Proposition 12

A metrizable space is a Lusin space if and only if it is the image of a Polish space under a continuous bijective mapping.

The condition is clearly necessary; let us show that it is also sufficient. If $f$ is a continuous bijection of a Lusin space $X$ onto a metrizable space $Y$, it follows from Definition 6 that $Y$ is a Lusin space. Hence we need only show that a Polish space is a Lusin space.

Notice first that, if $X$ is a Lusin space, every closed (resp. open) subspace $A$ of $X$ is a Lusin space (cf. no. 7, Theorem 3); for if $f$ is a continuous bijection of a zero-dimensional Polish space $P$ onto $X$, then $f^{-1}(A)$ is closed (resp. open) in $P$ and is therefore a zero-dimensional Polish subspace of $P$ (no. 1, Propositions 1 and 2).

Every countable product of Lusin spaces is a Lusin space; this follows from Proposition 1 of no. 1 and the fact that every product of zero-dimensional spaces is zero-dimensional. Every countable intersection of Lusin subspaces of a Hausdorff topological space is a Lusin subspace; this follows from the preceding remarks and from Lemma 1 of no. 1. Furthermore: