptyset$ whenever $x \in F_{i}$ , $y \in F_{j}$ with $i \neq j$ .

If $X$ is first countable (First countable space: a countable neighbourhood base at every point) and only PMEA- $σ$ is assumed, the same conclusion holds for countable families $U_{x}$ of open neighbourhoods of $x$ that are local bases.

Facts & Assumptions

Given: A normal space $X$ , a discrete family ${F_{i} : i \in I}$ , families $U_{x}$ of neighbourhoods of the points as in the statement, and a full extension $ν$ of the fair-coin product measure on $2^{I}$ (PMEA and PMEA-sigma).

[F1]

Under PMEA, $ν$ may be taken $c$ -additive; under PMEA- $σ$ it may be taken countably additive. Consequently, if ${A_{t}}$ is an upwards-directed family of subsets of $2^{I}$ of cardinality $< c$ in the first case and countable in the second, covering $2^{I}$ , then $\sup_{t} ν (A_{t}) = 1$ (Fremlin, Lemma 8E and the continuity-from-below consequence of PMEA and PMEA-sigma).

[F2]

$ν$ agrees with $μ_{I}$ on cylinders; in particular, for distinct $i, j \in I$ the event ${z : z (i) \neq z (j)}$ has measure $1 / 2$ (PMEA and PMEA-sigma).

[L1]

For each $z \in 2^{I}$ , the closures of $⋃_{z (i) = 1} F_{i}$ and $⋃_{z (i) = 0} F_{i}$ are disjoint. Indeed, if a point lay in both closures, a neighbourhood meeting at most one member of the discrete family would have to meet one member from each complementary subfamily, a contradiction. Normality therefore supplies disjoint open sets containing the two original subunions (Discrete families and $σ$ -locally-finite and $σ$ -discrete bases, Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly). No individual $F_{i}$ is asserted closed.

[L2]

Probabilities are monotone, and $ν (2^{I} ∖ A) = 1 - ν (A)$ (PMEA and PMEA-sigma).

Proof

technique · direct

1.1

For $z \in 2^{I}$ choose disjoint open $G_{z}, H_{z}$ containing respectively the closures of $⋃_{z (i) = 1} F_{i}$ and $⋃_{z (i) = 0} F_{i}$ , by normality via [L1]. In particular $F_{i} \subseteq G_{z}$ when $z (i) = 1$ and $F_{i} \subseteq H_{z}$ when $z (i) = 0$ .

givenL1

2.1

For $x \in F_{i}$ and $U \in U_{x}$ put $A (x, U) : = {z : z (i) = 1 and U \subseteq G_{z}, or z (i) = 0 and U \subseteq H_{z}}$ . If $V \subseteq U$ , then $A (x, V) \supseteq A (x, U)$ . Because $U_{x}$ is downwards directed, the events $A (x, U)$ are therefore upwards directed. They cover $2^{I}$ : for $z$ with $z (i) = 1$ we have $F_{i} \subseteq G_{z}$ , so the hypothesis on $U_{x}$ gives $U \subseteq G_{z}$ , and symmetrically for $z (i) = 0$ .

step 1.1given

3.1

For $x \in F_{i}$ there is $U^{x} \in U_{x}$ with $ν (A (x, U^{x})) > 3 / 4$ : the family ${A (x, U)}$ is upwards directed, of cardinality $< c$ , and covers $2^{I}$ , so its measures have supremum $1$ by [F1].

step 2.1F1

4.1

If $x \in F_{i}$ , $y \in F_{j}$ with $i \neq j$ , then $ν (A (x, U^{x}) \cap A (y, U^{y}) \cap {z : z (i) \neq z (j)}) > 0$ : the first two complements have measure strictly below $1 / 4$ by step 3.1, while the complement of the difference event has measure $1 / 2$ by [F2]. Hence the complement of the displayed intersection has measure strictly below $1 / 4 + 1 / 4 + 1 / 2 = 1$ by subadditivity, so the intersection has positive measure.

step 3.1F2L2

5.1

Choose $z$ in that intersection. Since $z (i) \neq z (j)$ , either $z (i) = 1$ , $z (j) = 0$ , so that $U^{x} \subseteq G_{z}$ , $U^{y} \subseteq H_{z}$ and $U^{x} \cap U^{y} \subseteq G_{z} \cap H_{z} = \emptyset$ , or the reverse. Hence $U^{x} \cap U^{y} = \emptyset$ .

step 4.1step 1.1

5.2

In the PMEA- $σ$ , first countable case the same argument applies with countable local bases. Such a base is downwards directed: for $U, V$ in the base, $U \cap V$ is a neighbourhood of $x$ , so some base member lies inside it. Thus the upwards-directed countable cover ${A (x, U) : U \in U_{x}}$ has a member of measure $> 3 / 4$ by countable additivity, and steps 4.1 and 5.1 are unchanged.

step 3.1step 4.1F1

6.1

Steps 3.1 and 5.1 give the required assignment under PMEA, and step 5.2 gives it under PMEA- $σ$ for first countable $X$ .

step 3.1step 5.1step 5.2∎

Remarks

The numbers. Two events of measure above $3 / 4$ overlap in measure above $1 / 2$ , and the difference event has measure exactly $1 / 2$ ; a point of the triple overlap separates the two chosen neighbourhoods. The companion page computes this arithmetic as an example.
Only two coordinates are used, through the measure of ${z : z (i) \neq z (j)}$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

PMEA makes normal low-character spaces collectionwise normal

Statement

Work in $Z F C$ (The Axiom of Choice), as on this page. Under PMEA every normal space of character below $c$ is collectionwise normal; under PMEA- $σ$ every first countable normal space is collectionwise normal (PMEA and PMEA-sigma, Normalized families and collectionwise normality, First countable space: a countable neighbourhood base at every point).

Facts & Assumptions

Given: A normal space $X$ ; under PMEA an open neighbourhood base $U_{x}$ at each $x$ of cardinality $χ (x, X) < c$ , and under PMEA- $σ$ with first countability a countable open local base $U_{x}$ at each $x$ ; and a discrete family $F = {F_{i} : i \in I}$ of closed subsets of $X$ . Such open bases may be used without increasing cardinality by replacing every base member $N$ with its interior, which is an open neighbourhood of $x$ contained in $N$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open).

[F1]

The three-quarter separation estimate: under PMEA with nonempty downwards-directed neighbourhood families $U_{x}$ of size less than $c$ satisfying the open-refinement hypothesis of [F2], and under PMEA- $σ$ for first countable $X$ with countable local bases, there is $x \mapsto U^{x}$ with $U^{x} \in U_{x}$ and $U^{x} \cap U^{y} = \emptyset$ for $x \in F_{i}$ , $y \in F_{j}$ , $i \neq j$ (The PMEA three-quarter separation estimate).

[F2]

A neighbourhood base at $x$ contains, for each open $G ∋ x$ , a member $U$ with $x \in U \subseteq G$ ; so the hypothesis of [F1] is met whenever $x \in F_{i} \subseteq G$ (First countable space: a countable neighbourhood base at every point, Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open).

[F3]

Collectionwise normality asks that every discrete family of closed sets be separated by pairwise disjoint open expansions (Normalized families and collectionwise normality, Discrete families and $σ$ -locally-finite and $σ$ -discrete bases).

Proof

technique · direct

1.1

If $X = \emptyset$ or the discrete family is empty, empty expansions suffice. Otherwise use AC to choose one local base of the stated size at each point, and replace its members by their interiors. This is an image of the original base, so its cardinality does not increase; each interior contains the point, and the image remains a local base. Fix the discrete family $F$ and the open neighbourhood bases $U_{x}$ from the Given line, with $∣ U_{x} ∣ < c$ , or with $∣ U_{x} ∣ \leq ω$ in the first countable case.

givenF2

2.1

Each $U_{x}$ is nonempty, by applying its base property to $X$ . For $U, V \in U_{x}$ , the open intersection $U \cap V$ contains $x$ , so [F2] supplies $W \in U_{x}$ with $W \subseteq U \cap V$ . This is precisely downward directedness; the original base need not be closed under finite intersections and is not enlarged. If $x \in F_{i} \subseteq G$ with $G$ open, [F2] likewise supplies a base member inside $G$ . Apply [F1] to $F$ and the bases $U_{x}$ , obtaining $U^{x} \in U_{x}$ with $U^{x} \cap U^{y} = \emptyset$ whenever $x \in F_{i}$ , $y \in F_{j}$ , $i \neq j$ .

step 1.1F1F2

3.1

For each $i$ put $G_{i} : = ⋃ {U^{x} : x \in F_{i}}$ . Each selected $U^{x}$ belongs to the open base $U_{x}$ , so $G_{i}$ is open; it contains $F_{i}$ because $x \in U^{x}$ , and distinct $G_{i}, G_{j}$ are disjoint by step 2.1. Hence $F$ is separated and $X$ is collectionwise normal.

givenstep 2.1F3∎

Remarks

Character, not weight. The hypothesis is pointwise, so the theorem applies to every normal Moore space once PMEA- $σ$ is available, since Moore spaces are first countable (Moore spaces and developments).
Fremlin's remark (b) after Theorem 8F. The proof needs only as much additivity as the size of the local bases, which is why the countably additive version suffices in the first countable case.

TheoremStatement: Literature-sourcedProof: Literature-sourcedaudited 2026-09-22‡ rests on unproved material (inherited)Open item page →

PMEA implies the normal Moore space conjecture

Statement

$Z F C + P M E A$ proves the normal Moore space conjecture: every normal Moore space is metrizable. Already PMEA- $σ$ suffices (PMEA and PMEA-sigma, Moore spaces and developments, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

Facts & Assumptions

Given: A normal Moore space $X$ and PMEA- $σ$ . (The formally stronger PMEA hypothesis in the first sentence of the Statement supplies PMEA- $σ$ by [F4].)

[F1]

Every Moore space is first countable: a development supplies at each point the countable star family as a local base (Moore spaces and developments).

[F2]

Under PMEA- $σ$ , every first countable normal space is collectionwise normal (PMEA makes normal low-character spaces collectionwise normal).

[F3]

Every collectionwise normal Moore space is metrizable (Collectionwise normal Moore spaces are metrizable).

[F4]

PMEA implies PMEA- $σ$ , since a $c$ -additive full extension is countably additive (PMEA and PMEA-sigma).

Proof

technique · direct

1.1

The space $X$ is first countable by [F1] and is normal and Moore by hypothesis.

givenF1

2.1

By [F2] the space $X$ is collectionwise normal.

step 1.1F2

3.1

By [F3] the collectionwise normal Moore space $X$ is metrizable; since PMEA implies PMEA- $σ$ by [F4], the argument used only PMEA- $σ$ .

step 2.1F3F4∎

Remarks

This is Nyikos' provisional solution in Fremlin's form. The measure-theoretic input is PMEA- $σ$ alone; the topological input is that a first countable normal space is collectionwise normal under it; the metrization of collectionwise normal Moore spaces is the separate local theorem of this page.

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22‡ rests on unproved material (inherited)Open item page →

A strongly compact cardinal gives the NMSC consistency upper bound

Statement

$Con (Z F C + there is a strongly compact cardinal)$ implies $Con (Z F C + N M S C)$ , where NMSC is the normal Moore space conjecture. No ground-model implication and no actual strongly compact cardinal is asserted (Strong compactness and the product-measure extension interface).

Facts & Assumptions

Given: The metatheoretic assumption $Con (Z F C + a strongly compact cardinal)$ .

[F1]

The published product-measure interface: $Con (Z F C + a strongly compact cardinal)$ implies $Con (Z F C + P M E A)$ , the PMEA sentence being exactly the full-domain extension axiom of PMEA and PMEA-sigma (Strong compactness and the product-measure extension interface).

