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Assuming dependent choice, every locally compact Hausdorff space is a Baire space

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

Let $(X, T)$ be a locally compact Hausdorff space (Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space, 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). Then $X$ is a Baire space (Baire space: a topological space in which every countable intersection of dense open subsets is dense): for every sequence $(U_{n})_{n \in N}$ of dense open subsets of $X$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets, Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure), the intersection $⋂_{n \in N} U_{n}$ is dense in $X$ .

Facts & Assumptions

Given: A locally compact Hausdorff space $(X, T)$ , a sequence $(U_{n})_{n \in N}$ of dense open subsets of $X$ , and the Axiom of Dependent Choice.

[L1]

$A \subseteq X$ is dense exactly when $\overline{A} = X$ , exactly when $A$ meets every nonempty open subset of $X$ ; and $X$ is a Baire space when every sequence of dense open sets has dense intersection (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets, Interior, closure, boundary, exterior, derived set and isolated point in a topological space, Baire space: a topological space in which every countable intersection of dense open subsets is dense).

[L2]

If $O \subseteq X$ is open and $x \in O$ , there is an open $V$ with $x \in V \subseteq \overline{V} \subseteq O$ and $\overline{V}$ a compact subset of $X$ (In a locally compact Hausdorff space every point has a neighbourhood base of compact sets, every open subspace and every closed subspace is locally compact, every open set around a point contains an open set with compact closure inside it, and every compact set sits inside an open set with compact closure, claim 3; Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space).

[L3]

In a Hausdorff space every compact subset is closed (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones, claim 3), and $A \subseteq \overline{A}$ with $\overline{A}$ closed (Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[L4]

The closed subsets of a subspace are the traces of the closed subsets of the ambient space, and a subset is a compact subset when the subspace it carries is a compact space (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, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

[L5]

A space is compact exactly when every family of its closed subsets with the finite intersection property has nonempty intersection (A space is compact exactly when every family of closed subsets with the finite intersection property has nonempty intersection).

[L6]

Assuming dependent choice: if $A$ is nonempty and $(R_{n})_{n \in N}$ are relations on $A$ such that every $u \in A$ has some $v \in A$ with $u R_{n} v$ , then for every $a_{0} \in A$ there is $a : N \to A$ with $a (0) = a_{0}$ and $a (n) R_{n} a (n + 1)$ for every $n$ (Dependent choice along a sequence of relations: if $R_{n}$ is entire on $A$ for every $n$ , then from any $a$ there is a sequence with $a_{n} R_{n} a_{n + 1}$ , Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure, The natural numbers $N$ (von Neumann)).

Proof

technique · direct

1.1L1suffices: every nonempty open W meets the intersection

By [L1] it suffices to show that every nonempty open $W \subseteq X$ meets $⋂_{n \in N} U_{n}$ ; if $X$ has no nonempty open subset the requirement is vacuous and there is nothing to prove, so fix a nonempty open $W$ .

2.1L1L2step 1.1

$W \cap U_{0}$ is open and nonempty, $U_{0}$ being dense and $W$ nonempty open; fixing a point of it, [L2] gives an open $V_{0}$ with that point in $V_{0}$ and $\overline{V_{0}} \subseteq W \cap U_{0}$ , with $\overline{V_{0}}$ a compact subset of $X$ . In particular $V_{0}$ is nonempty.

3.1L1step 2.1construct

Let $A$ be the set of nonempty open $V \subseteq X$ whose closure $\overline{V}$ is a compact subset of $X$ ; it contains $V_{0}$ by step 2.1, so it is nonempty. For $n \in N$ define a relation $R_{n}$ on $A$ by declaring $V R_{n} V^{'}$ to hold exactly when $\overline{V^{'}} \subseteq V \cap U_{n + 1}$ .

4.1L1L2step 3.1

Each $R_{n}$ is entire on $A$ : given $V \in A$ , the set $V \cap U_{n + 1}$ is open and nonempty, $U_{n + 1}$ being dense and $V$ nonempty open, so fixing a point of it and applying [L2] gives an open $V^{'}$ containing that point with $\overline{V^{'}} \subseteq V \cap U_{n + 1}$ and $\overline{V^{'}}$ compact; then $V^{'} \in A$ and $V R_{n} V^{'}$ . This is a pure existence statement and selects nothing, which is why dependent choice and nothing stronger is spent below.

