Let $\mathfrak{M}$ and $\mathfrak{N}$ be perfect Polish spaces, $P$ a nonempty perfect subset of $\mathfrak{M}$, and $f: \mathfrak{M} \rightarrow \mathfrak{N}$ a continuous surjection that's injective on $P$. Naively, the image $f[P]$ should be perfect. For if it has an isolated point $x$ with a neighborhood $X$ containing no other elements of $f[P]$, then by continuity and injectivity the preimage $f^{-1}[X]$ is an open set containing no points of $P$ other than the unique element $p$ mapped to $x$, contradicting perfection. However, someone suggested to me that this argument fails unless $P$ is compact. Is this true?
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A perfect subset of a space $X$ is required not only to have no isolated points but also to be closed in $X$. Compactness of $P$ is used to ensure that $f[P]$ is (compact and therefore) closed. |
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Your argument is fine. Perhaps someone didn't say "Polish" or something. |
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