Proposition 1

a) Every closed subspace of a Polish space is Polish. b) The product of a countable family of Polish spaces is Polish. c) The sum of a countable family of Polish spaces is Polish.

Every subspace of a metrizable space of countable type is metrizable of countable type, and every closed subspace of a complete space is complete; (Chapter II, § 3, no. 4, Proposition 8). Every countable product of metrizable spaces of countable type is again metrizable of countable type: ($§ 2$, no. 8), and every countable product of complete metric spaces is a complete metric space with respect to a metric compatible with its topology (Chapter II, § 3, no. 5, Proposition 10 and Chapter IX, § 2, no. 4, Theorem 1, Corollary 2). Finally, let $(X_n)$ be a sequence of non-empty Polish spaces, and consider the product space $Y = \mathbf{N} \times \prod_n X_n$, where $\mathbf{N}$ carries the discrete topology; $Y$ is a Polish space by what has already been proved. On the other hand, let $a_n$ be a point of $X_n$ for each $n$, and let $f_n$ be the mapping of $X_n$ into $Y$ such that for each $x \in X_n$ we have

where $y_p = a_p$ if $p \neq n$ and $y_n = x$. If $X$ is the topological sum of the $X_n$, it is clear that the mapping $f$ of $X$ into $Y$ which agrees with $f_n$ on $X_n$ for each $n$ is a homeomorphism of $X$ onto $f(X)$; also, for each $n$, $f_n(X_n)$ is closed in $Y$, and the family $(f_n(X_n))$ is locally finite because $\mathbf{N}$ is discrete; therefore $f(X) = \bigcup_n f_n(X_n)$ is closed in $Y$ (Chapter I, § 1, no. 5, Proposition 4), and thus $f(X)$ is a Polish space, by a).