[F2]

$Z F C + P M E A$ proves NMSC (PMEA implies the normal Moore space conjecture).

[F3]

If an arithmetic base $B$ verifies a total code map $r$ and $\forall p ({Prf}_{U} (p, ⌜ ⊥ ⌝) \to {Prf}_{T} (r (p), ⌜ ⊥ ⌝))$ , then $B ⊢ Con (T) \to Con (U)$ (Formal consistency transfer from a verified reduction).

[F4]

Both theories are formulated over $Z F C$ with $A C$ explicit (The Axiom of Choice).

Proof

technique · direct

1.1

Assume $Con (Z F C + a strongly compact cardinal)$ .

given

1.2

Put $T_{0} : = Z F C + P M E A$ and $U : = Z F C + N M S C$ . By [F2], fix a finite $T_{0}$ -proof $q$ of NMSC. Define $r$ on codes of $U$ -proofs by scanning the finite proof, copying logical and ZFC axiom lines and inference steps, and replacing every use of the added NMSC axiom by the fixed proof $q$ , with line references renumbered. This is a total primitive-recursive code map. The chosen arithmetic proof checker verifies by induction on the length of the input proof that every copied line remains valid and every replaced line is the conclusion of $q$ ; hence it verifies ${Prf}_{U} (p, ⌜ ⊥ ⌝) \to {Prf}_{T_{0}} (r (p), ⌜ ⊥ ⌝)$ for every $p$ .

F2F3F4construct

2.1

By [F1] the theory $Z F C + P M E A$ is consistent.

step 1.1F1

2.2

Apply [F3] to the verified map $r$ of [step 1.2]. It gives $Con (T_{0}) \to Con (U)$ , that is, $Con (Z F C + P M E A) \to Con (Z F C + N M S C)$ .

step 1.2F3

3.1

Chaining steps 1.1, 2.1 and 2.2 gives the displayed implication, under the metatheoretic consistency assumption only.

step 2.1step 2.2∎

Remarks

Three distinct claims are kept apart. "PMEA implies NMSC" is a theorem of $Z F C$ ; the consistency transfer is metatheoretic; and the strongly compact cardinal is assumed only inside the consistency hypothesis.
AC is used in the supplier, in the random-real construction and in the cardinal arithmetic; it is declared as a dependency and no choice-free reading is claimed.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-22Open item page →

Fleissner's HYP covering interface

Definition

Work in $Z F C$ (The Axiom of Choice). HYP is the assertion that there exist an infinite cardinal $κ$ , an increasing sequence of cardinals $(κ_{n})_{n \in ω}$ cofinal in $κ$ , and a set $E$ such that

$(1 a) \sup_{n} κ_{n} = κ; (1 b) 2^{κ_{n}} < κ for every n; (2) 2^{κ} = κ^{+};$ $(3 a) E \subseteq {δ < κ^{+} : cf (δ) = ω} and E is stationary in κ^{+}; (3 b) E \cap β is not stationary in β for every β < κ^{+} with cf (β) > ω .$

Clause $(1 b)$ together with $(1 a)$ says that $κ$ is a strong limit cardinal of countable cofinality; $(2)$ is the $κ$ -continuum hypothesis; $(3 a)$ and $(3 b)$ say that $E$ is a nonreflecting stationary subset of the ordinals below $κ^{+}$ of countable cofinality (Cardinal (initial ordinal) and cardinality, The successor cardinal $κ^{+}$ , the alephs $ℵ_{α}$ , the beths $ℶ_{α}$ , successor and limit cardinals, and the identifications $ℵ_{0} = ω$ and $ℵ_{1} = ω_{1}$ , Cofinality $cf (α)$ , and regular and singular cardinals, For every ordinal $α$ there is a least ordinal $β$ admitting a map $β \to α$ with cofinal range, and that map may always be taken strictly increasing, Closed unbounded subsets of ordinals, The club filter and nonstationary ideal, Cofinality strata, trace, and reflection).

Ladders. In a structure satisfying HYP fix, for each $δ \in E$ , an increasing sequence $(δ_{i})_{i \in ω}$ of nonlimit ordinals cofinal in $δ$ . For each $δ$ , For every ordinal $α$ there is a least ordinal $β$ admitting a map $β \to α$ with cofinal range, and that map may always be taken strictly increasing gives a strictly increasing cofinal sequence $c_{i} < δ$ ; replacing $c_{i}$ by $c_{i} + 1$ gives successor ordinals still strictly below the limit $δ$ , strictly increasing and cofinal. The axiom of choice then fixes one such sequence for every $δ \in E$ (The Axiom of Choice); nothing below depends on which ladders are chosen beyond the two properties just named.

Remarks

The CH case. Under $C H$ , take $κ = ω$ , $κ_{n} = n$ and $E$ the nonzero limit ordinals below $ω_{1}$ . The finite cardinals have supremum $ω$ and $2^{n} < ω$ , while CH gives $2^{ω} = ω_{1}$ . The set $E$ is stationary: given any club $C \subseteq ω_{1}$ , choose a strictly increasing sequence from $C$ . Its supremum is a countable nonzero limit ordinal, belongs to $C$ by closure, and belongs to $E$ . Every nonzero limit $β < ω_{1}$ has cofinality $ω$ , so the successor-sequence club described above avoids $E \cap β$ . In particular, both ${ω n : 1 \leq n < ω}$ and ${ω n + 1 : n < ω}$ are clubs in $ω^{2}$ and are disjoint. Containing a club at an ordinal of countable cofinality therefore does not imply meeting every club. This is precisely Fleissner's CH instance on printed p.367; ladder separation may also be proved directly in that case.
The separation function is not recorded here. Fleissner's Lemma 1 derives, from $(3 b)$ and the ladders, a function $m_{β}$ separating distinct ladders below $β$ ; that derivation is proved locally in Ladder separation from HYP, and nothing in this definition asserts it.
The large-cardinal reading is an input, not a consequence. That HYP is implied by the nonexistence of an inner model with a measurable cardinal is a theorem about the Dodd-Jensen core model, recorded in Dodd-Jensen covering supplies Fleissner HYP data and not folded into this definition.

LemmaStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

Ladder separation from HYP

Statement

Let $κ, (κ_{n})_{n \in ω}, E$ satisfy HYP and fix ladders $(δ_{i})_{i \in ω}$ for $δ \in E$ (Fleissner's HYP covering interface). Then for every $β < κ^{+}$ there is a function $m_{β} : E \cap β \to ω$ such that for all distinct $δ, η \in E \cap β$ , writing $m : = \max (m_{β} (δ), m_{β} (η))$ , one has $δ_{m} \neq η_{m}$ ; and in fact $δ_{i} \neq η_{i} for every i \geq \max (m_{β} (δ), m_{β} (η)) .$

Facts & Assumptions

Given: Witnesses $κ, (κ_{n}), E$ for HYP and ladders $(δ_{i})$ for $δ \in E$ ; the induction is on $β < κ^{+}$ .

[F1]

For every limit $β < κ^{+}$ there is a club $C \subseteq β$ disjoint from $E$ . If $cf (β) > ω$ , this is exactly HYP clause (3b). If $cf (β) = ω$ , fix a strictly increasing cofinal sequence $(b_{n})$ in $β$ and take $C = {0} \cup {b_{n} + 1 : n \in ω}$ . This set is unbounded; it has no limit point below $β$ (every proper initial segment of an increasing $ω$ -sequence is finite), hence is closed; and its nonzero members are successors, while every member of $E$ has cofinality $ω$ by clause (3a). These are all possible cofinalities of a nonzero limit ordinal in ZFC (Fleissner's HYP covering interface, Cardinal (initial ordinal) and cardinality).

[F2]

Every element of $E$ has cofinality $ω$ , so successor ordinals are not in $E$ . If $C \subseteq β$ is club, $0 \in C$ , and $δ < β$ is not in $C$ , then $γ = \sup (C \cap δ)$ belongs to $C$ and is below $δ$ , while $\min (C ∖ δ)$ exists and is strictly between $δ$ and $β$ (Cardinal (initial ordinal) and cardinality).

[L1]

Each ladder $(δ_{i})$ is increasing and cofinal in $δ$ , so for every $γ < δ$ there is a least $i$ with $δ_{i} > γ$ , and $δ_{i} > γ$ for all larger $i$ (Fleissner's HYP covering interface, The natural numbers $N$ (von Neumann)).

[L2]

Induction: if a statement about $m_{β}$ is proved for $β = 0$ , for $β = α + 1$ from $m_{α}$ , and for limit $β$ from all $m_{γ}$ with $γ < β$ , then it holds for every $β < κ^{+}$ (Cardinal (initial ordinal) and cardinality).

Proof

technique · induction

1.1

We build $m_{β}$ for all $β < κ^{+}$ by induction, maintaining the strengthened property that for distinct $δ, η \in E \cap β$ one has $δ_{i} \neq η_{i}$ for every $i \geq \max (m_{β} (δ), m_{β} (η))$ .

givenL2

2.1

For $β = 0$ take $m_{0} : = \emptyset$ ; the domain is empty and both properties are vacuous.

basestep 1.1

2.2

Successor case. Let $β = α + 1$ . If $α \notin E$ put $m_{β} : = m_{α}$ ; this keeps the domain and the strengthened property. If $α \in E$ , define, for $δ \in E \cap α$ , $m_{β} (δ) : = \max (m_{α} (δ), j (δ, α))$ where $j (δ, α)$ is the least $i$ with $α_{i} > δ$ , and put $m_{β} (α) : = 0$ .

step 1.1F2L1

2.3

Limit case. Let $β$ be a limit ordinal. By [F1] choose a club $C \subseteq β$ disjoint from $E$ and containing $0$ . For $δ \in E \cap β$ , [F2] gives $γ (δ) : = \sup (C \cap δ) \in C$ with $γ (δ) < δ$ and $\overset{ˉ}{γ} (δ) : = \min (C ∖ δ) \in C$ with $δ < \overset{ˉ}{γ} (δ) < β$ . Thus $m_{\overset{ˉ}{γ} (δ)}$ is defined by induction and contains $δ$ in its domain. Define $m_{β} (δ) : = \max (m_{\overset{ˉ}{γ} (δ)} (δ), i (δ))$ , where $i (δ)$ is the least $i$ with $δ_{i} > γ (δ)$ .

step 1.1F1F2L1

3.1

In the successor case, pairs inside $E \cap α$ keep the strengthened property because $m_{β} \geq m_{α}$ pointwise, so their separating level is not decreased. For $δ \in E \cap α$ and the new point $α$ : for $i \geq \max (m_{β} (δ), m_{β} (α)) = m_{β} (δ) \geq j (δ, α)$ we have $α_{i} \geq α_{j (δ, α)} > δ > δ_{i}$ , so $α_{i} \neq δ_{i}$ . Hence $m_{β}$ works.

step 2.2F2L1

3.2

In the limit case let $δ < η$ in $E \cap β$ . If $γ (δ) = γ (η)$ then also $\overset{ˉ}{γ} (δ) = \overset{ˉ}{γ} (η) = : \overset{ˉ}{γ}$ , and $m_{β} \geq m_{\overset{ˉ}{γ}}$ pointwise on ${δ, η}$ , so the strengthened property for $m_{\overset{ˉ}{γ}}$ , which is available by induction and holds at levels $\geq \max (m_{\overset{ˉ}{γ}} (δ), m_{\overset{ˉ}{γ}} (η))$ , transfers to $m_{β}$ .

