Under Dependent Choice, every completely metrizable subspace of a metric space is $G_{\delta }$

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).

Let $(X,\mathcal{T})$ be a topological space (def-topological-space) and let $A\subseteq X$. $A$ is a $G_{\delta }$ set of $X$ when there is a sequence $(V_n{)}_{n\in \mathbb{N}}$ of open subsets of $X$ with $A={\bigcap }_{n\in \mathbb{N}}{V}_n$, and an $F_{\sigma }$ set of $X$ when there is a sequence $(F_n{)}_{n\in \mathbb{N}}$ of closed subsets of $X$ with $A={\bigcup }_{n\in \mathbb{N}}{F}_n$. ($G_{\delta }$ and $F_{\sigma }$ subsets of a topological space, agreeing with the real-line notion).

Let $X$ be a set and let $R\subseteq X\times X$ be a binary relation on $X$. Call $R$ entire on $X$ when $\text{for every }x\in X\text{ there is }y\in X\text{ with }xRy\mathrm{.}$ The Axiom of Dependent Choice, written $\mathrm{D}\mathrm{C}$, is the following statement. The statement is: for every nonempty set $X$, every relation $R$ entire on $X$, and every $a\in X$, there is a sequence $x:\mathbb{N}\to X$ with $x_0=a$ and $x_{n}R{x}_{n+1}$ for every $n\in \mathbb{N}$. (The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain).

Let $(X,d)$ be a metric space (def-metric-space) and let $p,q\in X$ with $p\mathrm{\ne }q$. Put $r:=d(p,q)\mathrm{/}2$. Then $r>0$ and $B(p,r)\cap B(q,r)=\mathrm{\varnothing }\mathrm{.}$ Both sets are open (thm-metric-open-set-algebra) and contain $p$ respectively $q$ (def-metric-ball), so every metric space is Hausdorff: distinct points are separated by disjoint open sets (def-metric-topology). (Distinct points of a metric space have disjoint balls around them).