Under Dependent Choice, a subspace of a Polish space is Polish exactly when it is $G_{\delta }$

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

A topological space is Polish when it is separable (def-separable-space) and completely metrizable: its topology is induced by some complete metric (lem-complete-remetrisation). No particular compatible complete metric or countable dense subset is part of the structure. (Polish spaces are separable completely metrizable spaces).

Assume Dependent Choice. For a subspace $Y$ of a complete metric space $X$, $Y$ is completely metrizable if and only if $Y$ is a $G_{\delta }$ subset of $X$. (Alexandrov's theorem, under Dependent Choice: a subspace of a complete metric space is completely metrizable exactly when it is $G_{\delta }$).

Every subspace of a second countable space is second countable. (Second countability is hereditary).

Assuming ${\mathrm{A}\mathrm{C}}_{\omega }$, every second countable space is separable. (Assuming countable choice, every second countable space is separable).

Assume the Axiom of Countable Choice. For a completely metrizable space, separability is equivalent to second countability. Thus either countability convention gives the same notion of Polish space. (For completely metrizable spaces, the separable and second-countable definitions of Polish space agree under countable choice).