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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$\begingroup$ Suvrit, is that poor style? $\endgroup$– Cole LeahyCommented Aug 12, 2011 at 1:06
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$\begingroup$ Dear Cole, i meant it in jest ;-) $\endgroup$– SuvritCommented Aug 12, 2011 at 1:54
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$\begingroup$ I took no offense! I really wondered, since I'm not a mathematician. $\endgroup$– Cole LeahyCommented Aug 12, 2011 at 2:03
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2 Answers
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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.
answered Aug 12, 2011 at 2:08
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$\begingroup$ And the reason for that is that we want it to be a perfect Polish space, right? (with the induced topology) So that relates to what Gerald Edgar pointed out below. $\endgroup$– ftontiCommented Aug 28, 2011 at 10:44
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1$\begingroup$ Well, it would be a perfect Polish space even if it were only
$G_\delta$rather than closed (because a$G_\delta$subset is complete under a new metric that generates the same topology). But yes, if you merely prohibit isolated points then most of what one likes about perfect Polish spaces is lost; for example the space can be countable, like $\mathbb Q$. $\endgroup$ Commented Aug 28, 2011 at 21:46
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Your argument is fine. Perhaps someone didn't say "Polish" or something.
answered Aug 11, 2011 at 23:07
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$\begingroup$ Gerald, I oversimplified a bit. The someone is Yiannis Moschovakis, and he didn't quite suggest it to me personally. But the last line of the first paragraph on page 60 of the second edition of his book on descriptive set theory (sorry for the verbiage) is what led me to ask the question. $\endgroup$ Commented Aug 12, 2011 at 1:13