DC is equivalent to Baireness of compact-Hausdorff products

Assume DC and let $(X_i{)}_{i\in I}$, $(U_n)$ and $B$ be as in the assumptions; the task is to find a point of $B\cap {\bigcap }_n{U}_n$.

Assume instead that every product of compact Hausdorff spaces is Baire, and let $R$ be a serial relation on a nonempty set $A$; the task is to build an infinite $R$-chain.

Under step 1.1: if $X=\mathrm{\varnothing }$ then $X$ is Baire and the empty product is the one-point space, which is Baire, so assume $X\mathrm{\ne }\mathrm{\varnothing }$ and fix $x\in X$.

Under step 1.2: if $A$ is finite and nonempty, finite choice gives $f:A\to A$ with $aRf(a)$ for every $a$, and recursion on $\mathbb{N}$ gives the $R$-chain $a,f(a),f(f(a)),\dots$; so assume $A$ is infinite.

Under step 1.1 and step 2.1, let $Y$ be the set of quadruples $(n,F,(B_i{)}_{i\in F})$ where $n\in \mathbb{N}$, $F\subseteq I$ is finite, each $B_i\subseteq X_i$ is nonempty open, and ${\bigcap }_{i\in F}{\pi }_i^{-1}[B_i]\subseteq B\cap U_n$; here ${\pi }_i$ is the projection. Then $Y\mathrm{\ne }\mathrm{\varnothing }$: the set $B\cap U_0$ is nonempty open by [L1], so it contains a basic open cylinder, which is a finite intersection of coordinates.

Under step 2.2 and $A$ infinite, let $\alpha A:=A\cup \{\mathrm{\infty }\}$ be the one-point compactification of the discrete space $A$; by [F3] it is compact Hausdorff, and $A$ is open and dense in $\alpha A$ because the infinite discrete space $A$ is not compact.

Under step 3.1 define $\rho$ on $Y$ by $(n,F,(B_i))\rho (n^{\mathrm{\prime }},F^{\mathrm{\prime }},(B_i^{\mathrm{\prime }}))$ iff $n^{\mathrm{\prime }}=n+1$, $F\subseteq F^{\mathrm{\prime }}$, and $\overline{B_i^{\mathrm{\prime }}}\subseteq B_i$ for $i\in F$; the new open sets indexed by $F^{\mathrm{\prime }}\setminus F$ are unrestricted beyond the membership condition of $Y$. Then $\rho$ is entire on $Y$: given $(n,F,(B_i))\in Y$ the cylinder $C:={\bigcap }_{i\in F}{\pi }_i^{-1}[B_i]$ is a nonempty open subset of $B\cap U_n$, so $C\cap U_{n+1}$ is nonempty open because $U_{n+1}$ is dense; fix a point $w$ of it. Some basic open cylinder around $w$ lies in the open set $C\cap U_{n+1}$; intersecting it with $C$ and adding the coordinates of $C$ that it omits, we obtain a basic open cylinder ${\bigcap }_{i\in F^{†}}{\pi }_i^{-1}[B_i^{†}]$ with $F\subseteq F^{†}$, $B_i^{†}\subseteq B_i$ for $i\in F$, and $w\in {\bigcap }_{i\in F^{†}}{\pi }_i^{-1}[B_i^{†}]\subseteq C\cap U_{n+1}$. For each $i\in F^{†}$ the point $w_i$ lies in $B_i^{†}$, so by the regularity clause of [L2] applied in the compact Hausdorff space $X_i$ there is a nonempty open $B_i^{\mathrm{\prime }}$ with $w_i\in B_i^{\mathrm{\prime }}\subseteq \overline{B_i^{\mathrm{\prime }}}\subseteq B_i^{†}$; put $F^{\mathrm{\prime }}:=F^{†}$. Then ${\bigcap }_{i\in F^{\mathrm{\prime }}}{\pi }_i^{-1}[B_i^{\mathrm{\prime }}]\subseteq {\bigcap }_{i\in F^{\mathrm{\prime }}}{\pi }_i^{-1}[B_i^{†}]\subseteq B\cap U_{n+1}$, so $(n+1,F^{\mathrm{\prime }},(B_i^{\mathrm{\prime }}))\in Y$, and $\overline{B_i^{\mathrm{\prime }}}\subseteq B_i^{†}\subseteq B_i$ for every $i\in F$, so $(n,F,(B_i))\rho (n+1,F^{\mathrm{\prime }},(B_i^{\mathrm{\prime }}))$.

