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Let $X$ be a compact metric space and $Y$ any metric space. If $f:X \to Y$ is continuous, then $f(X)$ is compact (that is, continuous functions carry compact sets into compact sets).

Proof:

Consider an open cover of $f(X)$.

Then $f(X) \subset \bigcup_{\alpha \in A}V_\alpha$ where each $V_\alpha$ is open in $Y$.

$X \subset f^{-1}(f(X)) \subset f^{-1}\left(\bigcup_{\alpha \in A}V_\alpha\right) = \bigcup_{\alpha \in A}f^{-1}(V_\alpha)$.

Hence $\bigcup_{\alpha \in A}f^{-1}(V_\alpha)$ is an open cover of $X$. Since $X$ is compact then we can choose a finite subcover $\{V_i\}_{i=1}^n$ such that $X \subset \bigcup_{i=1}^n f^{-1}(V_i)$.

So then $f(X) \subset f\left(\bigcup_{i=1}^n f^{-1}(V_i)\right) = \bigcup_{i=1}^n f\left(f^{-1}(V_i)\right) \subset \bigcup_{i=1}^n V_i$, a finite subcover of $f(X)$. $\therefore f(X)$ is compact.

Does this proof have an error?

Thanks for your help.

  • 4
    What is your question?2012-05-31
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    Please don't write your titles in capital letters like this. It attracts attention, but not the good sort.2012-05-31
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    By "test" do you mean "proof?"2012-05-31
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    Where do you think there might be a mistake in the proof? ¿En dónde crees tú que existe un error en esa demostración?2012-05-31
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    @DonAntonio Perhaps you might be willing to invite him to post in Spanish, and then translate the question yourself? I and many others would much prefer well-translated intelligible questions to unintelligible questions, [as per this meta thread](http://meta.math.stackexchange.com/questions/1617/what-is-the-site-etiquette-about-i-asking-and-ii-answering-questions-in-a-la).2012-05-31
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    @Mark I think he's studying with a book in english as the language is pretty smooth (well, escept for the part he wrote by himself), but of course I wouldn't mind doing so.2012-05-31
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    The proof looks fine. Note that you never use the fact that these spaces are metric, so the first sentence could read, "Let $X$ and $Y$ be topological spaces, with $X$ compact."2012-05-31
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    The proof looks good.2012-05-31
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    so where did this proof use the fact that $f$ is continuous?2014-11-24

1 Answers 1

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The proof is not good. Knowing that $f$ is continuous does not say much about $f^{-1}$, while the proof assumes that it exists and is continuous. It was not given that $f$ is a homeomorphism.

The statement is still true though. Take any sequence $x_n$ in compact $X$. Then it has a convergent subsequence $x_q$ (by definition of compactness). Now take the sequence $f(x_n)$ and notice that it has a convergent subsequence $f(x_q)$ (by definition of continuity). Since for any point $y \in f(X)$ there exists $x \in X $ such that $f(x) = y$, every sequence if $f(X)$ can be written as $f(x_n)$ for some sequence $x_n \in X$ and we are done.

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    $f^{-1}$ is being used in [this sense](http://en.wikipedia.org/wiki/Image_(mathematics)#Inverse_image), I think.2012-05-31
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    You're right. I was worried about the pre-image of $f(x) = const.$, but that does work too.2012-05-31