CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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Refuted, assuming countable choice: every Hausdorff space built from ordinal spaces is normal. The deleted Tychonoff plank $((\omega_1 + 1) \times (\omega + 1)) \setminus \{(\omega_1, \omega)\}$ is Hausdorff and not normal

Statement refuted

False claim: every Hausdorff space obtained from ordinals with their order topologies (The order topology on an ordinal, with the half-open intervals $(\alpha, \beta]$ and the initial segments $[0, \beta]$ as a basis) by forming a product (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space) and passing to a subspace (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) is normal (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly).

The witness is the deleted Tychonoff plank. Write $W := \omega_1 + 1$ and $Z := \omega + 1$ , which by the successor clause of ordinal addition (Ordinal addition $\alpha + \beta$ ) are $\omega_1^{+}$ and $\omega^{+}$ ; give each its order topology and $W \times Z$ the product topology, and put

$T \;:=\; (W \times Z) \setminus \{(\omega_1, \omega)\}$

with the subspace topology. Then $T$ is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) and $T$ is not normal: the two sets

$A \;:=\; \{\, (\omega_1, n) : n \in \omega \,\}, \qquad B \;:=\; \{\, (\xi, \omega) : \xi \in \omega_1 \,\}$

are disjoint and closed in $T$ and have no disjoint open neighbourhoods.

Assuming the Axiom of Countable Choice (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )). The cost is inherited from Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$ , so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable, which is the single step of the argument that spends it; everything else below is a theorem of ZF, the ordinals $\alpha_n$ being defined as least elements rather than selected.

What is and is not claimed. $T$ is Hausdorff and not normal, and that is all. Nothing here asserts that $T$ is regular, nor that $W \times Z$ itself is normal, nor anything about which separation axioms are hereditary or productive; those questions need machinery this page does not have.

Facts & Assumptions

Given: $W = \omega_1 + 1$ and $Z = \omega + 1$ with their order topologies, the product $W \times Z$ , the subspace $T = (W \times Z) \setminus \{(\omega_1,\omega)\}$ , and the sets $A$ and $B$ above.

[A1]

The basic open sets of an ordinal $\gamma$ are $[0,\beta]$ for $\beta \in \gamma$ and $(\alpha,\beta]$ for $\alpha < \beta$ in $\gamma$ , and they form a basis (The order topology on an ordinal, with the half-open intervals $(\alpha, \beta]$ and the initial segments $[0, \beta]$ as a basis, Basis and subbasis for a topology, and the topology generated by a family of sets).

[A2]

For a binary product the basic product-open sets are exactly the boxes $P \times Q$ with $P$ and $Q$ open, and the boxes $P_0 \times Q_0$ with $P_0$ , $Q_0$ basic in the factors also form a basis: given $(x,y) \in P \times Q$ , basic $P_0 \ni x$ inside $P$ and $Q_0 \ni y$ inside $Q$ give $(x,y) \in P_0 \times Q_0 \subseteq P \times Q$ (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis, Basis and subbasis for a topology, and the topology generated by a family of sets).

[A3]

The open sets of $T$ are the traces $U' \cap T$ of the open sets of $W \times Z$ , the closed sets of $T$ are the traces of the closed sets, and the traces of a basis form a basis (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, For $A \subseteq S \subseteq X$ the closure of $A$ in $S$ is $\overline{A}^{X} \cap S$ , while the interior only contains $\operatorname{int}^{X}(A) \cap S$ , with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$ ).

[L1]

$\alpha + 1 = \alpha^{+}$ ; $\omega$ and $\omega_1$ are limit ordinals; $\omega_1$ is uncountable and every ordinal below it is at most countable (Ordinal addition $\alpha + \beta$ , $\omega$ is the least limit ordinal, Successor and limit ordinals, $\omega_1$ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF, The first uncountable ordinal $\omega_1 := \aleph(\omega)$ , Finite, countably infinite, countable, uncountable).

[L2]

For ordinals exactly one of $\xi < \eta$ , $\xi = \eta$ , $\eta < \xi$ holds, and every nonempty set of ordinals has a least element (Trichotomy and well-ordering of the ordinals, Basic closure properties of ordinals, Ordinal (von Neumann)).

[L3]

Assuming $\mathrm{AC}_\omega$ , every at most countable subset of $\omega_1$ has an upper bound $\alpha \in \omega_1$ , and no such subset is cofinal in $\omega_1$ (Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$ , so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable, Cofinal subset of an ordinal, The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )).

