Under Dependent Choice, every completely metrizable space is Baire

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

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).

Assume the Axiom of Dependent Choice ($\mathrm{D}\mathrm{C}$). If a nonempty metric space $X$ is complete, then it is not the union of a sequence of closed sets each having empty interior. Equivalently, the intersection of countably many open dense subsets of $X$ is dense. (Under Dependent Choice, a nonempty complete metric space is not a countable union of closed sets with empty interior).

A topological space $(X,\mathcal{T})$ (def-topological-space) is a Baire space when for every sequence $(U_n{)}_{n\in \mathbb{N}}$ of subsets of $X$ that are open and dense in $X$ (def-dense-top, def-sequence-convergence-top, def-natural-numbers), the intersection ${\bigcap }_{n\in \mathbb{N}}{U}_n$ is dense in $X$. (Baire space: a topological space in which every countable intersection of dense open subsets is dense).