step 2.3

3.3

In the limit case, if $γ (δ) < γ (η)$ then $\overset{ˉ}{γ} (δ) \leq γ (η)$ : otherwise $γ (η) < \overset{ˉ}{γ} (δ)$ would put both $δ$ and $η$ in the same gap of $C$ , forcing $γ (δ) = γ (η)$ . Then for $i \geq \max (m_{β} (δ), m_{β} (η)) \geq \max (i (δ), i (η))$ we have $δ_{i} < δ < \overset{ˉ}{γ} (δ) \leq γ (η) < η_{i}$ , so $δ_{i} \neq η_{i}$ .

step 2.3F2L1

4.1

Steps 2.1, 3.1, 3.2 and 3.3 establish the strengthened property for $m_{β}$ in all three cases of the induction, so by [L2] the functions $m_{β}$ exist for every $β < κ^{+}$ with the strengthened property. Since $δ_{m} \neq η_{m}$ for the single level $m = \max (m_{β} (δ), m_{β} (η))$ is the special case $i = m$ , the asserted functions exist.

step 2.1step 3.1step 3.2step 3.3L2discharge-induction∎

Remarks

Where (3b) is needed. At limit stages of uncountable cofinality it supplies a club disjoint from $E$ . At countable-cofinality stages such a club is automatic from clause (3a), by using a cofinal $ω$ -sequence of successors. In either case the club gaps put each $δ \in E$ below a smaller ordinal $\overset{ˉ}{γ} (δ)$ where the induction hypothesis separates ladders.
The strengthened form is not needed elsewhere, but it is what makes both the same-gap and the different-gap cases work at once; the paper's Lemma 1 is the special case of the single level $m$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-22Open item page →

The Dodd-Jensen covering and square package

Definition

Work in $Z F C$ and let $K$ denote the Dodd-Jensen core model, an inner model containing the constructible universe and contained in $V$ (Cardinal (initial ordinal) and cardinality). The Dodd-Jensen covering and square package for $K$ consists of the following interface statements.

Covering. $Cov (V, K)$ : every uncountable set $X$ of ordinals in $V$ is contained in a set $Y \in K$ with $∣ Y ∣ = ∣ X ∣$ (the covering lemma of Dodd-Jensen).
GCH in $K$ and square. $K$ satisfies the generalised continuum hypothesis, and for every infinite cardinal $λ$ of $K$ there is a square sequence $□_{λ}$ in $K$ : a sequence $⟨ C_{α} : α < λ^{+}, \lim (α) ⟩$ with each $C_{α}$ club in $α$ , $otp (C_{α}) < λ$ whenever $cf (α) < λ$ , and $C_{β} = β \cap C_{α}$ whenever $β < α$ is a limit point of $C_{α}$ (Cardinal (initial ordinal) and cardinality).
Singular strong limits. $Cov (V, K)$ implies that there is an uncountable strong limit cardinal $κ$ of countable cofinality with $2^{κ} = κ^{+}$ and with a square sequence $□_{κ}$ (Good, Lemma 8).
Weak diamond on the countable-cofinality points. Let $W : = {α < κ^{+} : cf (α) = ω}$ . From $□_{κ}$ and $2^{κ} = κ^{+}$ one has $♢_{κ^{+}} (W)$ : a sequence $⟨ S_{α} : α \in W ⟩$ with $S_{α} \subseteq α$ such that for every $X \subseteq κ^{+}$ the set ${α \in W : X \cap α = S_{α}}$ is stationary in $κ^{+}$ (Good, Lemma 11, after Devlin).
Nonreflecting stationary set. From $□_{κ}$ and $♢_{κ^{+}} (W)$ there is a stationary $E \subseteq W$ with $□_{κ} (E)$ , the assertion of Good's Definition 3: there is a sequence $⟨ C_{α} : α < κ^{+}, \lim (α) ⟩$ with each $C_{α}$ club in $α$ , $otp (C_{α}) < κ$ whenever $cf (α) < κ$ , and such that for every limit point $β$ of $C_{α}$ one has $β \notin E$ and $C_{β} = β \cap C_{α}$ ; the clause $β \notin E$ is the one from which Good derives that a stationary $E$ with $□_{κ} (E)$ is nonreflecting. In addition ${α \in E : X \cap α = S_{α}}$ is stationary for every $X \subseteq κ^{+}$ (Good, Lemma 12, exactly item IV.2.10 of Devlin's Constructibility).

Statement 1 is the covering theorem of Dodd-Jensen; statements 2, 4 and 5 are fine-structural consequences recorded with the exact citations above. The package is the input of Dodd-Jensen covering supplies Fleissner HYP data; the deep covering theorem itself is not reproved in this library, and no clause of the package is asserted to be a theorem of $Z F C$ without the hypothesis "no inner model with a measurable cardinal" that supplies it.

Remarks

No measurable cardinal is constructed here. The package is a conditional consequence of the nonexistence of inner models with measurable cardinals; the definition records the objects and their properties, not the inner-model construction that produces them.
Ordinals and clubs. Clubs and stationarity are taken in the ordinal spaces $λ^{+}$ with the order topology; "nonreflecting" means $E \cap β$ is nonstationary in $β$ for every $β < κ^{+}$ (Cardinal (initial ordinal) and cardinality).
AC is used throughout, in the comparison of cardinalities, the choice of nonlimit ladders and the standard cardinal arithmetic (The Axiom of Choice).

TheoremStatement: Literature-sourcedProof: Literature-sourcedaudited 2026-09-22Open item page →

Dodd-Jensen covering supplies Fleissner HYP data

Statement

In $Z F C$ , if there is no inner model with a measurable cardinal, then the Dodd-Jensen covering and square package (The Dodd-Jensen covering and square package) supplies a singular strong limit cardinal $κ$ of cofinality $ω$ with $2^{κ} = κ^{+}$ and a nonreflecting stationary set $E \subseteq {δ < κ^{+} : cf (δ) = ω}$ . Consequently HYP holds (Fleissner's HYP covering interface).

Facts & Assumptions

Given: The hypothesis that there is no inner model with a measurable cardinal, and the Dodd-Jensen covering and square package for the core model $K$ that this hypothesis supplies.

[F1]

The package: $Cov (V, K)$ ; GCH and square in $K$ ; an uncountable strong limit cardinal $κ$ of countable cofinality with $2^{κ} = κ^{+}$ and $□_{κ}$ ; $♢_{κ^{+}} (W)$ for $W = {α < κ^{+} : cf (α) = ω}$ ; and a stationary $E \subseteq W$ with $□_{κ} (E)$ (The Dodd-Jensen covering and square package).

[F2]

If $C$ is club in an ordinal $β$ of uncountable cofinality, then the set $acc (C)$ of its limit points is also club in $β$ ; closedness gives $acc (C) \subseteq C$ . The uncountable-cofinality qualification is essential: a club of order type $ω$ can have no limit points below its supremum. Here " $α$ is a limit point of $C_{β}$ " means $α = \sup (C_{β} \cap α)$ (Cardinal (initial ordinal) and cardinality).

[L1]

A cardinal $κ$ is a strong limit exactly when $2^{λ} < κ$ for every $λ < κ$ ; if in addition $cf (κ) = ω$ , then there is an increasing sequence of cardinals $(κ_{n})_{n \in ω}$ cofinal in $κ$ (Cardinal (initial ordinal) and cardinality, The successor cardinal $κ^{+}$ , the alephs $ℵ_{α}$ , the beths $ℶ_{α}$ , successor and limit cardinals, and the identifications $ℵ_{0} = ω$ and $ℵ_{1} = ω_{1}$ ).

[L2]

A $Σ_{1}$ -definable choice from the given data, e.g. $κ_{n} : = {(\sup_{m \leq n} 2^{γ_{m}})}^{+}$ for a fixed cofinal $ω$ -sequence $(γ_{n})$ in $κ$ , is legitimate, since a single sequence is chosen once and the rest is defined by a formula; only the initial choice of $(γ_{n})$ uses the axiom of choice (The Axiom of Choice).

Proof

technique · direct

1.1

Assume there is no inner model with a measurable cardinal; by [F1] the package provides $κ$ with $2^{κ} = κ^{+}$ , $□_{κ}$ , $♢_{κ^{+}} (W)$ and a stationary $E \subseteq W$ with $□_{κ} (E)$ .

givenF1

2.1

Clause (2) of HYP holds: $2^{κ} = κ^{+}$ by step 1.1.

step 1.1

2.2

Clause (1) of HYP holds: $κ$ is an uncountable strong limit of countable cofinality by step 1.1, so [L1] and [L2] give an increasing sequence $(κ_{n})$ of cardinals cofinal in $κ$ with $2^{κ_{n}} < κ$ for every $n$ .

step 1.1L1L2

2.3

$E$ is stationary in $κ^{+}$ : it is stationary in $κ^{+}$ as a subset of $W$ by step 1.1, and $E \subseteq W \subseteq κ^{+}$ . Hence clause (3a) of HYP holds.

step 1.1

2.4

Clause (3b) holds. Suppose towards a contradiction that $E \cap β$ is stationary in some $β < κ^{+}$ with $cf (β) > ω$ . Let $⟨ C_{α} ⟩$ witness $□_{κ} (E)$ . By [F2], $acc (C_{β})$ is club in $β$ , so stationarity gives $α \in E \cap acc (C_{β})$ . But clause (iii) of $□_{κ} (E)$ says that every limit point of $C_{β}$ lies outside $E$ , a contradiction. Hence $E \cap β$ is nonstationary for every such $β$ , exactly as required by the local HYP interface.

step 1.1F2

3.1

By steps 2.1, 2.2, 2.3 and 2.4 the objects $κ, (κ_{n})_{n \in ω}, E$ satisfy clauses (1a), (1b), (2), (3a) and (3b) of the local interface. Thus HYP holds, with $κ$ singular of cofinality $ω$ and $E$ nonreflecting at every uncountable-cofinality stage as asserted.

step 2.1step 2.2step 2.3step 2.4∎

Remarks

The covering theorem is a declared input. Statements 1-2 of the package are the Dodd-Jensen covering theorem and the fine-structure of $K$ ; this item derives the HYP clauses from them and does not reprove them. The exact citations are in The Dodd-Jensen covering and square package.
Where the conclusion is used. HYP is the hypothesis of the construction of a normal nonmetrizable Moore space recorded elsewhere on this page, and hence of the inner-model lower bound for the normal Moore space conjecture.

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

No inner measurable implies Fleissner's HYP

Statement

$Z F C$ plus "there is no inner model with a measurable cardinal" proves HYP with the parameters needed by Fleissner (Fleissner's HYP covering interface, Dodd-Jensen covering supplies Fleissner HYP data).

Facts & Assumptions

Given: The hypothesis that no inner model contains a measurable cardinal.

[F1]

That hypothesis supplies the Dodd-Jensen covering and square package and, from it, a singular strong limit $κ$ of cofinality $ω$ with $2^{κ} = κ^{+}$ and a nonreflecting stationary $E \subseteq {δ < κ^{+} : cf (δ) = ω}$ (Dodd-Jensen covering supplies Fleissner HYP data).

[F2]

HYP is the conjunction of clauses (1a), (1b), (2), (3a), (3b) for some $κ, (κ_{n}), E$ (Fleissner's HYP covering interface).

Proof

technique · direct

1.1

Assume no inner model contains a measurable cardinal. By [F1] there are $κ, (κ_{n})$ and $E$ with $2^{κ} = κ^{+}$ , $2^{κ_{n}} < κ$ for all $n$ , $\sup_{n} κ_{n} = κ$ , $E$ stationary in $κ^{+}$ inside ${δ < κ^{+} : cf (δ) = ω}$ , and $E$ nonreflecting.

givenF1

2.1

These objects satisfy every clause of [F2], so HYP holds with exactly the parameters, namely $κ, (κ_{n})_{n \in ω}$ and the nonreflecting stationary $E$ , that the Fleissner construction consumes.

step 1.1F2∎

Remarks

This is a repackaging item. It records that the parameters produced by the covering route are the parameters HYP asks for; no new mathematics beyond the identification of the clauses is claimed.

