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Continuous images of compact sets are compact

Theorem: Let V be a metric space. If $X\subseteq V$ and $f:X\rightarrow Y$ is continuous then $f(X)\subseteq Y$ is compact.

I am following a course on functional analysis and this theorem came about. I would like to know the proof. I have a feeling that the proof of this theorem involves topology, but maybe some of you know a proof without topology?

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marked as duplicate by rschwieb, Michael Greinecker, P.., Davide Giraudo, Henry T. Horton Jan 16 '13 at 19:55

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The terms "continuous" and "compact" are topological terms. What should a proof without topology be? –  Michael Greinecker Jan 16 '13 at 18:57
    
This is obviously false as it stands. Do you mean to make some assumptions about $X$, such as its being compact? –  Chris Eagle Jan 16 '13 at 19:14

2 Answers 2

Definitions are your friends when it comes to proofs like this.

$V$ is a metric space...what does that tell you about $\;X \subseteq V\,$?

Read your question again (from your text or problem sheet): I suspect that you are also given that $X$ is closed (and $V$ compact), and/or that $X$ is compact. Otherwise, I'm afraid, you face a losing battle.

What do you need to know whether a set is compact?

With that, what does it mean to know that $\;f:X \to Y\;$ is continuous?

What does this imply in terms of the properties of the image of $f$ in $Y\,$? (Image: the set $f(X) \subset Y$). What do you need to know about $X$ and $f$ to conclude that $f(X)$ it is compact in $Y$?

If you work through these questions, unpack the definitions, use some handy theorems, and answer the questions with these "tools", you'll pretty much have your proof.

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See Micheal O Searcoid, Metric Spaces, p210.

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