5.1L6step 3.1step 4.1

By [L6] applied to $A$ , the relations $R_{n}$ and the point $V_{0}$ , there is a sequence $(V_{n})_{n \in N}$ in $A$ with $V_{0}$ as given and $\overline{V_{n + 1}} \subseteq V_{n} \cap U_{n + 1}$ for every $n \in N$ .

6.1L3L4step 5.1

The sets $\overline{V_{n}}$ are nonempty, since $V_{n}$ is nonempty and $V_{n} \subseteq \overline{V_{n}}$ by [L3], and they decrease: $\overline{V_{n + 1}} \subseteq V_{n} \subseteq \overline{V_{n}}$ by step 5.1 and [L3]. Each is a compact subset of $X$ and hence closed in $X$ by [L3], so each is the trace of a closed set on $\overline{V_{0}}$ and therefore closed in the subspace $\overline{V_{0}}$ by [L4].

7.1L4L5step 6.1

The family ${\overline{V_{n}} : n \in N}$ of closed subsets of the compact space $\overline{V_{0}}$ has the finite intersection property: the intersection of the empty list is $\overline{V_{0}}$ , which is nonempty, and the intersection of a nonempty finite list is $\overline{V_{N}}$ for $N$ the greatest of the indices occurring, the sets being decreasing, and that is nonempty. So [L5] gives a point $x \in ⋂_{n \in N} \overline{V_{n}}$ .

8.1step 2.1step 5.1step 7.1

That point lies in $W \cap ⋂_{n \in N} U_{n}$ : from $x \in \overline{V_{0}} \subseteq W \cap U_{0}$ it lies in $W$ and in $U_{0}$ , and for every $n \in N$ it lies in $\overline{V_{n + 1}} \subseteq V_{n} \cap U_{n + 1} \subseteq U_{n + 1}$ , so it lies in $U_{m}$ for $m = 0$ and for every $m$ of the form $n + 1$ , that is in every $U_{m}$ .

9.1L1step 1.1step 8.1∎

So every nonempty open $W$ meets $⋂_{n \in N} U_{n}$ , which by [L1] makes that intersection dense; as $(U_{n})$ was an arbitrary sequence of dense open sets, $X$ is a Baire space.

Remarks

Dependent choice is sufficient here. The proof below records its exact use and makes no assertion that DC is necessary. For the compact Hausdorff subclass, the later Compact Hausdorff Baire is equivalent to DMC ↗ proves the precise equivalence with dependent multiple choice over ZF.

Where each hypothesis is spent. Local compactness and the Hausdorff condition enter only through [L2], the shrinking clause of In a locally compact Hausdorff space every point has a neighbourhood base of compact sets, every open subspace and every closed subspace is locally compact, every open set around a point contains an open set with compact closure inside it, and every compact set sits inside an open set with compact closure, which is used twice: once to start the construction and once to continue it. Compactness of $\overline{V_{0}}$ is used once, at step 7.1, to turn a decreasing sequence of nonempty closed sets into a common point; that is the finite intersection characterisation A space is compact exactly when every family of closed subsets with the finite intersection property has nonempty intersection and nothing else.

Why the stage-dependent form of dependent choice is needed. The $n$ -th shrinking must land inside $U_{n + 1}$ , so the admissible successors of $V$ change with $n$ ; that is a family of relations and not a relation, and The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain as stated applies to a single relation on a single set. Dependent choice along a sequence of relations: if $R_{n}$ is entire on $A$ for every $n$ , then from any $a$ there is a sequence with $a_{n} R_{n} a_{n + 1}$ is exactly the bridge, and it costs nothing beyond dependent choice itself.

Each $U_{n}$ is used exactly once. The base step consumes $U_{0}$ and the step from $V_{n}$ to $V_{n + 1}$ consumes $U_{n + 1}$ , so as $n$ ranges over $N$ every index is consumed and none twice. Since $N$ contains $0$ , dropping the base step would leave $U_{0}$ untouched and the conclusion false as stated; the accounting is checked at step 8.1, where membership in $U_{m}$ is established separately for $m = 0$ and for $m = n + 1$ .

Assuming dependent choice, every locally compact Hausdorff space is a Baire space

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources