PCF Scales and ZFC Dowker Spaces

Assume (i), and let $(U_n)$ be an increasing open cover. Take a locally finite open refining cover $\mathcal{R}$. For each $R\in \mathcal{R}$ let $a(R)$ be the least $n$ with $R\subseteq U_n$, and set $C_n=X\setminus \bigcup \{R:a(R)>n\}$. This is closed. If $x\in C_n$, some $R$ contains $x$ because $\mathcal{R}$ covers; necessarily $a(R)\le n$, so $x\in U_n$. If $x\in X$, choose an open neighborhood meeting only finitely many members of $\mathcal{R}$. At least one meets it because $x$ is covered. The maximum $N$ of their assigned indices exists, and the neighborhood is contained in $C_N$. Thus the interiors cover and (ii) holds. All index assignments are least natural numbers, without choice.

Assume (ii) and let $(O_n)$ be any countable open cover; finite nonempty covers can be extended by empty entries. Put $U_n={\bigcup }_{i\le n}{O}_i$, obtain closed $C_n\subseteq U_n$ as in (ii), and replace $C_n$ by $D_n={\bigcup }_{i\le n}{C}_i$. Then $D_n$ is closed and increasing, is contained in $U_n$, and its interiors cover. With $D_{-1}=\mathrm{\varnothing }$ put $V_n=U_n\setminus D_{n-1}$. For each $x$, the least $n$ with $x\in D_n$ gives $x\in V_n$, so these open sets cover. If a neighborhood $N_x$ lies in $D_N$, it misses every $V_n$ with $n\ge N+1$. The family $\{V_n\cap O_i:i\le n<\omega \}$ covers $X$, refines $(O_n)$, and near $x$ has at most ${\sum }_{n\le N}(n+1)$ possibly meeting indexed members. It is the locally finite open refining cover required for (i). If the original cover is empty, $X$ is empty and its empty refining cover suffices.

Complementation proves (ii) implies (iii): given decreasing closed $F_n$ with empty intersection, use $U_n=X\setminus F_n$, and put $G_n=X\setminus C_n$. Then $F_n\subseteq G_n$ and ${\bigcap }_n\overline{G_n}=X\setminus {\bigcup }_n\mathrm{int}{C}_n=\mathrm{\varnothing }$. Conversely, given increasing $U_n$ covering $X$, put $F_n=X\setminus U_n$, take the expansions of (iii), and set $C_n=X\setminus G_n$. These are closed subsets of $U_n$, and the same identity says their interiors cover. The identity uses $\overline{X\setminus C}=X\setminus \mathrm{int}C$, which follows because a point has a neighborhood contained in $C$ exactly when it is outside that closure.

Condition (iii) implies (iv), since $G_n\subseteq \overline{G_n}$. Suppose now that $X$ is normal and (iv) holds. For a given decreasing closed sequence with empty intersection, take its open expansions $G_n$ with empty intersection. For each $n$, the closed set $F_n$ lies in the open $G_n$, so normal shrinking provides an open $H_n$ with $F_n\subseteq H_n\subseteq \overline{H_n}\subseteq G_n$. AC selects these witnesses for all $n$ simultaneously. Hence ${\bigcap }_n\overline{H_n}\subseteq {\bigcap }_n{G}_n=\mathrm{\varnothing }$, proving (iii). Steps 1.1–1.3 prove the other equivalences; empty sets cause no exception to these inclusions. QED.

The interval $K$ is compact also for relative open covers: write each relative open set as $K\cap O$ with $O$ real-open, and apply F6 to the resulting real-open cover. One can use all possible $O$ with a specified trace, so this needs no choice. A closed subset $L$ of $K$ is compact: adjoin $K\setminus L$ to any relative open cover and then remove that member from a finite subcover. For any compact space $L$ and closed $C\subseteq X\times L$, its projection to $X$ is closed. Indeed, if $x$ is outside the projection, for each $t\in L$ the product basis gives a rectangle $U_t\times V_t$ about $(x,t)$ missing $C$. Using AC to select these rectangles, finitely many $V_t$ cover $L$, and their corresponding finite intersection of $U_t$ is an open neighborhood of $x$ missing the projection. If $L$ is empty, the projection is empty and the empty intersection is $X$.

