Under the ultrafilter lemma and the Axiom of Choice, every completely metrizable space is Čech-complete

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

A Tychonoff space $X$ is Čech-complete when there is a Hausdorff compactification $(K,i)$ of $X$ (def-compactification-of-a-tychonoff-space) for which $i[X]$ is a $G_{\delta }$ subset of $K$ (def-g-delta-and-f-sigma-in-a-topological-space). The definition asks for one compactification; thm-cech-completeness-is-independent-of-compactification proves the equivalent every-compactification form. (Čech-complete spaces as $G_{\delta }$ subspaces of Hausdorff compactifications).

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 Choice. Every metric space is paracompact. (Stone's theorem, under choice: every metric space is paracompact).

Consequently every metrizable space (def-metrizable-space) is Tychonoff and perfectly normal, and hence $T_6$, $T_5$, $T_4$, $T_{3\frac{1}{2}}$, $T_3$, $T_{2\frac{1}{2}}$, $T_2$, $T_1$ and $T_0$. (In a metric space every closed set is a zero set and a $G_{\delta }$, and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal).

Assume the ultrafilter lemma. If $X$ is Tychonoff, $e:X\to [0,1{]}^{C(X,[0,1])}$ is its full evaluation map, and $K=\overline{e[X]}$, then $(K,e)$ is a Hausdorff compactification of $X$. In particular every Tychonoff space has one. This statement uses the ultrafilter lemma only for compactness of the cube; it makes no assertion about dependent choice. (Assuming the ultrafilter lemma, every Tychonoff space has a Hausdorff compactification).

Assume the ultrafilter lemma. A Tychonoff space is a $G_{\delta }$ subset of some Hausdorff compactification if and only if it is a $G_{\delta }$ subset of every Hausdorff compactification. (Under the ultrafilter lemma and Dependent Choice, a Tychonoff space is $G_{\delta }$ in some Hausdorff compactification exactly when it is $G_{\delta }$ in every one).