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

Fleissner's construction of a normal nonmetrizable Moore space from level data

Statement

Let $κ$ be an infinite cardinal, let $(κ_{n})_{n \in ω}$ be increasing with $2^{κ_{n}} < κ$ for every $n$ and $\sup_{n} κ_{n} = κ$ , let $2^{κ} = κ^{+}$ , and let $E \subseteq {δ < κ^{+} : cf (δ) = ω}$ be stationary in $κ^{+}$ . Fix, for each $δ \in E$ , an increasing sequence $(δ_{i})_{i \in ω}$ of nonlimit ordinals cofinal in $δ$ (Fleissner's HYP covering interface), and assume the ladder separation conclusion of Fleissner's Lemma 1: for every $β < κ^{+}$ there is $m_{β} : E \cap β \to ω$ such that $δ_{i} \neq η_{i}$ for all distinct $δ, η \in E \cap β$ and all $i \geq \max (m_{β} (δ), m_{β} (η))$ .

Then there is a normal nonmetrizable Moore space (Moore spaces and developments, Metrizable spaces are collectionwise normal). The same space carries the stated uniform base, hence is metacompact.

Facts & Assumptions

Given: $κ, (κ_{n}), E$ and the ladders as in the Statement. By passing to a cofinal tail-subsequence and then prepending $0$ , we may and do assume that $κ_{0} = 0$ , that $κ_{m} \geq 1$ for $m \geq 1$ , and, when $κ > ω$ , that every $κ_{m}$ with $m \geq 1$ is infinite. These reindexings preserve the power bounds and the supremum $κ$ .

[F1]

$κ^{+}$ is a regular cardinal with $cf (κ^{+}) = κ^{+} > ω$ ; sums and products of infinite cardinals absorb, in particular $μ \cdot μ = μ$ . From the given $2^{κ} = κ^{+}$ and the cardinal exponent laws, for every cardinal $λ$ with $1 \leq λ \leq κ$ one has $(κ^{+})^{λ} = (2^{κ})^{λ} = 2^{κ \cdot λ} = 2^{κ} = κ^{+} .$ (Cardinal (initial ordinal) and cardinality, Cardinal sum $κ \oplus λ$ , product $κ \otimes λ$ and exponentiation $κ^{λ}$ , and why they are written apart from the ordinal operations, Absorption: for cardinals $κ, λ$ with $κ$ infinite and $λ \leq κ$ , $κ \oplus λ = κ$ , and $κ \otimes λ = κ$ when $λ \neq 0$ , Assuming the Axiom of Choice, $2^{κ} = ∣ P (κ) ∣$ , and Cantor's theorem in cardinal form: $κ < 2^{κ}$ ).

[F2]

Every set can be well-ordered, so a set of cardinality $μ$ can be enumerated as ${x_{α} : α < μ}$ (Well-order and well-ordered set, Injection, surjection, bijection, The Axiom of Choice).

[F3]

Clubs and stationary sets in $κ^{+}$ : a set is closed if it contains the sup of each of its bounded subsets, clubs are the closed unbounded sets, stationary means meeting every club; a club is stationary; supersets of stationary sets are stationary; a stationary set minus a non-stationary set is stationary; and a countable union of non-stationary subsets of $κ^{+}$ is non-stationary (Closed unbounded subsets of ordinals, The club filter and nonstationary ideal, Basic stationary-set calculus).

[F4]

Fodor's pressing-down lemma: if $S \subseteq κ^{+} ∖ {0}$ is stationary and $f : S \to κ^{+}$ is regressive ( $f (α) < α$ throughout), then some fibre of $f$ is stationary (Fodor’s pressing-down lemma, Regressive functions on ordinals).

[F5]

For $κ > ω$ the Erdős–Rado theorem gives $ℶ_{1} (κ_{n})^{+} = (2^{κ_{n}})^{+} ⟶ (κ_{n}^{+})_{κ_{n}}^{2}$ , and for $κ = ω$ the infinite Ramsey theorem gives $ω \to (ω)_{r}^{2}$ for every finite $r \geq 1$ ; the arrow means that every colouring of pairs admits a homogeneous set of the target size (Erdős–Rado for arbitrary infinite cardinals and finite arity, Infinite Ramsey theorem for fixed finite arity and colors, Partition arrows and homogeneous sets, Cardinal (initial ordinal) and cardinality).

[F6]

Moore spaces, developments, stars, and first countability; regular $T_{1}$ spaces; metrizable spaces are collectionwise normal; a metrizable space is collectionwise normal, so a non-collectionwise-normal Moore space is not metrizable (Moore spaces and developments, Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Normalized families and collectionwise normality, Metrizable spaces are collectionwise normal, Discrete families and $σ$ -locally-finite and $σ$ -discrete bases).

[F7]

Topological vocabulary: basis, open and closed sets, closure, discrete subspaces, subspace topology (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, Discrete families and $σ$ -locally-finite and $σ$ -discrete bases, Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly).

[F8]

A function with domain $ω$ and values in $E$ is exactly an element of $E^{ω}$ ; finite sequences from $E$ are functions on natural numbers with range in $E$ ; the wood $Σ$ below is the set of all finite such sequences (A function is a relation $f$ with $(a, b) \in f$ and $(a, c) \in f$ implying $b = c$ ; $f : A \to B$ , the value $f (a)$ , domain and codomain, The natural numbers $N$ (von Neumann)).

Proof

technique · construction, with the acceptability recursion in steps 2.2, 3.3, 4.3 and 5.2 and the later Ramsey/Erdős–Rado argument

1.1

Fix a well-ordering of the universe and enumerate the family $Z : = {Z \subseteq Σ : card Z \leq κ and Z \subseteq Σ_{k} for some k \in ω}$ as ${Z_{α} : α < κ^{+}}$ , arranged so that $Z_{0} = \emptyset$ . Here $Σ : = ⋃_{n \in ω} Σ_{n}$ with $Σ_{n} : = {f ↾ n : f \in E^{ω}}$ , and $[σ] : = {f \in F : σ \subseteq f}$ for $F : = E^{ω}$ ; write $a^{*}$ for the greatest ordinal in the range of a nonempty $a \in Σ$ , and put $Σ^{β} : = {σ \in Σ : σ = \emptyset or σ^{*} < β}$ and $Z (β) : = {Z_{α} : α < β}$ . Enumerating $Z$ is legitimate: stationarity gives $∣ E ∣ = κ^{+}$ , so [F1] with $λ = ℵ_{0}$ gives $∣ E^{ω} ∣ = ∣ Σ ∣ = κ^{+}$ ; and [F1] with $λ = κ$ gives at most $(κ^{+})^{κ} = κ^{+}$ subsets of size at most $κ$ on every finite level. Conversely, the singletons of $Σ_{1}$ already give $κ^{+}$ members of $Z$ .

givenF1F2F3F8

1.2

With $κ$ , $(κ_{n})$ and the enumeration ${Z_{α} : α < κ^{+}}$ fixed as above, let $level (Z)$ be the least $k$ with $Z \subseteq Σ_{k}$ ( $0$ for $Z = \emptyset$ ), and for nonempty $θ \in Σ$ of length $r$ put $F_{θ} : = Z (θ^{*}) \cap {Z : level (Z) < r}$ . This set has cardinality at most $κ$ . Using [F2], fix an enumeration $e_{θ} : μ_{θ} \to F_{θ}$ with $μ_{θ} \leq κ$ , arranged with $e_{θ} (0) = \emptyset$ ; the latter is possible because $Z_{0} = \emptyset \in F_{θ}$ . Let $P (θ, m) : = e_{θ} [\min (κ_{m}, μ_{θ})]$ and, for $σ \in Σ$ of length $ℓ$ , define $A (σ, m) : = ⋃ {P (θ, m) : \emptyset \neq θ \subseteq σ}$ , with $A (\emptyset, m) = A (σ, 0) = \emptyset$ . Then $⋃_{m} A (σ, m) = F_{σ}$ : an element of $F_{σ}$ has an index $ξ < μ_{σ} \leq κ$ in $e_{σ}$ , and cofinality of $(κ_{m})$ gives an $m$ with $ξ < κ_{m}$ . Also, if $σ \subseteq σ^{'}$ and $m \leq m^{'}$ , then $A (σ, m) \subseteq A (σ^{'}, m^{'})$ . Finally, $σ$ has only finitely many nonempty prefixes, so $∣ A (σ, m) ∣ \leq λ_{ℓ, m}$ , where $λ_{ℓ, m} : = κ_{m}$ if $κ_{m}$ is infinite and $λ_{ℓ, m} : = (ℓ + 1) κ_{m}$ if $κ_{m}$ is finite. Thus the source's special $κ = ω$ clause gives finite $A (σ, m)$ ; it does not assert the false bound $∣ A (σ, m) ∣ \leq κ_{m}$ .

givenF1F2

1.3

Construction requirement (well-definedness). The condition (12) below is imposed only at traces lying in $dom (g) = Z (ρ^{*})$ : for $σ \in Σ_{n}$ , a triple $⟨ g, ρ, τ ⟩$ satisfies (12) at index $σ$ iff $Z \in A (σ, n), Z \subseteq Σ_{m}, Z \cap Σ^{ρ (m)} \in Z (ρ^{*}) ⟹ g (Z \cap Σ^{ρ (m)}) = 1 ⟺ σ ↾ m \in Z .$ No instance of (12) is used anywhere below at a trace outside $Z (ρ^{*})$ , and the §6 instances are proved in Case 1 below, so this guard is a definition of the space, not an assumption. This is the documented discharge of the trace-domain obligation.

givenF8

1.4

Lemma 3(a). If $T \subseteq Σ$ satisfies $⋃ {[σ] : σ \in T} = F$ , then for some $n$ the set $T \cap Σ_{n}$ has a stafull subset, where $S \subseteq Σ_{n}$ is stafull iff for all $σ \in S$ and $j < n$ the set ${τ (j) : σ ↾ j \subset τ \in S}$ is stationary in $κ^{+}$ .

given

2.1

Let $Q_{k}$ be the set of triples $⟨ g, ρ, τ ⟩$ such that $ρ, τ \in Σ_{k}$ , $ρ (i) < τ (i) < ρ (i + 1) < τ (i + 1)$ for all $i < k - 1$ , and $g$ is a function from $Z (ρ^{*})$ to ${0, 1}$ (for $k = 0$ read $ρ^{*} = 0$ ), and let $Q : = ⋃_{k} Q_{k}$ . For $σ \in Σ_{n}$ let $C (σ)$ be the set of $⟨ g, ρ, τ ⟩ \in Q$ satisfying: $σ \subseteq ρ$ or $σ \subseteq τ$ ; $ρ (0)_{i} = τ (0)_{i}$ for all $i < n$ ; and the guarded (12) of step 1.3 at index $σ$ . Put $B (σ) : = [σ] \cup C (σ)$ , and let $X : = F \cup Q$ with basis ${B (σ) : σ \in Σ} \cup {{q} : q \in Q}$ .