Under step 3.2 let $X:=(\alpha A{)}^{\omega }$; it is a product of compact Hausdorff spaces, hence Baire by the hypothesis of step 1.2, and the sets $D_n:=\{g\in X:g(n)\in A\}$ are open and dense, so $A^{\omega }={\bigcap }_n{D}_n$ is a dense $G_{\delta }$ of $X$; a dense $G_{\delta }$ subspace of a Baire space is Baire, since the traces of countably many dense open sets of the bigger space witness the subspace condition.

Under step 1.1 and step 4.1, DC gives a sequence $y_n=(n,F_n,(B_i^n{)}_{i\in F_n})$ in $Y$ with $y_n\rho y_{n+1}$ for all $n$; in particular $F_n\subseteq F_{n+1}$ and, for every $i\in F_n$, the closures nest inside the previous open sets: $\overline{B_i^{n+1}}\subseteq B_i^n$.

Under step 4.2 the subspace $A^{\omega }$ is the discrete sequence space with its product topology, complete under the reciprocal first-difference metric by [F5]; so this complete metric space is Baire.

Under step 5.1 the set $F:={\bigcup }_n{F}_n$ is at most countable: it is a countable union of finite sets, and countable choice holds by [F2].

Under step 5.1, for each $i\in F$ let $n_i$ be least with $i\in F_{n_i}$; then the sets $\overline{B_i^n}$ for $n\ge n_i$ form a decreasing sequence of nonempty closed subsets of the compact space $X_i$, so their intersection is nonempty and is contained in ${\bigcap }_{n\ge n_i}{B}_i^n$, since $\overline{B_i^{n+1}}\subseteq B_i^n$ for every $n$.

Under step 5.2 the sets $V_i:=\{f\in A^{\omega }:\mathrm{\exists }jf(i)Rf(j)\}$ are open and dense in $A^{\omega }$ by [F5], so their intersection is dense and hence nonempty; fix $f\in {\bigcap }_i{V}_i$.

Under step 6.1, step 6.2 and countable choice, choose $b_i\in {\bigcap }_{n\ge n_i}\overline{B_i^n}$ for each $i\in F$ — the set is nonempty by step 6.2, where compactness gives a point of the intersection of the nested closed sets, and step 6.2 also identifies the intersection as a subset of ${\bigcap }_{n\ge n_i}{B}_i^n$ — and define $y\in X$ by $y_i:=b_i$ for $i\in F$ and $y_i:=x_i$ for $i\mathrm{\notin }F$.

Under step 6.3 define $q(i):=\min \{j\in \mathbb{N}:f(i)Rf(j)\}$, which exists by [F6] because $f\in V_i$, and $k(0):=0$, $k(n+1):=q(k(n))$, a definition by recursion; then $a(n):=f(k(n))$ satisfies $a(n)=f(k(n))Rf(q(k(n)))=f(k(n+1))=a(n+1)$ for every $n$, so $a$ is an infinite $R$-chain.

Under step 7.1, $y\in {\bigcap }_{i\in F_n}{\pi }_i^{-1}[B_i^n]\subseteq B\cap U_n$ for every $n$, because $b_i\in B_i^n$ for every $i\in F_n$ and the inclusion is the defining property of $Y$; hence $B\cap {\bigcap }_n{U}_n\mathrm{\ne }\mathrm{\varnothing }$, so every product of compact Hausdorff spaces is Baire under DC.

Under step 2.2 and step 7.2 every serial relation on a nonempty set admits an infinite chain, which is the starting-point-free form of DC; the prescribed-start form follows by [F6], so the hypothesis of step 1.2 implies DC.

Step 8.1 proves that DC implies Baireness of every product of compact Hausdorff spaces, and step 8.2 proves the converse; the two implications are the displayed equivalence.