Under the ultrafilter lemma, every metrizable Čech-complete space is completely metrizable

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).

Assume the ultrafilter lemma and Dependent Choice. 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).

Let $(X,d)$ be a metric space (def-metric-space) and let $\mathcal{C}$ be the set of all Cauchy sequences in $X$ (def-cauchy-in-metric). Then: 1. For all $x=(x_n)$ and $y=(y_n)$ in $\mathcal{C}$ the real sequence $(d(x_n,y_n){)}_n$ converges, so $\rho (x,y):={\lim }_{n}d(x_n,y_n)$ is a single well-determined real (thm-cauchy-criterion-via-lub, lem-limit-unique). 2. The relation $x\sim y:⟺\rho (x,y)=0$ is an equivalence relation on $\mathcal{C}$. Write $\widehat{X}:=\mathcal{C}\mathrm{/}\text{\hspace{0.17em}⁣}\sim$ for the set of its classes and $[x]$ for the class of $x$. 3. $\widehat{d}([x],[y]):=\rho (x,y)$ does not depend on the chosen representatives, and $\widehat{d}$ is a metric on $\widehat{X}$. 4. The map $\iota :X\to \widehat{X}$ sending $p$ to the class of the constant sequence at $p$ is an isometric embedding with dense image (def-isometry-and-metric-embedding, def-metric-interior-closure-boundary). 5. $(\widehat{X},\widehat{d})$ is complete. Consequently $((\widehat{X},\widehat{d}),\iota )$ is a completion of $(X,d)$ (def-metric-completion), and every metric space has a completion. The notation is kept honest. A Cauchy sequence in $X$ need not converge in $X$, so no symbol ${\lim }_n{x}_n$ appears anywhere below; the only limits taken are limits of real sequences, and each is written only after its existence has been proved. The equivalence relation is defined and verified here rather than cited, as was done for def-integers, so that the construction is self-contained and its transitivity argument is visible at the point of use. (Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences).

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 Axiom of Countable Choice. If $(X,d)$ is a complete metric space and $Y\subseteq X$ is $G_{\delta }$ in $X$, then the subspace $Y$ is completely metrizable. (Under the Axiom of Countable Choice, every $G_{\delta }$ subspace of a complete metric space is completely metrizable).

Let $(X,d)$ be a metric space with its metric topology. Then $X$ is Tychonoff and perfectly normal; in particular every metric space is a Tychonoff space (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).