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This was left as an unproved theorem in our class:

Theroem: If $X$ is a finite dimensional normed vector space then each subset $M$ of $X$ is compact if and only if $M$ is closed and bounded.

How do I prove it? I know that $M$ is also finte dimensional and hence complete. So it is bounded.

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If $X$ has dimension $n$, then $X$ is isomorphic to Euclidean $n$-space. Use the Heine-Borel Theorem. –  David Mitra Apr 10 '12 at 23:19
    
In your theorem, did you want to say: "If X is a finite dimensional normed vector space then a subset $M$ of $X$ is compact if and only if $M$ is closed and bounded."? –  David Mitra Apr 10 '12 at 23:23
    
@DavidMitra: Would the proof then be a one line proof? –  Nano Apr 10 '12 at 23:23
    
Yes, more or less, as long as you've already proven the two ingredients (Heine-Borel for $\Bbb R^n$ and the fact that two finite dimensional normed spaces of the same dimension are isomorphic). –  David Mitra Apr 10 '12 at 23:25
    
Yes.${}{}{}{}{}{}$ –  Nano Apr 10 '12 at 23:25

1 Answer 1

$Hint$ If $X$ has dimension $n$ then $X$ is linearly homeomorphic to $\mathbb R^n$.

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