Compact Hausdorff Baire implies DMC

Assume every compact Hausdorff space is Baire, let $T$ be a pruned tree of height $\omega$ with nonempty levels on a set $A$, and let $\mathrm{\infty }$ be a point outside $T$; it suffices by [F1] to produce a subtree of $T$ with nonempty finite levels.

Give $T$ the discrete topology and let $T^{\ast }:=T\cup \{\mathrm{\infty }\}$ be its one-point compactification; by [F3] the discrete $T$ is locally compact Hausdorff, so $T^{\ast }$ is compact Hausdorff by [F2], and it is nonempty because $\mathrm{\infty }\in T^{\ast }$. In the discrete space a compact subset is finite: its singleton cover has a finite subcover. Conversely finite subsets are compact by finite choices from a cover. Thus the neighbourhoods of $\mathrm{\infty }$ are exactly the complements of finite subsets of $T$.

Let $X:=(T^{\ast }{)}^{\omega }$ with the product topology; by [F3] $X$ is Hausdorff, and $X$ is nonempty because the constant-$\mathrm{\infty }$ function is a member.

Let $K\subseteq X$ be the set of those $x$ with: for all $m<n$, either $x(m)=\mathrm{\infty }$, or $x(n)=\mathrm{\infty }$, or $x(m),x(n)\in T$ and $x(m)$ is a proper initial segment of $x(n)$.

$K$ is closed in $X$: its complement is the union, over $m<n$ and $s,t\in T$ for which $s$ is not a proper initial segment of $t$, of the two-coordinate cylinders $\{x:x(m)=s,x(n)=t\}$. Each such cylinder is open because every $s\in T$ is an isolated point of $T^{\ast }$, so the complement of $K$ is open. Hence $K$ is a closed subspace of the Hausdorff space $X$, and is Hausdorff by [F3]; it is nonempty because the constant sequence with value $\mathrm{\infty }$ lies in $K$.

For $n\in \mathbb{N}$ put $G_n:=\{x\in K:x(i)\mathrm{\ne }\mathrm{\infty }\text{ for some }i\ge n\}$, a subset of $K$.

Each $G_n$ is open in $K$: it is the union over $i\ge n$ and $t\in T$ of the sets $K\cap \{x:x(i)=t\}$, and each $\{x:x(i)=t\}$ is a basic open set of the product because $\{t\}$ is open in the discrete space $T$.

Case B: $K$ is not compact. Choose an open cover of $K$ with no finite subcover, and refine it by taking all canonical basic product cylinders $W\subseteq X$ whose trace $K\cap W$ is contained in a member of the cover; here the coordinate restrictions in $W$ may be taken to be a singleton $\{t\}$, $t\in T$, or a cofinite neighbourhood of $\mathrm{\infty }$. This is still a cover of $K$ with no finite subcover. Let $\mathcal{E}$ consist of all finite intersections of $X$, the closed cylinder complements $X\setminus W$, and all constraint complements $X\setminus \{x:x(m)=s,x(n)=t\}$ for $m<n$ and $s,t\in T$ with $s$ not a proper initial segment of $t$. Every such generator is a closed member of the finite-coordinate cylinder algebra. Every finite intersection is nonempty: choose in $K$ a point outside the finitely many $W$'s, which automatically satisfies every constraint complement. The intersection of all members is empty because the constraint complements cut the intersection down to $K$ and the $W$'s cover $K$. Thus $\mathcal{E}$ is a downwards-directed family of nonempty closed subsets of $X$, contains $X$ and every constraint complement, has empty intersection, and every member belongs to the finite-coordinate cylinder algebra.

Each $G_n$ is dense in $K$: let $N$ be a nonempty relatively open subset of $K$, and choose a basic product-open $W$ with $\mathrm{\varnothing }\mathrm{\ne }K\cap W\subseteq N$, restricting only the coordinates in a finite set $J$; choose $u\in K\cap W$. If $u(i)\mathrm{\ne }\mathrm{\infty }$ for some $i\ge n$ then $u\in N\cap G_n$ and we are done. Otherwise every non-$\mathrm{\infty }$ value of $u$ occurs at an index $<n$; let $t$ be a value of $u$ of maximal length among those, or the empty node if $u$ has no non-$\mathrm{\infty }$ value. Choose $N_1\ge n$ larger than every index in $J$ and larger than $n$, and let $s\in T$ be a proper extension of $t$ of length at least $N_1$, which exists by finitely iterating pruning and restricting to the required length if needed (choose a length at least $\max (N_1,\mathrm{dom}(t)+1)$). Define $y$ by: $y(i):=u(i)$ for $i\in J$; $y(i):=\mathrm{\infty }$ if $i\mathrm{\notin }J$ and $i>N_1$; $y(N_1):=s$; and $y(i):=\mathrm{\infty }$ if $i\mathrm{\notin }J$ and $i<N_1$. Then $y\in K\cap W$, because the only non-$\mathrm{\infty }$ values are the values of $u$ at indices in $J$ together with $s$ at $N_1$, all of which are comparable by the maximality of $t$ and the choice of $s$; and $y(N_1)=s\mathrm{\ne }\mathrm{\infty }$ with $N_1\ge n$ gives $y\in G_n$.

