FALSE: every metrizable space is Čech-complete

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

$\mathbb{Q}$, with its usual subspace topology from $\mathbb{R}$, is not a Baire space. The cited item is a false-statement item: the sentence displayed under its Statement heading is the claim it refutes, and what it establishes is that negation (FALSE: the rational numbers form a Baire space).

Assume Dependent Choice. Every Čech-complete space is a Baire space. (Under Dependent Choice, every Čech-complete space is Baire).

Assume the ultrafilter lemma and the Axiom of Choice. A metrizable space is Čech-complete if and only if it is completely metrizable. (Under the ultrafilter lemma and the Axiom of Choice, a metrizable space is Čech-complete exactly when it is completely metrizable).

Write ${\mathbb{Q}}_{\mathbb{R}}$ for the image of $\mathbb{Q}$ in $\mathbb{R}$ under the canonical embedding $q\mapsto \widehat{q}$ (lem-rat-embeds-dense), the set usually written $\mathbb{Q}$ once the identification is made, and put $X:=\mathbb{R}\setminus {\mathbb{Q}}_{\mathbb{R}}$ for the irrationals. Then: 1. ${\mathbb{Q}}_{\mathbb{R}}$ is an $F_{\sigma }$ set (def-f-sigma-g-delta) and is meager (def-nowhere-dense-meager); 2. $X$ is a $G_{\delta }$ set and is residual; 3. ${\mathbb{Q}}_{\mathbb{R}}$ is not a $G_{\delta }$ set, and $X$ is not an $F_{\sigma }$ set. Claims 1 and 2 are bookkeeping. Claim 3 is the substance and is exactly where thm-baire-category-r is spent: no argument from the algebra of open and closed sets alone can reach it, since ${\mathbb{Q}}_{\mathbb{R}}$ and $X$ are interchanged by complementation while $F_{\sigma }$ and $G_{\delta }$ are, so any such argument would prove the same thing about both sets and about neither. ($\mathbb{Q}$ is $F_{\sigma }$, meager and not $G_{\delta }$, while the irrationals are $G_{\delta }$, residual and not $F_{\sigma }$).