Can we conclude that an analytic subset $A$ of a Polish space $X$ is also Polish?
Let $\mathcal{M}(R^d)$ denotes the family of Borel probability measures on $R^d$ equipped with the Lévy-Prokhorov metric. Is it true that $\mathcal{M}(R^d)$ is an analytic space?
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8$\begingroup$ For 1, see en.wikipedia.org/wiki/Polish_space#Properties : only $G_\delta$ subsets are Polish. $\endgroup$– Emil JeřábekCommented Jul 29, 2023 at 10:15
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2$\begingroup$ For 2, $\mathcal{M}(R^d)$ is even Polish. $\endgroup$– Michael GreineckerCommented Jul 29, 2023 at 10:17
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$\begingroup$ $M(R^d)$ is analytic? $\endgroup$– B-SCommented Jul 29, 2023 at 10:26
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4$\begingroup$ Yes, every Polish space is analytic. $\endgroup$– Michael GreineckerCommented Jul 29, 2023 at 11:29
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