In case B, construct families ${\mathcal{E}}_n$ of closed subsets of $X$ by ${\mathcal{E}}_0:=\mathcal{E}$ and: if $\mathrm{\infty }\in {\pi }_n[E]$ for every $E\in {\mathcal{E}}_n$, let ${\mathcal{E}}_{n+1}$ be ${\mathcal{E}}_n$ together with $E\cap {\pi }_n^{-1}[\{\mathrm{\infty }\}]$ for $E\in {\mathcal{E}}_n$, and otherwise let ${\mathcal{E}}_{n+1}:={\mathcal{E}}_n$; here ${\pi }_n$ is the $n$-th projection. The added sets are nonempty by the condition triggering the first alternative, and downward directedness follows because a lower bound $D\subseteq E\cap E^{\mathrm{\prime }}$ in ${\mathcal{E}}_n$ gives the lower bound $D\cap {\pi }_n^{-1}[\{\mathrm{\infty }\}]$ whenever one or both of the two sets carries the new coordinate constraint. Thus each ${\mathcal{E}}_n$ is downwards directed, consists of nonempty closed sets, and has empty intersection because it contains $\mathcal{E}$. Inductively every member lies in the finite-coordinate cylinder algebra generated by the sets ${\pi }_i^{-1}[\{t\}]$, $i\in \mathbb{N}$, $t\in T$, and ${\pi }_i^{-1}[\{\mathrm{\infty }\}]$, $i<n$. For such an $E$, take one finite Boolean expression in the listed generators and let $H\subseteq T$ be the finite set of node labels tested at coordinate $n$. No test $x(n)=\mathrm{\infty }$ occurs, since all infinity tests have indices less than $n$. If $x\in E$ and $x(n)\mathrm{\notin }H$, replacing only $x(n)$ by $\mathrm{\infty }$ preserves every test and hence membership in $E$. Therefore $\mathrm{\infty }\mathrm{\notin }{\pi }_n[E]$ implies ${\pi }_n[E]\subseteq H$, so that projection is finite. This uses no compactness of an infinite or finite product.

Case A: $K$ is compact. Then $K$ is a compact Hausdorff space, so by the hypothesis assumed in step 1.1 it is Baire, and by step 6.1 and step 7.1 the sets $G_n$ are dense open, so their intersection is dense and in particular nonempty; fix $x\in {\bigcap }_n{G}_n$. Let $D:=\{i\in \mathbb{N}:x(i)\mathrm{\ne }\mathrm{\infty }\}$.

In case B, put ${\mathcal{E}}^{\ast }:={\bigcup }_n{\mathcal{E}}_n$ and $D:=\{n:{\pi }_n^{-1}[\{\mathrm{\infty }\}]\mathrm{\notin }{\mathcal{E}}^{\ast }\}$. If $n\in D$, then some $E_0\in {\mathcal{E}}_n$ has $\mathrm{\infty }\mathrm{\notin }{\pi }_n[E_0]$: otherwise the first alternative in step 7.2, applied also to $X\in {\mathcal{E}}_n$, would put $X\cap {\pi }_n^{-1}[\{\mathrm{\infty }\}]={\pi }_n^{-1}[\{\mathrm{\infty }\}]$ in ${\mathcal{E}}_{n+1}$, contrary to $n\in D$. The finite-test argument in step 7.2 makes ${\pi }_n[E_0]$ finite. Put $F_n:=\bigcap \{{\pi }_n[E]:E\in {\mathcal{E}}^{\ast }\}$. Downward directedness makes the projected family finitely intersecting, so its intersection inside the finite set ${\pi }_n[E_0]$ is nonempty; hence $F_n$ is a nonempty finite subset of $T$. Moreover some $E_n^0\in {\mathcal{E}}^{\ast }$ satisfies ${\pi }_n[E_n^0]=F_n$: using [F5] after fixing a listing of the finite set ${\pi }_n[E_0]\setminus F_n$, for each such $t$ take a member whose projection omits $t$, and take a common lower bound with $E_0$. Its projection is contained in $F_n$, while the definition of $F_n$ gives the reverse inclusion.

In case A, $D$ is infinite: for each $n$ the membership $x\in G_n$ gives some $i\ge n$ with $x(i)\mathrm{\ne }\mathrm{\infty }$, so $D$ contains arbitrarily large naturals and is therefore infinite.