[L4]

The image of $\mathbb{N}$ under a function is nonempty and at most countable (A nonempty set is at most countable iff it is a surjective image of $\mathbb{N}$ , Finite, countably infinite, countable, uncountable).

[L5]

Every ordinal with its order topology is $T_1$ , Hausdorff and regular (Every ordinal with its order topology has a basis of clopen sets, and is $T_1$ , Hausdorff and regular, $T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces).

[L6]

Normality: two disjoint closed sets have disjoint open supersets. Hausdorff: distinct points have disjoint open neighbourhoods. A set is closed exactly when its complement is open (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, Interior, closure, boundary, exterior, derived set and isolated point in a topological space, 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).

Counterexample

technique · contradiction

1.1

$\omega_1$ is open in $W$ and $\omega$ is open in $Z$ : for $\xi < \omega_1$ the basic set $[0,\xi]$ lies inside $\omega_1$ , and for $n < \omega$ the basic set $[0,n]$ lies inside $\omega$ .

A1L2

1.2

$A$ and $B$ are subsets of $T$ and are disjoint: every point of $A$ has second coordinate in $\omega$ and every point of $B$ has second coordinate $\omega$ , and $\omega \notin \omega$ .

L1L2

1.3

$T$ is Hausdorff: let $(\xi,\eta) \ne (\xi',\eta')$ in $T$ . If $\xi \ne \xi'$ , [L5] gives disjoint open $P, P'$ in $W$ containing them, and $(P \times Z) \cap T$ and $(P' \times Z) \cap T$ are disjoint open sets of $T$ containing the two points by [A2] and [A3]; if $\eta \ne \eta'$ the same argument runs in the second factor.

A2A3L5L6

2.1

$T \setminus A = (\omega_1 \times Z) \cap T$ : a point $(\xi,\eta) \in T$ with $\xi = \omega_1$ has $\eta \ne \omega$ , hence $\eta \in \omega$ and $(\xi,\eta) \in A$ ; so the points of $T$ outside $A$ are exactly those with $\xi \in \omega_1$ .

step 1.2L1L2

2.2

$T \setminus B = (W \times \omega) \cap T$ : a point $(\xi,\eta) \in T$ with $\eta = \omega$ has $\xi \ne \omega_1$ , hence $\xi \in \omega_1$ and $(\xi,\eta) \in B$ ; so the points of $T$ outside $B$ are exactly those with $\eta \in \omega$ .

step 1.2L1L2

3.1

By steps 1.1, 2.1 and 2.2 and [A2] and [A3], the sets $T \setminus A$ and $T \setminus B$ are open in $T$ , so $A$ and $B$ are closed in $T$ .

step 1.1step 2.1step 2.2A2A3L6

4.1

Suppose $U$ and $V$ are disjoint open subsets of $T$ with $A \subseteq U$ and $B \subseteq V$ .

step 3.1L6assume-contra

5.1

Fix $n \in \omega$ . Then $(\omega_1, n) \in U$ , so by [A2] and [A3] there are basic $P$ in $W$ and $Q$ in $Z$ with $(\omega_1,n) \in (P \times Q) \cap T \subseteq U$ .

step 4.1A2A3

6.1

Under step 5.1: $P$ contains $\omega_1$ , and $\omega_1$ is the largest element of $W$ , so $P$ is $[0,\omega_1] = W$ or $(\alpha,\omega_1]$ with $\alpha \in \omega_1$ ; in either case $(\alpha, \omega_1] \subseteq P$ for some $\alpha \in \omega_1$ , taking $\alpha := 0$ in the first case.

step 5.1A1L1L2

7.1

Under step 5.1: $(\alpha,\omega_1] \times \{n\} \subseteq (P \times Q) \cap T \subseteq U$ , since $n \in Q$ and since every point of $(\alpha,\omega_1] \times \{n\}$ has second coordinate $n \ne \omega$ and so lies in $T$ .

step 5.1step 6.1L1L2

8.1

For each $n \in \omega$ the set of $\alpha \in \omega_1$ with $(\alpha,\omega_1] \times \{n\} \subseteq U$ is nonempty by step 7.1, so it has a least element $\alpha_n$ by [L2]; this defines $\alpha_n$ from $n$ and $U$ alone and selects nothing.