givenF6F7

2.2

Suppose, for contradiction, that $T \cap Σ_{n}$ has no stafull subset for every $n$ ; then every nonempty $S \subseteq T \cap Σ_{n}$ is non-stafull, because a stafull subset of $S$ would be a stafull subset of $T \cap Σ_{n}$ . Call a finite $ρ \in Σ$ acceptable iff for every $n > ∣ ρ ∣$ and every nonempty $S \subseteq T \cap Σ_{n}$ with $σ \supseteq ρ$ for all $σ \in S$ , there are $σ \in S$ and $j$ with $∣ ρ ∣ \leq j < n$ such that ${τ (j) : σ ↾ j \subset τ \in S}$ is non-stationary. Then $\emptyset$ is acceptable by non-stafullness of $S$ itself. We construct $f \in F$ whose every finite prefix is acceptable; then $f ↾ 0 = \emptyset \notin T$ (else $T \cap Σ_{0}$ would be stafull) and $f ↾ (i + 1) \notin T$ for all $i$ , so $f \notin ⋃ {[σ] : σ \in T}$ , contradicting the covering hypothesis.

givenstep 1.4F3

2.3

Lemma 3(b) in the form used. If $S \subseteq Σ_{n}$ is stafull and $h (σ) = σ (0)_{i}$ for a fixed $i < n$ , then there is a stafull $S^{'} \subseteq S$ on which $h$ is constant. Indeed $A_{i} : = {τ (0) : τ \in S} \cap E$ is stationary, $h$ restricted to $A_{i}$ is regressive (ladder values are below their ordinal), so Fodor's lemma (F4) provides a stationary $A^{'} \subseteq A_{i}$ on which $h$ is constant; then $S^{'} : = {τ \in S : τ (0) \in A^{'}}$ is stafull: for $j < n$ the branching set ${τ (j) : σ ↾ j \subset τ \in S^{'}}$ equals the corresponding branching set of $S$ when $j \geq 1$ (the first coordinate is determined by $σ ↾ j$ ) and equals $A^{'}$ when $j = 0$ . Iterating for $i = 0, \dots, n - 1$ gives a stafull $S^{'} \subseteq S$ with $σ (0)_{i} = σ^{'} (0)_{i}$ for all $σ, σ^{'} \in S^{'}$ and all $i < n$ .

step 1.4F4given

2.4

Lemma 3(c). If $S^{'} \subseteq Σ_{n}$ is stafull and $β < κ^{+}$ , then there is $W = {σ_{α} : α < β} \subseteq S^{'}$ with $σ_{α} (i) < σ_{α^{'}} (i^{'})$ iff $i < i^{'}$ or ( $i = i^{'}$ and $α < α^{'}$ ). Construct the coordinates level by level: having chosen levels $< i$ so that all level- $(i - 1)$ values lie below a common bound $< κ^{+}$ and each prefix $σ_{α} ↾ i$ extends to a member of $S^{'}$ , stafullness of $S^{'}$ makes each set $A_{α} : = {τ (i) : σ_{α} ↾ i \subset τ \in S^{'}}$ stationary, hence unbounded above the bound; choose $σ_{α} (i) \in A_{α}$ strictly increasing in $α$ and above the sup of the level- $(i - 1)$ values, possible because that sup is $< κ^{+}$ by regularity (F1) since $β < κ^{+}$ ; the final level's choice lies in $S^{'}$ by the definition of $A_{α}$ .

step 1.4F1given

3.1

The family ${B (σ)}$ together with the singletons is a basis: for $x \in F$ and $x \in B (σ) \cap B (σ^{'})$ the sequences $σ, σ^{'}$ are both initial segments of $x$ , hence comparable, and the longer one $σ^{''}$ satisfies $B (σ^{''}) \subseteq B (σ) \cap B (σ^{'})$ because $C (σ^{''}) \subseteq C (σ) \cap C (σ^{'})$ : the conditions (10) and (11) transfer downwards, and a guarded (12)-instance at $σ^{''}$ restricts to the corresponding instance at $σ$ since $A (σ, ∣ σ ∣) \subseteq A (σ^{''}, ∣ σ^{''} ∣)$ by step 1.2 and $σ^{''} ↾ m = σ ↾ m$ for $m < ∣ σ ∣$ . For $x \in Q$ the singleton ${x}$ is a basis element contained in every basis element containing $x$ .

step 1.2step 2.1F7

3.2

$X$ is $T_{1}$ , and $Q$ is open, hence $F = X ∖ Q$ is closed. To see the $T_{1}$ property, distinct branches in $F$ have incompatible finite initial segments, while every $q \in Q$ has the open singleton ${q}$ . Conversely, for fixed $q = ⟨ g, ρ, τ ⟩ \in Q_{k}$ and $f \in F$ , choose $n > k$ ; then $q \notin B (f ↾ n)$ because condition (10) cannot make a length- $n$ sequence an initial segment of either length- $k$ coordinate. Thus every point other than $q$ has a neighbourhood avoiding $q$ , so ${q}$ is also closed.

step 2.1F6F7

3.3

Successor step of the acceptability recursion. Let $ρ$ be acceptable with $∣ ρ ∣ = i$ . Call $S$ dangerous for $v$ if, for some $n > i + 1$ , $S \subseteq T \cap Σ_{n}$ is nonempty, every $σ \in S$ extends $ρ \hat{} v$ , and ${τ (j) : σ ↾ j \subset τ \in S}$ is stationary for all $σ \in S$ and $i + 1 \leq j < n$ ; let $B_{1} : = {v \in E :$ some $S$ is dangerous for $v}$ , and $B_{2} : = {v \in E : ρ \hat{} v \in T \cap Σ_{i + 1}}$ . Then $B_{1} \cup B_{2} \neq E$ .

step 2.2

4.1

The branch neighbourhoods have the star property needed below. If $x \in F$ and open $D ∋ x$ , choose $σ_{0} \subseteq x$ with $B (σ_{0}) \subseteq D$ . For $n \geq ∣ σ_{0} ∣$ , the unique length- $n$ initial segment $σ = x ↾ n$ satisfies $B (σ) \subseteq B (σ_{0})$ by step 3.1, so every length- $n$ basic member containing $x$ lies in $D$ . At a point of $Q$ the singleton is a basic neighbourhood.

step 2.1step 3.1F7

4.2

For disjoint closed $H, K \subseteq X$ , it is enough to find disjoint open $U, V$ with $H \cap F \subseteq U$ and $K \cap F \subseteq V$ . Indeed $U^{'} : = (U ∖ K) \cup (H \cap Q), V^{'} : = (V ∖ H) \cup (K \cap Q)$ are open, contain $H, K$ respectively, and are disjoint: $Q$ is discrete open, removing a closed set preserves openness, and each added isolated part lies in its own closed set and outside the other. Since $F$ is closed by step 3.2, the two traces are closed subsets of $F$ .

step 3.2F7

4.3

Suppose $B_{1} \cup B_{2} = E$ . If $B_{1}$ is stationary, choose for each $v \in B_{1}$ the least level $n_{v}$ of a dangerous $S_{v}$ and a least such $S_{v}$ in the fixed well-order; some $E^{'} : = {v \in B_{1} : n_{v} = n_{0}}$ is stationary by the countable case split, and $V : = ⋃_{v \in E^{'}} S_{v} \subseteq T \cap Σ_{n_{0}}$ is nonempty with all members extending $ρ$ . Acceptability of $ρ$ gives $σ \in V$ and $j \in [i, n_{0})$ with $B (σ, j, V)$ non-stationary. Pick $v \in E^{'}$ with $σ \in S_{v}$ . If $j \geq i + 1$ then $B (σ, j, V) \supseteq B (σ, j, S_{v})$ is stationary because $S_{v}$ is dangerous, a contradiction; if $j = i$ then $B (σ, i, V) = {τ (i) : τ \in V} \supseteq E^{'}$ , because $S_{v} ∋ σ$ witnesses $v \in B (σ, i, V)$ for every $v \in E^{'}$ , so $B (σ, i, V)$ is stationary, again a contradiction. If $B_{1}$ is non-stationary then $E ∖ B_{1}$ is stationary and contained in $B_{2}$ , so $B_{2}$ is stationary; applying acceptability of $ρ$ to the nonempty $S : = {ρ \hat{} v \in T \cap Σ_{i + 1} : v \in E}$ and $n = i + 1$ gives $B (σ, i, S)$ non-stationary for some $σ \in S$ , but $B (σ, i, S) = {τ (i) : τ \in S} \supseteq B_{2}$ is stationary. Both cases are contradictory, so $B_{1} \cup B_{2} \neq E$ .

step 3.3step 2.2F3

5.1

Define $G_{n} : = {B (σ) : σ \in Σ_{n}} \cup {{q} : q \in Q}$ . Each $G_{n}$ is an open cover: a point $f \in F$ lies in $[f ↾ n] \subseteq B (f ↾ n)$ , and every $q \in Q$ lies in its singleton. At $x \in F$ , step 4.1 gives $St (x, G_{n}) \subseteq D$ at every sufficiently large level. If $x = q \in Q_{k}$ , then for $n > k$ no $B (σ)$ with $∣ σ ∣ = n$ contains $q$ , because condition (10) would require $σ$ to be an initial segment of one of the length- $k$ sequences $ρ, τ$ ; thus $St (q, G_{n}) = {q}$ . Moreover $⋃_{n} G_{n}$ is a uniform base in the source's sense. A point $q \in Q_{k}$ belongs only to its singleton and to the finitely many $B (σ)$ whose $σ$ is an initial segment of its two length- $k$ coordinates, so it cannot lie in the intersection of an infinite subfamily. If an infinite subfamily $R$ has $x \in ⋂ R$ , then $x \in F$ has a branch $ρ$ , and the members of $R$ are the sets $B (ρ ↾ k)$ at arbitrarily large levels. Given open $D ∋ x$ , choose $n$ with $B (ρ ↾ n) \subseteq D$ ; any member $B (ρ ↾ k)$ of $R$ with $k \geq n$ lies in $D$ . Hence $R$ is a neighbourhood base at $x$ . By the Aleksandrov-Arhangel'skij equivalence of uniform bases with metacompact Moore spaces recorded in the source's Section 2, $X$ is a regular Moore space and is metacompact.

step 4.1step 2.1step 3.2F6

5.2

Choosing $v \in E ∖ (B_{1} \cup B_{2})$ gives that $ρ \hat{} v$ is acceptable — any nonempty $S \subseteq T \cap Σ_{n}$ above $ρ \hat{} v$ failing the acceptability witness would be dangerous for $v$ — and $ρ \hat{} v \notin T$ . The recursion produces an infinite sequence $f \in E^{ω}$ with every prefix acceptable and $f ↾ (i + 1) \notin T$ for all $i$ ; since $f \in [σ]$ means $σ = f ↾ ∣ σ ∣ \in T$ , the covering hypothesis is contradicted. Therefore some $T \cap Σ_{n}$ has a stafull subset, proving Lemma 3(a).

step 2.2step 4.3contradiction

6.1

For $n \in ω$ and $Z \subseteq Σ_{n}$ put $H_{Z} : = ⋃ {[σ] : σ \in Z}$ and $K_{Z} : = F ∖ H_{Z}$ . It suffices to separate $H_{Z}$ and $K_{Z}$ by disjoint open subsets of $X$ for every $n, Z$ (the source's Lemma 2). Here is the countable reduction. For disjoint closed $H, K \subseteq F$ , let $H_{n} = ⋃ {[σ] : ∣ σ ∣ = n, [σ] \cap H = \emptyset}, K_{n} = ⋃ {[σ] : ∣ σ ∣ = n, [σ] \cap K = \emptyset} .$ Then $K \subseteq ⋃_{n} H_{n}$ and $H \subseteq ⋃_{n} K_{n}$ , because the cylinders are a base for $F$ . By the assumed $H_{Z}$ -- $K_{Z}$ separation, choose disjoint open $O_{n}, P_{n}$ containing $H_{n}, F ∖ H_{n}$ , and disjoint open $R_{n}, S_{n}$ containing $K_{n}, F ∖ K_{n}$ . Thus $\overline{O_{n}} \cap H = \emptyset$ and $\overline{R_{n}} \cap K = \emptyset$ , since the corresponding $P_{n}, S_{n}$ are open and disjoint. The open sets $U = ⋃_{n} (R_{n} ∖ ⋃_{i \leq n} \overline{O_{i}}), V = ⋃_{n} (O_{n} ∖ ⋃_{i \leq n} \overline{R_{i}})$ contain $H, K$ respectively and are disjoint: if a point lay in the $n$ th piece of $U$ and the $m$ th piece of $V$ , the case $n \leq m$ contradicts removal of $\overline{R_{n}}$ from the latter, and $m \leq n$ contradicts removal of $\overline{O_{m}}$ from the former. Step 4.2 then handles isolated points.

