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Separable complete metric spaces are Baire in ZF

Statement

In $Z F$ , every separable (Separability: the existence of an at most countable dense subset) complete (Complete metric space: every Cauchy sequence converges in the space) metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) is a Baire space (Baire space: a topological space in which every countable intersection of dense open subsets is dense).

The form of the conclusion used below. A space $X$ is Baire exactly when $W \cap ⋂_{n} U_{n} \neq \emptyset$ for every sequence $(U_{n})$ of dense open subsets and every nonempty open $W \subseteq X$ . No choice principle is spent: separability supplies a single at most countable dense set, the padding lemma turns it into a fixed sequence, and the recursion below selects a least index-radius pair at each stage, a definable operation.

Facts & Assumptions

Given: A separable complete metric space $(X, d)$ ; a sequence $(U_{n})_{n \in N}$ of dense open subsets of $X$ ; a nonempty open $W \subseteq X$ .

[F1]

$X$ is separable when it has an at most countable dense subset; $A$ is dense in $X$ when $B (x, r) \cap A \neq \emptyset$ for every $x$ and every $r > 0$ (Separability: the existence of an at most countable dense subset, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, Open ball, closed ball and sphere in a metric space).

[F2]

$U$ is open when every $u \in U$ has $r > 0$ with $B (u, r) \subseteq U$ ; open balls are open, and open sets are closed under finite intersections and arbitrary unions (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed).

[F3]

Every nonempty at most countable set is the range of a sequence $N \to D$ (A nonempty countable set has a padded enumeration in ZF, Finite, countably infinite, countable, uncountable).

[L1]

Completeness of $(X, d)$ means every Cauchy sequence in $X$ converges in $X$ (Complete metric space: every Cauchy sequence converges in the space).

[L2]

Cauchy sequences and convergence are tested by arbitrarily small positive distances; real and rational epsilon tests agree (Cauchy sequence in a metric space, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

[L3]

Triangle inequality, symmetry and separation are the metric axioms; nonnegativity follows from them (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Nonnegativity of a metric is a consequence of the other axioms, not an axiom).

[L4]

For every positive real $ε$ there exists an integer $m \geq 1$ with $1 / m < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L5]

A specified self-map of a set with an initial state defines a unique natural-number sequence by recursion (The recursion theorem).

Proof

technique · cases, on whether the ambient space is empty

1.1givenF1

Assume $(X, d)$ is separable and complete, let $(U_{n})$ be dense open subsets of $X$ , and let $W$ be a nonempty open subset of $X$ ; the task is to produce a point of $W \cap ⋂_{n} U_{n}$ .

1.2assume-caseassume-case

Case A: $X = \emptyset$ . Case B: $X \neq \emptyset$ .

2.1step 1.2F1

In case A the conclusion is vacuous and no sequence is constructed: a space with empty underlying set has no nonempty open subset, so there is no $W$ to test, and in particular no sequence into $X$ is invented.

2.2step 1.2F1F3

In case B, separability gives a dense at most countable $D \subseteq X$ , and $D \neq \emptyset$ because the dense set $D$ meets the nonempty open set $X$ ; by [F3] fix a sequence $s : N \to D$ whose range is $D$ .

3.1step 2.2F1F2L3L4

In case B, continuing, for every nonempty open $G \subseteq X$ and every real $ρ > 0$ the set of pairs $(k, m) \in N \times N$ with $m \geq 1$ , $1 / m \leq ρ$ and $\overset{ˉ}{B} (s (k), 1 / m) \subseteq G$ is nonempty. Fix $y \in G$ and $r > 0$ with $B (y, r) \subseteq G$ by [F2], choose $m \geq 1$ with $1 / m < \min (r / 3, ρ)$ by [L4], and use density of $D$ to fix $k$ with $s (k) \in B (y, 1 / m)$ . If $z \in \overset{ˉ}{B} (s (k), 1 / m)$ , then $d (z, y) \leq d (z, s (k)) + d (s (k), y) < 2 / m < r,$ so $z \in B (y, r) \subseteq G$ . Thus the required closed ball, not merely its open subball, lies in $G$ .

3.2step 2.2F1F2

In case B, continuing, put $G_{0} : = W \cap U_{0}$ ; this set is nonempty because $W$ is nonempty open and $U_{0}$ is dense, and it is open by [F2].

