Every open subspace of a completely metrizable space is completely metrizable

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 $(X,d)$ be a metric space (def-metric-space), let $A\subseteq X$ be nonempty and let $x,y\in X$. Then $\mathrm{\mid }d(x,A)-d(y,A)\mathrm{\mid }\le d(x,y),$ with $d(\cdot ,A)$ the distance to a nonempty set (def-metric-bounded-diameter). Thus the real-valued function $u\mapsto d(u,A)$ changes by at most $d(u,v)$ between $u$ and $v$: it is $1$-Lipschitz. ($\mathrm{\mid }d(x,A)-d(y,A)\mathrm{\mid }\le d(x,y)$, so the distance to a fixed nonempty set is $1$-Lipschitz).

Let $(X,d)$ be a metric space (def-metric-space) and let $A\subseteq X$ carry the subspace metric $d_A$ (def-isometry-and-metric-embedding). Then: 1. If $(A,d_A)$ is complete (def-complete-metric-space), then $A$ is closed in $(X,d)$ (def-metric-topology). No hypothesis on $X$ is needed. 2. If $(X,d)$ is complete and $A$ is closed in $(X,d)$, then $(A,d_A)$ is complete. Consequently, for a complete $(X,d)$ a subset $A\subseteq X$ is complete if and only if it is closed. (A subspace of a complete metric space is complete iff it is closed, and a complete subspace of any metric space is closed).