step 4.2step 5.1F7

6.2

Non-metrizability. For $δ \in E$ put $Y_{δ} : = {f \in F : f (0) = δ}$ , a closed discrete family in $X$ . If $X$ were metrizable it would be collectionwise normal (F6), so there would be pairwise disjoint open sets $U_{δ} \supseteq Y_{δ}$ .

step 5.1F6

7.1

Fix $n \geq 1$ and $Z \subseteq Σ_{n}$ with $H_{Z}$ as in step 6.1, and let $C : = {γ < κ^{+} : if β < γ then Z \cap Σ^{β} = Z_{α} for some α < γ}$ . Then $C$ is a club: it is closed because for a limit $γ$ of $C$ -points and $β < γ$ some $γ^{'} \in C$ has $β < γ^{'} < γ$ , so $Z \cap Σ^{β} = Z_{α}$ with $α < γ^{'} < γ$ ; and it is unbounded because the assignment $γ \mapsto \sup {α + 1 : β < γ, Z \cap Σ^{β} = Z_{α}}$ can be iterated countably many times, remains below $κ^{+}$ by regularity, and its limit lies in $C$ . This uses $card Z \leq κ$ , so that at most $κ$ distinct truncations occur, and $cf (κ^{+}) > ω$ .

step 1.1step 1.2F1F3

7.2

Assume such ${U_{δ}}$ exist. Put $T : = {σ \in Σ : B (σ) \subseteq U_{σ (0)}}$ . Then $⋃ {[σ] : σ \in T} = F$ : for $f \in F$ one has $f \in Y_{f (0)} \subseteq U_{f (0)}$ , and since $f \notin Q$ some basic neighbourhood $B (σ) ∋ f$ lies in the open set $U_{f (0)}$ ; then $f \in [σ] \subseteq F$ , so $σ \subseteq f$ and $σ (0) = f (0)$ , giving $B (σ) \subseteq U_{σ (0)}$ and $f \in [σ]$ with $σ \in T$ .

step 6.2step 2.1F7

8.1

Let $γ (β)$ be the least element of $C$ greater than $β$ , and for $σ \in Σ_{n + 3}$ choose $j (σ) : = 1 + \max (n + 3, m_{γ (σ (n + 1))} (σ (0)), m^{*} (σ)),$ where $m^{*} (σ)$ is the least $m$ with $Z \cap Σ^{σ (n)} \in A (σ, m)$ if $γ (σ (n)) < σ (n + 2)$ , and $m^{*} (σ) : = 0$ otherwise; this is a definable choice from the given data. Then $j (σ) > m_{γ (σ (n + 1))} (σ (0))$ strictly, and whenever $γ (σ (n)) < σ (n + 2)$ the transfer of step 1.2 gives $Z \cap Σ^{σ (n)} \in A (σ, j (σ))$ (and even in $A (σ, j (σ) - 1)$ , which is what the applications at index $ρ^{'} ⊋ σ$ of level $j (σ)$ need). Put $W (σ) : = ⋃ {B (ρ) : σ \subseteq ρ \in Σ_{j (σ)}}$ .

step 1.2step 7.1given

8.2

By Lemma 3(a) proved above there are $n \geq 1$ and a stafull $S \subseteq T \cap Σ_{n}$ . Apply step 2.3 $n$ times to get a stafull $S^{'} \subseteq S$ with $σ (0)_{i} = σ^{'} (0)_{i}$ for all $σ, σ^{'} \in S^{'}$ and $i < n$ , and apply step 2.4 with $β : = κ$ to obtain $W = {σ_{α} : α < κ} \subseteq S^{'}$ with the interleaving property. Put $λ_{n} : = κ_{n}$ when $κ > ω$ and $λ_{n} : = (n + 1) κ_{n}$ when $κ = ω$ . For $σ \in W$ , step 1.2 gives $0 < ∣ A (σ, n) ∣ \leq λ_{n}$ : nonemptiness follows from $n \geq 1$ , $κ_{n} \geq 1$ , and $e_{σ} (0) = \emptyset$ . Enumerate it, repeating entries if necessary, as ${Z (σ, δ) : δ < λ_{n}}$ . Notice that $λ_{n} = κ_{n}$ is infinite when $κ > ω$ , while $λ_{n}$ is finite when $κ = ω$ .

step 4.3step 2.3step 2.4step 1.2step 7.2

9.1

Claim (Case 1). Let $σ, ν \in Σ_{n + 3}$ and suppose $σ ↾ n \in Z$ , $ν ↾ n \notin Z$ , and $γ (ν (n)) < σ (n + 2)$ ; assume $σ (0) < ν (0)$ . Then $W (σ) \cap W (ν) = \emptyset$ .

step 8.1

9.2

For $ρ, τ \in W$ with $ρ (0) < τ (0)$ , the triple $⟨ g, ρ, τ ⟩$ for any $g : Z (ρ^{*}) \to {0, 1}$ satisfies (7), (8), (10), (11) of step 2.1: $ρ, τ \in Σ_{n}$ ; the interleaving of step 2.4 gives (8); $ρ \subseteq ρ$ gives (10); and (11) is exactly the constancy of $σ (0)_{i}$ from step 2.3. Since $B (ρ) ∋ ⟨ g, ρ, τ ⟩$ would put it into $U_{ρ (0)} \cap U_{τ (0)} = \emptyset$ (as $ρ (0) \neq τ (0)$ ), the guarded (12) must fail at index $ρ$ or at index $τ$ for every $g$ .

step 8.2step 6.2step 2.3step 2.1

10.1

Let $⟨ g, ρ, τ ⟩ \in W (σ) \cap W (ν)$ : then there are $ρ^{'} \supseteq σ$ , $ν^{'} \supseteq ν$ with $∣ ρ^{'} ∣ = j (σ)$ , $∣ ν^{'} ∣ = j (ν)$ and $⟨ g, ρ, τ ⟩ \in C (ρ^{'}) \cap C (ν^{'})$ , because $[\cdot] \subseteq F$ and $F \cap Q = \emptyset$ . By condition (10) at both indices the cases $(ρ^{'} \subseteq ρ$ and $ν^{'} \subseteq ρ)$ and $(ρ^{'} \subseteq τ$ and $ν^{'} \subseteq τ)$ force $σ (0) = ν (0)$ , so after interchanging $σ$ and $ν$ if necessary we have $ρ^{'} \subseteq ρ$ and $ν^{'} \subseteq τ$ ; hence $ρ (0) = σ (0)$ , $τ (0) = ν (0)$ , and $ρ (i) = σ (i)$ for $i < j (σ)$ , $τ (i) = ν (i)$ for $i < j (ν)$ .

step 9.1step 2.1

10.2

Claim (Case 2). If $σ (n + 2) \leq γ (ν (n))$ under the hypotheses of step 9.1 (same membership assumptions), then $W (σ) \cap W (ν) = \emptyset$ .

step 9.1

10.3

For a fixed pair $ρ, τ$ consider the two systems of constraints on $g$ , with the exponent always the triple's first component $ρ$ , and include only guarded instances whose trace lies in $Z (ρ^{*})$ : the first system requires $g (Z \cap Σ^{ρ (m)}) = [ρ ↾ m \in Z]$ for $Z = Z (ρ, δ) \in A (ρ, n)$ of level $m$ , and the second requires the analogous value $[τ ↾ m \in Z]$ for $Z = Z (τ, η) \in A (τ, n)$ . Each guarded system is internally consistent: by step 2.3 and $(ρ ↾ m)^{*} = ρ (m - 1) < ρ (m)$ , the required value depends only on the trace $Z \cap Σ^{ρ (m)}$ ; the same holds for the $τ$ -system because $τ (m - 1) < ρ (m)$ by interleaving. No single total $g : Z (ρ^{*}) \to {0, 1}$ satisfies both guarded systems, by step 9.2. Hence a conflict exists: there are $δ, η < λ_{n}$ and $m < n$ whose common trace belongs to $Z (ρ^{*})$ and for which $Z (ρ, δ) \cap Σ^{ρ (m)} = Z (τ, η) \cap Σ^{ρ (m)}$ but $[ρ ↾ m \in Z (ρ, δ)] \neq [τ ↾ m \in Z (τ, η)]$ . Otherwise the function assigning each constrained trace its required value and $0$ to every other member of $Z (ρ^{*})$ would satisfy both systems. This is (17) in trace form together with (18a) or (18b).

step 9.2step 8.2step 2.3step 1.3

11.1

Condition (8) at the coordinates below $j (σ) \leq ∣ ρ ∣$ and $j (ν) \leq ∣ τ ∣$ , together with step 10.1, gives $σ (n) < ν (n) < σ (n + 1) \leq σ (n + 2)$ . Before applying guarded (12), verify its domain condition. Since $γ (σ (n))$ and $γ (ν (n))$ are club points and $σ (n) < ν (n)$ , step 7.1 gives indices $index (Z \cap Σ^{σ (n)}) < γ (σ (n)) \leq γ (ν (n)), index (Z \cap Σ^{ν (n)}) < γ (ν (n)) .$ The Case-1 hypothesis gives $γ (ν (n)) < σ (n + 2) \leq ρ^{*}$ , so both traces belong to $Z (ρ^{*})$ . Step 8.1 now puts $Z \cap Σ^{σ (n)}$ in $A (ρ^{'}, j (σ))$ and $Z \cap Σ^{ν (n)}$ in $A (ν^{'}, j (ν))$ . Their condition-(12) traces both reduce to $Z \cap Σ^{ρ (n)}$ because $ρ (n) = σ (n) < ν (n)$ , while $ρ^{'} ↾ n \in Z$ and $ν^{'} ↾ n \notin Z$ . Thus guarded (12) yields $1 = g (Z \cap Σ^{ρ (n)}) = 0$ , a contradiction. Hence Case 1 holds.

step 7.1step 8.1step 10.1step 1.3

11.2

Assume $⟨ g, ρ, τ ⟩ = : q \in W (σ) \cap W (ν)$ and argue as in step 10.1 to get $ρ^{'} \subseteq ρ$ , $ν^{'} \subseteq τ$ ; then (8) gives the chain $ν (n) < σ (n + 1) < ν (n + 1) < σ (n + 2) \leq γ (ν (n))$ , so both $σ (n + 1)$ and $ν (n + 1)$ lie in the interval $[ν (n), γ (ν (n)))$ , which contains no $C$ -point by the minimality of $γ (ν (n))$ ; applying the definition of $γ$ inside that gap gives $γ (σ (n + 1)) = γ (ν (n + 1)) = γ (ν (n))$ . By step 8.1, $j (σ) > m_{γ (σ (n + 1))} (σ (0)), j (ν) > m_{γ (ν (n + 1))} (ν (0)) .$ Put $m : = \max {j (σ), j (ν)}$ and $m^{*} : = \max {m_{γ (σ (n + 1))} (σ (0)), m_{γ (ν (n + 1))} (ν (0))}$ ; then $m^{*} < m$ , and condition (11) at the indices $ρ^{'}$ , $ν^{'}$ gives $ρ (0)_{i} = τ (0)_{i}$ for every $i < m$ .