4.1step 3.1step 3.2F1F2L5

In case B, continuing, define by recursion on $n \in N$ : given the nonempty open $G_{n}$ , let $(k_{n}, m_{n})$ be the least element of the admissible set of step 3.1 for $(G_{n}, 2^{- (n + 2)})$ , put $x_{n} : = s (k_{n})$ , $r_{n} : = 1 / m_{n}$ and $G_{n + 1} : = B (x_{n}, r_{n}) \cap U_{n + 1}$ ; use lexicographic order, taking first the least admissible $k$ and then the least admissible $m$ for that $k$ , and $G_{n + 1}$ is nonempty open because the nonempty open ball $B (x_{n}, r_{n})$ meets the dense set $U_{n + 1}$ and both sets are open.

5.1L5step 3.1step 3.2step 4.1

This recursion is a set recursion: use states $(n, G)$ with $G$ a nonempty open subset of $X$ , together with one default state. The least-pair rule defines the successor on every such state, by step 3.1 and density of $U_{n + 1}$ ; let the default state map to itself. Apply [L5] with initial state $(0, G_{0})$ . Projections and the uniquely defined least-pair function give $x_{n}, r_{n}$ . This uses no choice function.

5.2step 3.1step 4.1F2L3

In case B, continuing, put $C_{n} : = \overset{ˉ}{B} (x_{n}, r_{n})$ . By the admissibility condition of step 3.1 used at stage $n$ , $C_{n} \subseteq G_{n}$ . Hence $C_{n + 1} \subseteq G_{n + 1} = B (x_{n}, r_{n}) \cap U_{n + 1} \subseteq C_{n} .$ Each $C_{n}$ contains its already defined center $x_{n}$ , and is closed by [F2]. If $a, b \in C_{n}$ , then $d (a, b) \leq 2 r_{n} \leq 2^{- (n + 1)}$ by [L3].

6.1step 5.1step 5.2L1L2L4

For $j, k \geq N$ , nesting gives $x_{j}, x_{k} \in C_{N}$ , so $d (x_{j}, x_{k}) \leq 2^{- (N + 1)}$ . These bounds tend to zero: induction gives $2^{N + 1} \geq N + 1$ , and [L4] makes $1 / (N + 1)$ eventually smaller than any positive $ε$ . Thus $(x_{n})$ is Cauchy by [L2], and completeness [L1] gives a limit $p \in X$ . No points are selected from arbitrary sets; the sequence of centers was already defined in step 5.1.

7.1step 5.2step 6.1F2L2L3

For each fixed $N$ , every $x_{j}$ with $j \geq N$ belongs to the closed set $C_{N}$ . If $p \notin C_{N}$ , its open complement contains a ball $B (p, ε)$ by [F2], but convergence [L2] puts some $x_{j}$ , $j \geq N$ , in that ball, a contradiction. Hence $p \in ⋂_{N} C_{N}$ . If $q$ is another point of the intersection, step 5.2 gives $d (p, q) \leq 2^{- (N + 1)}$ for all $N$ , whence $d (p, q) = 0$ and $p = q$ by [L3].

8.1step 3.2step 5.2step 7.1

In case B, continuing, $p \in C_{0} \subseteq G_{0} = W \cap U_{0}$ . For every $n$ , step 5.2 also gives $p \in C_{n + 1} \subseteq G_{n + 1} \subseteq U_{n + 1}$ . Therefore $p \in W \cap ⋂_{n} U_{n}$ .

9.1step 2.1step 8.1step 4.1cases-exhaustive∎

Either the ambient space is empty, in which case step 2.1 gives the Baire condition vacuously, or it is nonempty, in which case steps 4.1 to 8.1 produce the required point of $W \cap ⋂_{n} U_{n}$ ; the two cases exhaust the possibilities, so $(X, d)$ is Baire and the only objects used were the supplied dense set, its enumeration, and least-element selections on $N \times N$ .

Remarks

Where the choice would have been, and why it is not spent. The classical proof of the complete-metric Baire theorem chooses a ball inside $G_{n} \cap U_{n}$ at every stage, which is dependent choice. Here the centre is forced to be the least index of a fixed enumeration of one dense set and the radius is forced to be the least admissible value, so each stage is a definable function of the previous one and no selection principle is invoked.
Completeness is used once. It supplies the limit of the explicitly constructed center sequence in step 6.1. Closedness puts that limit in every nested ball. No general intersection theorem for arbitrary nonempty sets is invoked.

Depends on

Dependency tree · 2 levels

Sources

Marianne Morillon, Synthese