Suppose $X\times K$ is normal. The slice $X\times \{0\}$ is closed and its projection to $X$ is a homeomorphism: a relative basic rectangle on the slice is either empty or $U\times \{0\}$, with $U$ open in $X$. By closed heredity $X$ is normal. Given closed $F_n\supseteq F_{n+1}$ with ${\bigcap }_{n\ge 1}{F}_n=\mathrm{\varnothing }$, put $D={\bigcup }_{n\ge 1}(F_n\times [2^{-n},1])$. This is closed: at a point outside $D$, choose $N$ with $x\mathrm{\notin }{F}_N$. The neighborhood $(X\setminus F_N)\times K$ misses every strip with $n\ge N$; intersect it with the complements of the finitely many earlier closed strips. This remains an open neighborhood avoiding $D$. Also $D$ misses the closed slice at zero. Normal shrinking gives an open $V$ containing that slice with $\bar{V}\cap D=\mathrm{\varnothing }$.

In the setting of step 1.2, define $G_n=\{x:(x,2^{-n})\mathrm{\notin }\bar{V}\}$. The fixed-height inclusion is continuous by the rectangle basis, so $G_n$ is open, and $F_n\subseteq G_n$ because $F_n\times \{2^{-n}\}\subseteq D$. For each $x$, a rectangle about $(x,0)$ inside $V$ contains $(x,t)$ for all sufficiently small $t\ge 0$. The inequality $2^n\ge n+1$ follows by induction, and F8 therefore gives $2^{-n}\to 0$. Consequently $(x,2^{-n})\in V$ for all sufficiently large $n$, whence $x\mathrm{\notin }{G}_n$ then. Thus ${\bigcap }_n{G}_n=\mathrm{\varnothing }$, and F1 applied to the normal space from step 1.2 gives countable paracompactness.

We next prepare the converse. Dyadic intervals form a countable real basis, directly from F8. Given $x\in \mathbb{R}$ and $\epsilon >0$, choose $n$ with $2^{-n}<\epsilon \mathrm{/}4$, using $2^n\ge n+1$ and F8. There is an integer $k$ with $k\le 2^{n}x<k+1$: F8 gives integer bounds on $2^{n}x$ in both directions, and the finite set of integers between those bounds and at most $2^{n}x$ has a greatest element. Then $x$ lies in $((k-1)2^{-n},(k+2)2^{-n})$, whose closed endpoints are within $\epsilon$ of $x$. All dyadic intervals with endpoints $k{2}^{-n}$ and $l{2}^{-n}$, where $k<l$ are integers, form a countable family: enumerate triples $(n,k,l)$ by increasing $n+\mathrm{\mid }k\mathrm{\mid }+\mathrm{\mid }l\mathrm{\mid }$, each level finite. Finite subsets of this enumerated family are countable by encoding the finite index set $E$ as ${\sum }_{i\in E}{2}^i$. Enumerate as $(B_j{)}_{j<\omega }$ all finite unions of their traces on $K$, including $\mathrm{\varnothing }$ and $K$. If disjoint closed sets $P,Q\subseteq K$ are given, each $t\in P$ has a relative interval neighborhood whose closure misses $Q$: choose $\epsilon >0$ with $(t-\epsilon ,t+\epsilon )\cap K\subseteq K\setminus Q$, and use the dyadic interval just constructed, whose closure lies inside that neighborhood. Compactness of $P$ from step 1.1 gives finitely many such intervals covering it. Their union is some $B_j$ with $P\subseteq B_j$ and $\overline{B_j}\cap Q=\mathrm{\varnothing }$. If $P$ is empty use $B_j=\mathrm{\varnothing }$; if $Q$ is empty one may use $B_j=K$.

Suppose now that $X$ is normal and countably paracompact, and let $F,G$ be disjoint closed subsets of $X\times K$. Their fibers $F_x,G_x$ are closed in $K$ because fixed-coordinate inclusions are continuous by F2. Put $O_j=\{x:F_x\subseteq B_j\text{ and }{G}_x\cap \overline{B_j}=\mathrm{\varnothing }\}$. The complement of $O_j$ is the union of the projections of $F\cap (X\times (K\setminus B_j))$ and $G\cap (X\times \overline{B_j})$. Both projections are closed by step 1.1, since their compact factors are closed in $K$. Hence $O_j$ is open; the fiber separation in step 2.2 shows these sets cover $X$. Take a locally finite open refining cover $\mathcal{R}$ and assign each $R$ its least index $a(R)$ with $R\subseteq O_{a(R)}$. Set $V_j=\bigcup \{R:a(R)=j\}$. These sets form a countable open cover and $V_j\subseteq O_j$. Near a point only finitely many $R$ meet a fixed neighborhood, so only their finitely many assigned $V_j$ can meet it. Thus $(V_j)$ is locally finite even as an indexed family.

Shrink this locally finite cover as follows. After $W_i$ for $i<j$ have been defined, put $A_j=X\setminus ({\bigcup }_{i<j}{W}_i\cup {\bigcup }_{i>j}{V}_i)$. Inductively the earlier $W_i$ together with $V_i$ for $i\ge j$ cover $X$, initially because $(V_i)$ covers, so the closed set $A_j$ lies in $V_j$. Choose open $W_j$ with $A_j\subseteq W_j\subseteq \overline{W_j}\subseteq V_j$ by F4. This replacement preserves that covering invariant. AC fixes a choice function on all nonempty sets of open shrinking witnesses in $X$, so ordinary recursion makes the choices simultaneously. If a point missed every $W_j$, the finite nonempty set $\{j:x\in V_j\}$ would have a largest index $j$; that point would lie in $A_j\subseteq W_j$, a contradiction. Thus $(W_j)$ covers. Because $W_j\subseteq V_j$, this family and, by F5, its closures are locally finite.

Put $H={\bigcup }_j(W_j\times B_j)$. It is open by the rectangle basis. If $(x,t)\in F$, choose $j$ with $x\in W_j$; then $x\in O_j$ and $t\in F_x\subseteq B_j$, so $(x,t)\in H$. The rectangles form a locally finite family because a neighborhood $N$ in $X$ meeting only finitely many $W_j$ gives $N\times K$ with the same property. Also $\overline{W_j\times B_j}=\overline{W_j}\times \overline{B_j}$: every basic rectangle about a pair of closure points meets the product, whereas outside either closure a rectangle misses it. By F5, $\bar{H}={\bigcup }_j(\overline{W_j}\times \overline{B_j})$. If $(x,t)$ belongs to one of these products, then $x\in V_j\subseteq O_j$ and $t\in \overline{B_j}$, so $(x,t)\mathrm{\notin }G$. Hence $H$ and $(X\times K)\setminus \bar{H}$ are disjoint open neighborhoods of $F$ and $G$, proving normality. If $X$ is empty, both sides of the asserted equivalence hold by the empty covers and empty separation; the constructions also allow empty fibers and shrinking sets. Together with step 2.1 this proves both directions. QED.