step 8.1step 10.1step 10.2

11.3

Define $U : = ⋃ {W (σ) : σ \in Σ_{n + 3}, σ ↾ n \in Z}, V : = ⋃ {W (ν) : ν \in Σ_{n + 3}, ν ↾ n \notin Z} .$ These sets are open and contain $H_{Z}, K_{Z}$ respectively: extend the length- $n$ initial segment of any branch to length $n + 3$ and use $B (σ) \subseteq W (σ)$ . If Cases 1 and 2 both hold, every cross-pair of their displayed constituents is disjoint, according as $γ (ν (n)) < σ (n + 2)$ or $σ (n + 2) \leq γ (ν (n))$ (interchange the two sequences first when their zeroth coordinates are reversed). Hence the two cases imply $U \cap V = \emptyset$ .

step 8.1step 9.1step 10.2

11.4

Colour each pair ${ρ, τ} \subseteq W$ with $ρ (0) < τ (0)$ by the least conflict witness $(δ, η, m, a l t)$ in a fixed well-ordering of $λ_{n} \times λ_{n} \times n \times 2$ . If $κ = ω$ , this is a finite colour set and the infinite Ramsey theorem (F5) gives an infinite homogeneous set. If $κ > ω$ , then $λ_{n} = κ_{n}$ is infinite and, since $2^{κ_{n}} < κ$ , the Erdős–Rado theorem (F5) applied to a subset of $W$ of size $(2^{κ_{n}})^{+}$ gives a homogeneous set of size $κ_{n}^{+}$ . Either way there are $ρ, σ, τ \in W$ with $ρ (0) < σ (0) < τ (0)$ and one common alternative of (18), common indices $δ, η$ , and a common $m$ such that (17) holds for each of the three pairs.

step 10.3step 2.4F5F2

12.1

The trace-domain verification in step 11.1 is symmetric for the two listed sets: their enumeration indices are below $γ (σ (n))$ and $γ (ν (n))$ , respectively, and both club successors lie below $ρ^{*}$ in Case 1. Thus every evaluation of $g$ made there is inside $Z (ρ^{*})$ , as required by step 1.3.

step 7.1step 8.1step 10.1step 11.1

12.2

Since $σ (0) \neq ν (0)$ (they are distinct elements of the interleaved family $W$ ) and both lie in $E \cap γ (σ (n + 1))$ , the ladder separation hypothesis applied at $β = γ (σ (n + 1))$ gives $(σ (0))_{i} \neq (ν (0))_{i}$ for every $i \geq m^{*}$ , in particular at $i = m^{*} < m$ . This contradicts the agreement of step 11.2 at level $m^{*}$ , because $ρ (0) = σ (0)$ and $τ (0) = ν (0)$ . Hence Case 2 holds.

step 11.2given

13.1

Step 11.1 proves Case 1 and step 12.2 proves Case 2, so step 11.3 gives disjoint open sets separating $H_{Z}$ and $K_{Z}$ for every $n$ and $Z \subseteq Σ_{n}$ . The countable reduction of step 6.1 then separates arbitrary disjoint closed subsets of $F$ , and step 4.2 adds their isolated parts. Hence $X$ is normal.

step 4.2step 6.1step 11.1step 11.3step 12.2

14.1

With the triple of step 11.4 the printed chain computes: from (17) for the pairs $(ρ, σ)$ , $(ρ, τ)$ and the monotonicity $Σ^{ρ (m)} \subseteq Σ^{σ (m)}$ from the interleaving, $Z (σ, η) \cap Σ^{ρ (m)} = Z (ρ, δ) \cap Σ^{ρ (m)} = Z (τ, η) \cap Σ^{ρ (m)} = Z (σ, δ) \cap Σ^{ρ (m)} .$ Under the common alternative (18a), the pair $(σ, τ)$ gives $σ ↾ m \in Z (σ, δ)$ and the pair $(ρ, σ)$ gives $σ ↾ m \notin Z (σ, η)$ ; since $σ ↾ m \in Σ^{ρ (m)}$ by step 2.3, the chain transfers membership across $Σ^{ρ (m)}$ and yields $σ ↾ m \in Z (σ, η)$ , a contradiction. Under (18b) the same two lines run with the membership signs exchanged. This contradiction establishes that ${U_{δ} : δ \in E}$ is not disjoint, so $X$ is not collectionwise normal and hence not metrizable; with step 13.1 it is a normal nonmetrizable Moore space.

step 11.4step 10.3step 2.3step 6.2step 13.1discharge-contradiction∎

Remarks

Two documented readings of the printed notation. (12) is used with the bound $ρ (m)$ for a level- $m$ set, which is how the source's own Case 1 display on printed p. 370 uses it; the printed notation sentence after (12) is off by one restriction step. (17) is used in trace form $Z (ρ, δ) \cap Σ^{ρ (m)} = Z (τ, η) \cap Σ^{ρ (m)}$ , which is what the printed four-term chain displays. Both are recorded as local repairs, not source attributions.
The two local repairs to the §6 parameters. $j (σ)$ is chosen strictly above the printed lower bounds, and the entry level of $Z \cap Σ^{σ (n)}$ is arranged one step below $j (σ)$ ; without the strictness the printed Case 2 does not close. Recorded as a local repair.
The trace-domain condition of step 1.3 is the guarded reading of (12); §6's instances are proved in step 12.1, and §7's instances are the applicable ones by definition, so no trace outside $Z (ρ^{*})$ is ever evaluated.
The finite-cardinal clause in (4). When $κ = ω$ , the source asks only that each $A (σ, m)$ be finite. Step 1.2 supplies the uniform finite bound $(∣ σ ∣ + 1) κ_{m}$ , and step 11.4 uses that finite bound as the Ramsey colour set. No absorption identity is applied to a finite $κ_{m}$ .
AC is used in the enumeration of $Z$ and of $E \cap β$ , in the choice of the ladders and the $j (σ)$ , in the countable recursion of step 3.3, and in the Ramsey/Erdős–Rado step; it is declared as a dependency and no choice-free reading is claimed.

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

CH yields a normal nonmetrizable Moore space

Statement

$Z F C + C H$ proves that a normal nonmetrizable Moore space exists. By Metrizable spaces are collectionwise normal such a space is the CH instance of the failure of the normal Moore space conjecture.

Facts & Assumptions

Given: The continuum hypothesis $C H$ (The continuum hypothesis, and what this page does not prove): every uncountable set of reals is equinumerous with $R$ ; equivalently here $2^{ℵ_{0}} = ℵ_{1}$ , i.e. $2^{ω} = ω_{1}$ in the von Neumann ordinals (Cardinal (initial ordinal) and cardinality, The natural numbers $N$ (von Neumann)).

[F1]

$ω = ℵ_{0}$ is the least infinite cardinal and $ω_{1} = ω^{+}$ is the least uncountable cardinal; a countable ordinal is one below $ω_{1}$ , and the nonzero limit ordinals below $ω_{1}$ are exactly the countable ordinals of cofinality $ω$ (The natural numbers $N$ (von Neumann), Cardinal (initial ordinal) and cardinality).

[F2]

Clubs and stationarity in $ω_{1}$ : the set of nonzero limit ordinals below $ω_{1}$ contains the club of all limit ordinals and is therefore stationary (Closed unbounded subsets of ordinals, Basic stationary-set calculus).

[F3]

Fleissner's construction (Fleissner's construction of a normal nonmetrizable Moore space from level data): if $κ$ is an infinite cardinal, $(κ_{n})_{n \in ω}$ is an increasing sequence of cardinals with $\sup_{n} κ_{n} = κ$ and $2^{κ_{n}} < κ$ for every $n$ , $2^{κ} = κ^{+}$ , $E \subseteq {δ < κ^{+} : cf (δ) = ω}$ is stationary, and some fixed ladders admit the separation functions $m_{β}$ for every $β < κ^{+}$ , then a normal nonmetrizable Moore space exists (Fleissner's HYP covering interface, Moore spaces and developments).

[F4]

In ZFC, for every at-most-countable set $A$ there is an index set $I_{A}$ which is either a finite initial segment of $ω$ or all of $ω$ , and a bijection $I_{A} \to A$ ; choice permits these bijections to be fixed simultaneously for all $β < ω_{1}$ (The Axiom of Choice, A function is a relation $f$ with $(a, b) \in f$ and $(a, c) \in f$ implying $b = c$ ; $f : A \to B$ , the value $f (a)$ , domain and codomain).

Proof

technique · direct; instantiate the construction of [F3]

1.1

Work in $Z F C + C H$ and put $κ : = ω$ , $κ_{n} : = n$ for $n \in ω$ , and $E : = {δ < ω_{1} : δ$ is a nonzero limit ordinal $}$ . Then $E \subseteq {δ < ω_{1} : cf (δ) = ω}$ by [F1], and $E$ is stationary in $ω_{1}$ by [F2].

givenF1F2

2.1

The parameters of step 1.1 satisfy the numerical hypotheses of [F3]: $\sup_{n} κ_{n} = \sup_{n} n = ω = κ$ ; for every $n$ the ordinal $2^{n}$ is a finite ordinal, hence $2^{n} < ω = κ$ ; and $2^{κ} = 2^{ω} = ω_{1} = κ^{+}$ is $C H$ .

step 1.1givenF3

2.2

Fix, for each $δ \in E$ , an increasing sequence $(δ_{i})_{i \in ω}$ of nonlimit ordinals cofinal in $δ$ ; such a sequence exists because $δ$ has cofinality $ω$ , and all the sequences are chosen simultaneously by choice.

step 1.1F4

3.1

For every $β < ω_{1}$ there is a function $m_{β} : E \cap β \to ω$ with $δ_{i} \neq η_{i}$ whenever $δ \neq η$ are in $E \cap β$ and $i \geq \max (m_{β} (δ), m_{β} (η))$ . Indeed $E \cap β \subseteq β$ is at most countable, so by [F4] fix a bijection $k \mapsto δ^{k}$ from an index set $I$ , where $I$ is either a finite initial segment of $ω$ or all of $ω$ , onto $E \cap β$ . For $k < n$ in $I$ the set $F (k, n) : = {i \in ω : (δ^{k})_{i} = (δ^{n})_{i}}$ is finite, because if the two increasing ladders agreed at infinitely many levels then those common values would be cofinal in both $δ^{k}$ and $δ^{n}$ , forcing $δ^{k} = δ^{n}$ . Recursively for $n \in I$ , define $m (n) : = 1 + \max ({0} \cup {m (k) : k < n} \cup ⋃_{k < n} F (k, n))$ . The set maximized over is finite (also when $I$ is finite or empty), so the recursion is well defined. If $k < n$ are in $I$ , then $m (n) > m (k)$ and every $i \in F (k, n)$ satisfies $i < m (n)$ ; hence no coincidence level of the pair reaches $\max (m (k), m (n)) = m (n)$ . Thus $m_{β} (δ^{k}) : = m (k)$ is well defined by injectivity of the enumeration and has the required separation property.

step 2.2F1F4

4.1

Steps 2.1, 2.2 and 3.1 put exactly the hypotheses of [F3] at $κ = ω$ , and ( $κ_{0} = 0$ already) applying it yields a normal nonmetrizable Moore space.

step 2.1step 3.1F3∎

Remarks

The CH instance satisfies the local HYP interface. For every $β < ω_{1}$ one has $cf (β) \leq ω$ , so the interface's clause (3b), which concerns only $cf (β) > ω$ , is vacuous. At $β = ω^{2}$ , the two disjoint sets ${ω n : 1 \leq n < ω}$ and ${ω n + 1 : n < ω}$ are both clubs; thus $E \cap ω^{2}$ contains a club but is nevertheless nonstationary. Step 3.1 independently proves the countable ladder separation that the construction consumes, as in the source's parenthetical countable-case argument.
Where CH is used. Only in $2^{ω} = ω_{1}$ (step 2.1). The stationarity of $E$ and the countable ladder separation are theorems of $Z F C$ .

CorollaryStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

V=L refutes the normal Moore space conjecture

Statement

$Z F C + V = L$ refutes the normal Moore space conjecture: it proves that there is a normal nonmetrizable Moore space (CH yields a normal nonmetrizable Moore space, Moore spaces and developments).

Facts & Assumptions

Given: The axiom $V = L$ over $Z F C$ .

[F1]

$V = L$ implies $A C + G C H$ (The generalized continuum hypothesis holds in L).

[F2]

GCH gives the continuum hypothesis at $ω$ : the instance at the infinite cardinal $ℵ_{0}$ says $2^{ℵ_{0}} = ℵ_{1}$ , since $ℵ_{0}^{+} = ℵ_{1}$ and $P (ω)$ has cardinality $2^{ℵ_{0}}$ (Cantor's theorem: $A ≺ P (A)$ , Fleissner's HYP covering interface); this is the form of CH consumed by CH yields a normal nonmetrizable Moore space.

[F3]

$Z F C + C H$ proves that a normal nonmetrizable Moore space exists (CH yields a normal nonmetrizable Moore space).

Proof

technique · direct

1.1

Assume $V = L$ . By [F1] both AC and GCH hold.

givenF1

2.1

By [F2], GCH gives $2^{ℵ_{0}} = ℵ_{1}$ , i.e. CH.

step 1.1F2

3.1

By [F3], applied under CH, there is a normal nonmetrizable Moore space.

step 2.1F3

4.1

A normal Moore space that is not metrizable is a counterexample to the normal Moore space conjecture, so $V = L$ refutes it.

step 3.1given∎

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

HYP produces a normal nonmetrizable Moore space

Statement

$Z F C + H Y P$ proves that there is a normal nonmetrizable Moore space (Fleissner's HYP covering interface, Fleissner's construction of a normal nonmetrizable Moore space from level data).

Facts & Assumptions

Given: Witnesses $κ, (κ_{n})_{n \in ω}, E$ for HYP and fixed ladders $(δ_{i})_{i \in ω}$ for $δ \in E$ (Fleissner's HYP covering interface).

[F1]

HYP asserts that $κ$ is an infinite cardinal, that $(κ_{n})_{n \in ω}$ is an increasing sequence of cardinals, and the conjunction of clauses (1a), (1b), (2), (3a), (3b): $\sup_{n} κ_{n} = κ$ ; $2^{κ_{n}} < κ$ for every $n$ ; $2^{κ} = κ^{+}$ ; $E \subseteq {δ < κ^{+} : cf (δ) = ω}$ is stationary in $κ^{+}$ ; and $E \cap β$ is not stationary in $β$ for every $β < κ^{+}$ with $cf (β) > ω$ (Fleissner's HYP covering interface, Cardinal (initial ordinal) and cardinality).

[F2]

Lemma 1 from the same hypotheses: for every $β < κ^{+}$ there is $m_{β} : E \cap β \to ω$ such that $δ_{i} \neq η_{i}$ for all distinct $δ, η \in E \cap β$ and all $i \geq \max (m_{β} (δ), m_{β} (η))$ (Ladder separation from HYP).

[F3]

The construction of Fleissner's construction of a normal nonmetrizable Moore space from level data converts exactly the data of [F1] together with [F2] into a normal nonmetrizable Moore space (Moore spaces and developments, Metrizable spaces are collectionwise normal).

Proof

technique · direct

1.1

Assume HYP. Then $κ$ is infinite, $(κ_{n})$ is increasing, and clauses (1a), (1b), (2), (3a) of [F1] hold for the given $κ$ , $(κ_{n})$ , and $E$ .

givenF1

2.1

The separation conclusion of [F2] holds for the fixed ladders. Its proof uses clause (3b) at limit ordinals of uncountable cofinality and uses clause (3a) to obtain an automatic successor club at countable-cofinality stages.

step 1.1F1F2

3.1

Steps 1.1 and 2.1 put all hypotheses of the construction of [F3] at the given parameters, so there is a normal nonmetrizable Moore space.

step 1.1step 2.1F3∎

Remarks

HYP is used only through [F2]. The source says so explicitly ("we will not use (3b) directly, but rather the following consequence"), and the construction of [F3] is stated with the separation conclusion as an hypothesis, so the formal dependency is exact.
The case $κ = ω$ is covered by the same item. When HYP holds with $κ = ω$ the clause (1b) is literal cardinal arithmetic on finite ordinals and the separation conclusion is supplied by Ladder separation from HYP like every other instance; the CH case with nonreflecting failure is treated separately in CH yields a normal nonmetrizable Moore space.

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

NMSC gives an inner model with a measurable cardinal

Statement

$Z F C + N M S C$ proves that there is an inner model with a measurable cardinal, where NMSC is the assertion that every normal Moore space is metrizable and an inner model is a transitive class model of $Z F C$ containing all ordinals (Semantic and formal inner-model theorem for L).

Facts & Assumptions

Given: The hypothesis NMSC over $Z F C$ .

[F1]

If there is no inner model with a measurable cardinal, then $Z F C$ proves HYP with the parameters the construction consumes (No inner measurable implies Fleissner's HYP, Fleissner's HYP covering interface).

[F2]

HYP proves that there is a normal nonmetrizable Moore space (HYP produces a normal nonmetrizable Moore space, Moore spaces and developments).

[F3]

Such a space is a counterexample to NMSC. [given]

Proof

technique · contrapositive

1.1

Argue contrapositively inside $Z F C$ : assume there is no inner model with a measurable cardinal.

contrapositive-reduceassume-hypgiven

2.1

By [F1] the assumption implies HYP with a fixed parameter triple $κ, (κ_{n}), E$ and fixed ladders.

step 1.1F1

3.1

By [F2] applied to those parameters there is a normal nonmetrizable Moore space.

step 2.1F2

4.1

That space violates NMSC by [F3]. Hence the assumption of step 1.1 implies the negation of NMSC; contraposition gives that NMSC implies the existence of an inner model with a measurable cardinal.

step 3.1F3discharge-contrapositive∎

Remarks

The inner model is not constructed by the topological argument. The topological half produces a counterexample to NMSC from HYP; the existence of the inner model is the contrapositive of the covering-theoretic half (No inner measurable implies Fleissner's HYP). No claim is made here that the space or its construction yields a measurable cardinal.

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Metatheoretic consistency lower bound for NMSC

Statement

Externally, in the metatheory $Z F C$ , $Con (Z F C + N M S C)$ implies $Con (Z F C + there is a measurable cardinal)$ . Here each consistency assertion is evaluated on the standard natural-number proof codes using the arithmetic formula fixed in The standard certified provability predicate. No claim is made that an unspecified arithmetic base proves the displayed implication.

Facts & Assumptions

Given: The metatheory $Z F C$ ; the fixed arithmetization of the calculus of The standard certified provability predicate.

[F1]

$Z F C + N M S C$ proves that there is an inner model with a measurable cardinal (NMSC gives an inner model with a measurable cardinal). In the first-order class convention this has the following finite-fragment meaning: for each externally fixed finite set $Δ$ of target axioms, one uses a single class-defining formula (with its fixed parameters) for the asserted inner model, and the source theory proves nonemptiness of that class and every $σ^{M}$ for $σ \in Δ$ . This is separate relativization for each fixed formula, not quantification over a class truth predicate (Relativization to sets and definable classes).

[F2]

Every actual derivation is finite. If an actual $U$ -refutation uses the finite set $Δ$ of nonlogical axioms, apply [F1] only to that $Δ$ . Relativization to its one nonempty class predicate is an interpretation of the finite theory $Δ$ in the source theory, so the finite derivation translates to a source refutation (Interpretation transports derivations and inconsistency, Finite-fragment model transfer proves relative consistency).

[F3]

This per-refutation, externally selected finite translation proves only the external consistency implication. It does not provide one fixed interpretation of all of $U$ , an effective selector of class predicates from proof codes, or a base-verifiable total refutation-code map. Any assertion that an arithmetic base $B$ proves the implication would require exactly such additional uniform data (Formal consistency transfer from a verified reduction).

Proof

technique · direct

1.1

Let $T : = Z F C + N M S C$ and $U : = Z F C + there is a measurable cardinal$ . Suppose, contrapositively, that an actual finite $U$ -refutation $p$ exists, and let $Δ$ be the finite set of nonlogical $U$ -axioms occurring in $p$ .

givenF2

2.1

Apply the finite-fragment reading of the inner-model theorem [F1] to this particular $Δ$ . It supplies one definable nonempty class $M$ and $T$ -proofs of $σ^{M}$ for every $σ \in Δ$ . With membership and equality unchanged, these finitely many obligations make relativization to $M$ an interpretation of the finite theory $Δ$ in $T$ .

step 1.1F1F2

3.1

Translate the fixed refutation $p$ through that finite interpretation. By [F2], its translated logical steps and the finitely many proofs from step 2.1 assemble into an actual $T$ -refutation. Thus every actual $U$ -refutation entails an actual $T$ -refutation, so absence of a $T$ -refutation entails absence of a $U$ -refutation. Under the standard-natural-number convention in the Statement, this is $Con (T) \to Con (U)$ .

step 1.1step 2.1F2F3∎

Remarks

What is and is not used. The proof supplies the external syntactic consistency implication by selecting a definable-class relativization after a particular finite refutation is fixed. It neither claims one fixed global interpretation nor that a named arithmetic base proves the implication, and it does not build or assume a transitive set model of the source theory.
AC. $Z F C$ is part of both theories; the relativization of AC to the inner model is part of [F1], and no additional choice principle is used in the transfer (The Axiom of Choice).

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22‡ rests on unproved material (inherited)Open item page →

The consistency-strength sandwich for NMSC

Statement

$Con (Z F C + N M S C)$ implies $Con (Z F C + there is a measurable cardinal)$ , while $Con (Z F C + there is a strongly compact cardinal)$ implies $Con (Z F C + N M S C)$ (A strongly compact cardinal gives the NMSC consistency upper bound, Metatheoretic consistency lower bound for NMSC).

Facts & Assumptions

Given: The two metatheoretic consistency hypotheses.

[F1]

The lower bound: $Con (Z F C + N M S C)$ implies $Con (Z F C + a measurable cardinal)$ (Metatheoretic consistency lower bound for NMSC).

[F2]

The upper bound: $Con (Z F C + a strongly compact cardinal)$ implies $Con (Z F C + N M S C)$ (A strongly compact cardinal gives the NMSC consistency upper bound).

[F3]

The internal assertion " $Z F C + N M S C$ proves that there is an inner model with a measurable cardinal" (NMSC gives an inner model with a measurable cardinal) is a distinct claim from both consistency implications. [given]

Proof

technique · direct

1.1

Assume $Con (Z F C + N M S C)$ : by [F1], $Con (Z F C + a measurable cardinal)$ follows.

givenF1

1.2

Assume $Con (Z F C + a strongly compact cardinal)$ : by [F2], $Con (Z F C + N M S C)$ follows.

givenF2

2.1

Steps 1.1 and 1.2 are the two implications of the Statement, and [F3] keeps the internal measurable-inner-model consequence separate from them.

step 1.1step 1.2F3∎

Remarks

Three distinct claims. The internal theorem, the lower consistency implication and the forcing/measure upper implication are kept as three separate assertions; no converse of the upper bound is claimed here (A strongly compact cardinal gives the NMSC consistency upper bound).

5 · Examples, counterexamples and false statements

None yet.