Dowker product characterization

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.