The relation ${=}_J$ is reflexive and symmetric. Its transitivity follows from $\{i:f(i)\mathrm{\ne }h(i)\}\subseteq \{i:f(i)\mathrm{\ne }g(i)\}\cup \{i:g(i)\mathrm{\ne }h(i)\}$. The corresponding failure set for a composite weak or strict inequality is likewise contained in the union of its two failure sets: outside that union the ordinal inequalities compose, strictly if either is strict. This proves transitivity of both comparisons, and the two mixed composition laws. Replacing either function by an ${=}_J$ equivalent one changes any comparison set only on the union of the equality-exception sets, so truth of the comparisons is unchanged. Mutual weak inequalities mean equality outside their union of exceptions; thus ${\le }_J$ on the quotient is antisymmetric. For proper $J$, $f{<}_{J}f$ fails because its exception set is $I$. Finally $f(i)+1<h(i)$ for limit-valued factors, so $f+1$ is in the product and strictly exceeds $f$. Weak cofinality and strict cofinality agree: dominate $f+1$ weakly to dominate $f$ strictly.

For proper $J$, its dual $J^{\ast }=\{I\setminus E:E\in J\}$ is a filter: it contains $I$, omits $\mathrm{\varnothing }$, and complementing finite unions and inclusions gives the other filter axioms. For an ultrafilter $D$, exactly one of the three sets where $f<g$, $f=g$, or $f>g$ belongs to $D$. At least one must belong, since otherwise their three complements would belong and have empty intersection; two cannot belong since they are disjoint. Hence the quotient is linearly ordered, and step 1.1 and the successor operation show it has no last element. The product has its zero function, so this order is nonempty.

More generally, suppose a product $P$ modulo a proper ideal has a strict cofinal regular-$\lambda$ chain $(s_{\alpha })$. For a family $H$ of size less than $\lambda$, assign to each $h$ its least chain index ${\alpha }_h$ weakly dominating it. Regularity bounds these indices below $\lambda$; a later chain term strictly dominates every $h$. Thus $P$ is strictly $\lambda$-directed and has no cofinal subset of size less than $\lambda$: a strict bound of a cofinal family would weakly lie below one of its members, contradicting irreflexivity and mixed composition in step 1.1. Its least cofinal cardinal is therefore exactly $\lambda$, proving uniqueness of true cofinality.

Let $L$ be one of these linear orders and $\nu$ its least cofinal cardinal. It exists because $L$ is a set and AC makes its subsets well-orderable. It is infinite: a finite cofinal family would have a maximum, hence give a last element. If $\nu$ were singular, enumerate a cofinal family $(q_{\alpha }{)}_{\alpha <\nu }$, and take a cofinal set $E\subseteq \nu$ of size $\mathrm{cf}(\nu )<\nu$. Each $\{q_{\alpha }:\alpha <\eta \}$ for $\eta \in E$ is noncofinal by minimality of $\nu$. In a linear order noncofinality gives a strict upper bound, since a point witnessing failure of cofinality is greater than every member. AC selects these bounds $b_{\eta }$. They form a cofinal family: for each $q_{\alpha }$ choose $\eta \in E$ with $\eta >\alpha$, so $q_{\alpha }<b_{\eta }$. This contradicts the least size $\nu$. Therefore $\nu$ is regular. Recursively choose $r_{\alpha }$ strictly above $q_{\alpha }$ and all earlier $r_{\beta }$; fewer than $\nu$ terms are noncofinal, so such a bound exists. Fix a choice function on all nonempty subsets of $L$ by AC before applying recursion. The resulting strict cofinal $\nu$-chain proves the existence of a scale in $L$.

Let $E:Q\to P$ be an embedding of the weak and strict comparisons with cofinal image. A strict cofinal chain in $Q$ maps to one in $P$. Conversely suppose $P$ has true cofinality $\lambda$. The image is strictly $\lambda$-directed: bound fewer than $\lambda$ image elements strictly in $P$ by step 2.2, then move weakly above that bound into the image. Choose an image point above each member of a fixed cofinal $\lambda$-chain in $P$, giving an image cofinal family $(q_{\alpha })$ of length $\lambda$. Recursively choose an image point strictly above $q_{\alpha }$ and all earlier selected points, using directedness at each $\alpha <\lambda$. AC fixes both the first family of witnesses and a choice function for the recursive bounds, and F6 gives the chain. Pull it back along $E$ to obtain a strict cofinal $\lambda$-chain in $Q$. Its cofinality is $\lambda$ by step 2.2.

