FALSE: every Baire space is completely metrizable

Given: The objects, hypotheses, and choice principles stated above.

Assume the Axiom of Dependent Choice (def-dependent-choice). Let $(X,\mathcal{T})$ be a locally compact Hausdorff space (def-locally-compact-space, def-hausdorff-space, def-topological-space). Then $X$ is a Baire space (def-baire-space): for every sequence $(U_n{)}_{n\in \mathbb{N}}$ of dense open subsets of $X$ (def-dense-top, def-sequence-convergence-top), the intersection ${\bigcap }_{n\in \mathbb{N}}{U}_n$ is dense in $X$. Dependent choice is sufficient here and no claim of necessity is made. The several statements that go by the name "Baire category theorem" have different choice-theoretic statuses over ZF. The compact Hausdorff version is equivalent to dependent multiple choice; DC implies DMC in ZF, the reversal remains open in ZF, and DMC does not imply DC in ZFA. That account is rem-baire-category-choice-strength, which this library states and does not prove. Nothing below asserts that dependent choice is needed for the statement above. (Assuming dependent choice, every locally compact Hausdorff space is a Baire space).

Assume the Axiom of Choice (def-axiom-of-choice). Let $I$ be a set and let $(X_i,{\mathcal{T}}_i{)}_{i\in I}$ be a family of compact topological spaces (def-compact-space, def-topological-space). Then the product $P:={\prod }_{i\in I}{X}_i$ with the product topology (def-product-topology) is compact. The Axiom of Choice is spent twice, and both uses are flagged below. Once inside thm-alexander-subbase-lemma, through Zorn's lemma (thm-zorn), and once directly at step 2.1, to produce a point of a product of nonempty sets. (Tychonoff's theorem: an arbitrary product of compact spaces is compact in the product topology, assuming the Axiom of Choice).

A topological space $(X,\mathcal{T})$ (def-topological-space) is first countable if every point of $X$ has an at most countable neighbourhood base: for each $x\in X$ there is a family ${\mathcal{B}}_x\subseteq \mathcal{N}(x)$ that is at most countable (def-countable, def-equinumerous) and such that every neighbourhood of $x$ contains a member of ${\mathcal{B}}_x$ (def-neighbourhood-top). (First countable space: a countable neighbourhood base at every point).

Assume the Axiom of Countable Choice. Let $(A_n{)}_{n\in \mathbb{N}}$ be a family of at most countable sets indexed by $\mathbb{N}$. Then $U={\bigcup }_{n\in \mathbb{N}}{A}_n$ is at most countable (Countable unions of at most countable sets, assuming ${\mathrm{A}\mathrm{C}}_{\omega }$).

Let $(X,d)$ be a metric space (def-metric-space) and let ${\mathcal{T}}_d$ be its metric topology (def-metric-topology). Call ${\mathcal{T}}_d$ completely metrizable if some metric $\rho$ on $X$ is topologically equivalent to $d$, that is ${\mathcal{T}}_{\rho }={\mathcal{T}}_d$ (def-equivalent-metrics), and makes $(X,\rho )$ complete (def-complete-metric-space). Then: 1. Homeomorphism invariance. Let $(Y,e)$ be a metric space and let $h:X\to Y$ be a bijection (def-injection-surjection-bijection) such that $h$ and $h^{-1}$ are continuous (def-metric-continuity). If ${\mathcal{T}}_d$ is completely metrizable then so is ${\mathcal{T}}_e$. 2. Closed subspaces. If ${\mathcal{T}}_d$ is completely metrizable and $A\subseteq X$ is closed in $(X,d)$, then ${\mathcal{T}}_{d_A}$ is completely metrizable, $d_A$ being the subspace metric (def-isometry-and-metric-embedding). 3. The property is strictly weaker than completeness. Let $P:=(0,\mathrm{\infty })\subseteq \mathbb{R}$ (def-interval) carry $d(x,y):=\mathrm{\mid }x-y\mathrm{\mid }$ (lem-real-line-is-a-metric-space). Then $(P,d)$ is not complete, while ${\rho }_P(x,y):=\mathrm{\mid }x-y\mathrm{\mid }+\mid \frac{1}{x}-\frac{1}{y}\mid$ is a complete metric on $P$ with ${\mathcal{T}}_{{\rho }_P}={\mathcal{T}}_d$. So ${\mathcal{T}}_d$ is completely metrizable although no completeness assumption holds for $d$ itself. Complete metrizability is a condition on the collection of open sets alone: the metric is quantified over and does not survive into the statement. That is exactly what completeness fails to be, and claim 3 shows the two conditions are genuinely different rather than merely stated differently. (Complete metrizability: admitting a topologically equivalent complete metric is preserved by homeomorphism and by closed subspaces, and $(0,\mathrm{\infty })$ has it without being complete).