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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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Suvrit, is that poor style? –  Cole Leahy Aug 12 '11 at 1:06
    
Dear Cole, i meant it in jest ;-) –  Suvrit Aug 12 '11 at 1:54
    
I took no offense! I really wondered, since I'm not a mathematician. –  Cole Leahy Aug 12 '11 at 2:03

2 Answers 2

up vote 4 down vote accepted

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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Duh! Thanks, Andreas. –  Cole Leahy Aug 12 '11 at 2:17
    
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. –  ftonti Aug 28 '11 at 10:44
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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 $\mathbb Q$. –  Andreas Blass Aug 28 '11 at 21:46

Your argument is fine. Perhaps someone didn't say "Polish" or something.

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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. –  Cole Leahy Aug 12 '11 at 1:13

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