If $\prod A\mathrm{/}J$ has a scale of length $\lambda$, extend the dual filter from step 2.1 to an ultrafilter $D$ using F4. Every strict comparison holding modulo $J$ holds modulo $D$, since its good set lies in $J^{\ast }\subseteq D$. The same functions still form a strict cofinal $\lambda$-chain, as each function was already weakly dominated modulo $J$. By step 2.2 its ultraproduct has cofinality exactly $\lambda$, so $\lambda \in \mathrm{pcf}(A)$.

If $D$ is an ultrafilter on $A$ and $B\in D$, restriction gives the ultrafilter $D_B=\{E\subseteq B:E\in D\}$ on $B$. The filter axioms and complementary-pair test follow from F2–F3 and $B\in D$. Restriction of product functions gives an order isomorphism of ultraproducts: comparison sets belong to $D$ exactly when their intersections with $B$ do; every function on $B$ extends by zero outside $B$. Conversely an ultrafilter on $B\subseteq A$ extends to $\{E\subseteq A:E\cap B\in D_B\}$, and the same computation gives the inverse isomorphism. Principal ultrafilters at $a\in A$ identify the ultraproduct with the ordinal $a$, whose cofinality is $a$ by its assumed regularity. Hence $A\subseteq \mathrm{pcf}(A)$, and extension from a support proves monotonicity.

In $\prod A$, every family $\mathcal{H}$ of size less than $\min A$ has the pointwise strict bound $b(a)=\sup \{f(a)+1:f\in \mathcal{H}\}<a$. Indeed regularity of $a$ forbids a cofinal subset of size less than $a$; the empty family gives $b=0$. Thus no such family can be cofinal in a proper ultraproduct, by step 1.1. Any cofinal family of quotient classes of size less than $\min A$ would, using AC to choose representatives, contradict this bound. Consequently $\nu \ge \min A$. AC also chooses representatives of all quotient classes, injecting the quotient into the product, so $\nu \le \mathrm{\mid }\prod A\mathrm{\mid }$. Ultrafilters form a subset of $\mathcal{P}(\mathcal{P}(A))$, and Replacement sends them to their cofinalities, proving that $\mathrm{pcf}(A)$ is a set of infinite regular cardinals.

For nonzero limit-valued $h$, choose strictly increasing cofinal maps $e_i:\mathrm{cf}(h(i))\to h(i)$. To construct one, enumerate a cofinal subset as $(t_{\xi }{)}_{\xi <\mu }$, where $\mu =\mathrm{cf}(h(i))$. At stage $\xi <\mu$, the previously chosen values and $t_{\xi }$ have cardinality less than $\mu$ and so are bounded below $h(i)$ by F5. Their supremum plus one is still below the limit $h(i)$; take it as $e_i(\xi )$. F6 gives a strictly increasing cofinal map. AC selects the initial enumerations for all coordinates. The map $f\mapsto (i\mapsto e_i(f(i)))$ preserves and reflects all coordinate comparisons, hence their versions modulo $J$. Its image is cofinal: given $g\in \prod h$, at each coordinate take the least ordinal $\xi$ with $g(i)\le e_i(\xi )$, which exists by cofinality. This defines an image bound without further choice. Step 3.2 proves the asserted transfer; no claim that the ceiling operation preserves strict inequalities is used.

For the repetition map, $J_c$ is an ideal because inverse images preserve empty sets, inclusions and unions, and it is proper since $c^{-1}[B]=I\mathrm{\notin }J$. For $e\in \prod B$ set $E(e)=e\circ c$. Every comparison's exception set is the inverse image under $c$ of the corresponding exception set on $B$, so this induces an embedding of equality and both orders. Given $t\in {\prod }_{i\in I}c(i)$, put $e(b)=\sup \{t(i)+1:c(i)=b\}$. Each fiber has size at most $\mathrm{\mid }I\mathrm{\mid }<b$, so regularity gives $e(b)<b$, and $E(e)(i)>t(i)$ for all $i$. Thus the image is cofinal, and step 3.2 proves transfer in both directions. Only an isomorphism onto the image is claimed.