step 7.1L2

9.1

The set $S := \{\, \alpha_n : n \in \omega \,\}$ is an at most countable subset of $\omega_1$ by [L4], so [L3] gives $\alpha \in \omega_1$ with $\alpha_n \le \alpha$ for every $n \in \omega$ .

step 8.1L3L4

10.1

Hence $(\alpha,\omega_1] \subseteq (\alpha_n,\omega_1]$ for every $n$ , and therefore $(\alpha,\omega_1] \times \{n\} \subseteq U$ for every $n \in \omega$ .

step 8.1step 9.1L2

10.2

$\alpha^{+} \in \omega_1$ , since $\omega_1$ is a limit ordinal and $\alpha \in \omega_1$ ; so $(\alpha^{+}, \omega) \in B \subseteq V$ .

step 4.1step 9.1L1L2

11.1

By [A2] and [A3] there are basic $P'$ in $W$ and $Q'$ in $Z$ with $(\alpha^{+},\omega) \in (P' \times Q') \cap T \subseteq V$ ; and $Q'$ contains $\omega$ , the largest element of $Z$ , so $Q'$ is $[0,\omega] = Z$ or $(m,\omega]$ with $m \in \omega$ , and in either case $(m,\omega] \subseteq Q'$ for some $m \in \omega$ .

step 10.2A1A2A3L1L2

12.1

Put $z := (\alpha^{+}, m^{+})$ . Then $m^{+} \in \omega$ because $\omega$ is a limit ordinal, and $m < m^{+} \le \omega$ , so $m^{+} \in (m,\omega] \subseteq Q'$ ; also $\alpha^{+} \in P'$ ; and $z \in T$ because its first coordinate is $\alpha^{+} \ne \omega_1$ . Hence $z \in (P' \times Q') \cap T \subseteq V$ .

step 11.1L1L2

13.1

Also $\alpha < \alpha^{+} \le \omega_1$ , so $\alpha^{+} \in (\alpha,\omega_1]$ , and $m^{+} \in \omega$ ; hence $z \in (\alpha,\omega_1] \times \{m^{+}\} \subseteq U$ by step 10.1.

step 10.1step 12.1L2

14.1

Steps 12.1 and 13.1 put $z$ in $U \cap V$ , contradicting the disjointness assumed in step 4.1; so no such $U$ and $V$ exist, the disjoint closed sets $A$ and $B$ of step 3.1 cannot be separated, and $T$ is not normal by [L6]. With step 1.3 the space $T$ is Hausdorff and not normal, which refutes the claim.

step 3.1step 1.3step 4.1step 12.1step 13.1L6discharge-contradiction∎

Remarks

Where the uncountability of $\omega_1$ is spent, and where it is not. The whole argument turns on step 9.1: a countable family of ordinals below $\omega_1$ is bounded below $\omega_1$ , so a single $\alpha$ works for every $n$ at once. Nothing analogous holds in the second factor, and nothing analogous is needed: $\omega$ enters only through the fact that $m^{+}$ is again below $\omega$ .
Why the point $(\omega_1,\omega)$ must be deleted. With that corner present, the set $B$ would not be closed in the ambient space in the form used here, and the two sets $A$ and $B$ would both have the corner in their closures; deleting it is exactly what makes them disjoint closed sets with no room between them.
The ordinals $\alpha_n$ are defined, not chosen. Taking the least $\alpha$ that works is what keeps the construction free of dependent choice; the only choice principle in the argument is the countable one inside Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$ , so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable, and it is declared in the statement.
The classical use of this witness is not made here. The plank is the standard example showing that normality is neither hereditary nor productive; both readings need the normality of $W \times Z$ itself, which this page does not prove and does not assert (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly).

Depends on

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Sources

Tychonoff plank (Wikipedia) (standard reference, not scraped)
Normal space (Wikipedia) (standard reference, not scraped)
L. Steen and J. Seebach, Counterexamples in Topology, §86-87 (standard reference, not scraped)
First uncountable ordinal (Wikipedia) (standard reference, not scraped)

Refuted, assuming countable choice: every Hausdorff space built from ordinal spaces is normal. The deleted Tychonoff plank ((ω1+1)×(ω+1))∖{(ω1,ω)}((\omega_1 + 1) \times (\omega + 1)) \setminus \{(\omega_1, \omega)\}((ω1​+1)×(ω+1))∖{(ω1​,ω)} is Hausdorff and not normal