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" are inequivalent over ZF, and the choice principles they correspond to differ; that account, including the fact that the compact Hausdorff version is equivalent to a principle strictly weaker than dependent choice, 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).