An ultrafilter on $A\cup B$ contains $A$ or $B$: if it omits $A$, it contains $(A\cup B)\setminus A\subseteq B$, and upward closure applies. Step 3.4 then gives $\mathrm{pcf}(A\cup B)\subseteq \mathrm{pcf}(A)\cup \mathrm{pcf}(B)$, and monotonicity gives the reverse inclusion. A singleton support has only its principal ultrafilter, with cofinality its coordinate cardinal. Finite induction using the union formula therefore yields $\mathrm{pcf}(A)=A$ for every finite $A$, including the empty case: an ultrafilter on the empty set would both contain and omit the empty set by F2. These proofs cover all infinite regular coordinates, including $\omega$, and in particular every progressive set. QED.

Since $\mathrm{pcf}(\mathrm{\varnothing })=\mathrm{\varnothing }$, the empty set belongs to $J_{<\lambda }[A]$. If $E$ belongs and $H\subseteq E$, monotonicity gives $\mathrm{pcf}(H)\subseteq \mathrm{pcf}(E)\subseteq \lambda$. If $E,H$ both belong, finite-union preservation gives $\mathrm{pcf}(E\cup H)=\mathrm{pcf}(E)\cup \mathrm{pcf}(H)\subseteq \lambda$. Thus it is an ideal; the same argument with ${\lambda }^{+}$ gives $J_{\le \lambda }$. Larger cutoffs enlarge the ideals because the corresponding ordinal intervals are nested. For $E\subseteq B\subseteq A$, the statement $\mathrm{pcf}(E)\subseteq \lambda$ does not depend on the ambient set, proving the restriction identity in both directions.

By the singleton calculation, $\mathrm{pcf}(\{a\})=\{a\}$, so its membership is exactly $a<\lambda$, including failure at the endpoint $a=\lambda$. If $\lambda \in \mathrm{pcf}(A)$, then $\mathrm{pcf}(A)\subseteq ̸\lambda$, so $A\mathrm{\notin }{J}_{<\lambda }[A]$. If $A$ is empty, its only ideal here is $\{\mathrm{\varnothing }\}$, improper on the empty support. For nonempty $A$ and $\lambda =0$ or $1$, the singleton calculation and monotonicity show that only the empty subset is null, since all coordinates are infinite. For a finite $A$, the same calculation gives exactly $J_{<\lambda }[A]=\mathcal{P}(\{a\in A:a<\lambda \})$.

Suppose first that $E\in J_{<\lambda }[A]$ and $D$ is an ultrafilter on $A$ containing $E$. Restrict $D$ to $E$; by the support isomorphism its product cofinality is unchanged and belongs to $\mathrm{pcf}(E)$, hence is below $\lambda$. Conversely, if every such $D$ has cofinality below $\lambda$, any ultrafilter on $E$ extends to one on $A$ supported on $E$ with the same cofinality. Thus every member of $\mathrm{pcf}(E)$ is below $\lambda$, proving membership. For $E=\mathrm{\varnothing }$ both the universal ultrafilter assertions are vacuous and membership was proved in step 1.1.

A set $H\subseteq A$ belongs to every ultrafilter of cofinality at least $\lambda$ exactly when none of those ultrafilters contains $A\setminus H$, by the complementary-pair law in F2. By step 2.1 this is exactly $A\setminus H\in J_{<\lambda }[A]$, or $H\in J_{<\lambda }[A{]}^{\ast }$. This proves both inclusions in the dual identity. If there are no high-cofinality ultrafilters, step 2.1 with $E=A$ puts $A$ in the ideal, so the ideal and its dual are both $\mathcal{P}(A)$; the empty-intersection convention gives the same result. No witnesses are chosen in these computations beyond the AC-dependent facts already proved in F1. QED.