In case B, $D$ is infinite. Suppose instead that it is finite. Successively for $n\in D$, one can adjoin a cylinder ${\pi }_n^{-1}[\{s_n\}]$, $s_n\in F_n$, while preserving the finite-intersection property: at that stage include the member $E_n^0$ from step 8.2, whose $n$-th projection is the finite set $F_n$; if no one of the finitely many cells ${\pi }_n^{-1}[\{s\}]$, $s\in F_n$, preserved the finite-intersection property, finitely many witnessing failures, obtained using [F5], would have a common lower bound meeting $E_n^0$ but none of those cells, a contradiction. Closing under finite intersections gives a downwards-directed family ${\mathcal{E}}_1^{\ast }$ of nonempty closed sets which contains ${\mathcal{E}}^{\ast }$ and the chosen cylinders. Define $z(n):=s_n$ for $n\in D$ and $z(n):=\mathrm{\infty }$ otherwise. Every basic neighbourhood of $z$ meets every $E\in {\mathcal{E}}_1^{\ast }$: intersect $E$ with the finitely many chosen singleton cylinders for its coordinates in $D$ and, for coordinates outside $D$, with the cylinders ${\pi }_n^{-1}[\{\mathrm{\infty }\}]\in {\mathcal{E}}^{\ast }$. Therefore $z$ lies in the closure of every member, hence in every member because they are closed, contradicting the empty intersection of ${\mathcal{E}}^{\ast }\subseteq {\mathcal{E}}_1^{\ast }$.

In case B, let $m<n$ lie in $D$ and $s\in F_m$. Then some $t\in F_n$ properly extends $s$. Otherwise every $t\in F_n$ gives a constraint complement $C_t:=X\setminus \{x:x(m)=s,x(n)=t\}\in {\mathcal{E}}^{\ast }$ by step 6.2. Take $E_n^0\in {\mathcal{E}}^{\ast }$ with ${\pi }_n[E_n^0]=F_n$ from step 8.2, and use downward directedness and the finiteness of $F_n$ to obtain $E^{\mathrm{\prime }}\in {\mathcal{E}}^{\ast }$ with $E^{\mathrm{\prime }}\subseteq E_n^0\cap {\bigcap }_{t\in F_n}{C}_t$. Since $s\in F_m\subseteq {\pi }_m[E^{\mathrm{\prime }}]$, choose $x\in E^{\mathrm{\prime }}$ with $x(m)=s$. But $x(n)\in {\pi }_n[E^{\mathrm{\prime }}]\subseteq F_n$; taking $t=x(n)$ contradicts $x\in C_t$.

In case A, the values $x(i)$ for $i\in D$ form a chain in $T$ under initial segment, strictly increasing in length, by the definition of $K$ in step 4.1 since no two of them are $\mathrm{\infty }$.

In case B, enumerate $D$ increasingly as $m_0<m_1<\cdots$ and put $F_0^{\mathrm{\prime }}:=F_{m_0}$ and $F_{i+1}^{\mathrm{\prime }}:=\{t\in F_{m_{i+1}}:s\text{ is a proper initial segment of }t\text{ for some }s\in F_i^{\mathrm{\prime }}\}$. By step 9.3, every $s\in F_i^{\mathrm{\prime }}$ has an extension in $F_{i+1}^{\mathrm{\prime }}$; by definition every member of $F_{i+1}^{\mathrm{\prime }}$ has a predecessor in $F_i^{\mathrm{\prime }}$. Thus every $F_i^{\mathrm{\prime }}$ is nonempty finite, and every one of its nodes has length at least $i$. Let $T^{\mathrm{\prime }}$ be the set of all initial segments of nodes in ${\bigcup }_i{F}_i^{\mathrm{\prime }}$. It is a subtree and has a node at every level $r$, because any member of $F_r^{\mathrm{\prime }}$ has length at least $r$. Its level $r$ is finite: if $u\in T^{\mathrm{\prime }}$ has length $r$, choose $t\in F_j^{\mathrm{\prime }}$ with $u$ an initial segment of $t$. If $j<r$, extend $t$ through the successive $F_i^{\mathrm{\prime }}$ to a member of $F_r^{\mathrm{\prime }}$; if $j>r$, follow predecessors down to a member of $F_r^{\mathrm{\prime }}$. In either case, because initial segments of the same node are comparable and every member of $F_r^{\mathrm{\prime }}$ has length at least $r$, $u$ is the length-$r$ initial segment of some member of the finite set $F_r^{\mathrm{\prime }}$. Hence the level has at most $\mathrm{\mid }{F}_r^{\mathrm{\prime }}\mathrm{\mid }$ elements.

In case A, let $T^{\mathrm{\prime }}$ be the set of all initial segments of the nodes $x(i)$, $i\in D$. Then $T^{\mathrm{\prime }}$ is a subtree of $T$: it is contained in $T$ because $T$ is closed under initial segments, and it is closed under initial segments by construction. Its level $m$ consists of the initial segments of length $m$ of the nodes $x(i)$ with $i\in D$; for $i\le j$ in $D$ the nodes $x(i),x(j)$ are comparable and $x(i)$ is an initial segment of $x(j)$, so all nodes $x(i)$ with length at least $m$ have the same initial segment of length $m$, and this common node is unique; since lengths in $D$ are unbounded there is such an $i$, so level $m$ of $T^{\mathrm{\prime }}$ is a singleton. Hence $T^{\mathrm{\prime }}$ has nonempty finite levels, as required.

In either case the pruned tree $T$ has a subtree with nonempty finite levels: case A by step 11.1, case B by step 10.2. This is the tree form of DMC, so by [F1] DMC in successor-menu form holds, and since $T$ was an arbitrary pruned tree with nonempty levels, the hypothesis that every compact Hausdorff space is Baire implies DMC.