Write $p_{\alpha }$ for the projection of $f_{\alpha }$. It suffices to rule out the assumption that none strictly bounds the sequence. Under that assumption, for each $\alpha$ let $\eta$ be the least index for which $P=\{a:p_{\alpha }(a)\le f_{\eta }(a)\}$ is $I$-positive, since $f_{\eta }{<}_I{p}_{\alpha }$ fails. Take $b(\alpha )$ to be the least index greater than $\alpha$ and $\eta$. Outside the small failure set for $f_{\eta }{<}_I{f}_{b(\alpha )}$, the set $P$ witnesses $p_{\alpha }<f_{b(\alpha )}$, so $\{a:p_{\alpha }(a)<f_{b(\alpha )}(a)\}$ remains positive. Removing a small set cannot destroy positivity: otherwise its union with the removed set would put $P$ in $I$. Every later index has the same positive comparison, by composition with another strict comparison and the same finite-union argument. No completeness of $I$ is used.

Recursively choose strictly increasing indices $u_{\xi }<\lambda$ for $\xi <\lambda$ so that $u_{\xi }>b(u_{\zeta })$ whenever $\zeta <\xi$. At each stage the earlier indices and witness indices form a set of cardinality less than $\lambda$, so F2 bounds them below $\lambda$; choose the least larger index. F3 gives the sequence. Its range $U$ is unbounded, because a bounded subset of the cardinal $\lambda$ has cardinality less than $\lambda$ and cannot contain a strictly increasing sequence of length $\lambda$. By step 1.1, $\{a:p_{\alpha }(a)<f_{\beta }(a)\}$ is positive for all $\alpha <\beta$ in $U$. Apply $(\ast {)}_{\kappa }$ to obtain indices $(v_i{)}_{i<\kappa }$ in $U$ and strong-increase witnesses $Z_i\in I$.

Put $E_i=\{a:f_{v_i}(a)\ge s(a)\}\in I$. For each $i<\kappa$, the positive set $\{a:p_{v_i}(a)<f_{v_{i+1}}(a)\}$ remains positive after removing $Z_i\cup Z_{i+1}\cup E_i\cup E_{i+1}$. Here $i+1<\kappa$ because $\kappa$ is infinite regular. In particular the remaining set is nonempty; AC selects a coordinate $a_i$ in it. Some fixed $a$ is selected for $\kappa$ many indices. Indeed, if every fiber $\{i:a_i=a\}$ had size less than $\kappa$, it would be bounded in $\kappa$ by F2; there are $\tau <\kappa$ such bounds, so F2 would bound their union, contradicting that the fibers cover $\kappa$.

Let $T=\{i:a_i=a\}$ have size $\kappa$. For $i<j$ in $T$, the chosen coordinate gives $p_{v_i}(a)<f_{v_{i+1}}(a)$. If $i+1=j$ the next comparison is equality; otherwise strong increase gives $f_{v_{i+1}}(a)<f_{v_j}(a)$, since $a\mathrm{\notin }{Z}_{i+1}\cup Z_j$. Finally $a\mathrm{\notin }{E}_j$, so the genuine ceiling rule gives $f_{v_j}(a)\le p_{v_j}(a)\in S(a)$. Also $p_{v_i}(a)\in S(a)$. Hence $i\mapsto p_{v_i}(a)$ is a strictly increasing injection of $T$ into $S(a)$, contradicting $\mathrm{\mid }T\mathrm{\mid }=\kappa >\mathrm{\mid }S(a)\mathrm{\mid }$. The assumption in step 1.1 is impossible. For every admitted family $S$ some projection therefore strictly bounds the sequence, exactly the asserted property. The finite-ideal countable case satisfies all these hypotheses with $\tau ={\mathrm{\aleph }}_0$. QED.