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Prove that a compact subset of a normed vector space is complete .

My thought process: Suppose S is a compact subset of $(V,\parallel . \parallel)$. Since S is compact, it is closed and bounded, so every sequence in S should converge to a point in S so if we fix a Cauchy sequence in S, it has to converge to a point in S thereby implying completeness.

Any help would be appreciated, Thanks

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"Every sequence in S should converge to a point in S": What if the sequence doesn't converge at all? –  Nate Eldredge Feb 13 '13 at 3:45

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In this case, every sequence will have a convergent subsequence since we're in a compact subset of a normed linear space. Fix a Cauchy sequence, say $\{x_n\}_{n=1}^\infty$. Since it is a sequence, it has a convergent subsequence, that is, $$\lim_{k\to\infty}x_{n_k}=x_0.$$By definition, this means that for every $\varepsilon>0$, there exists an $N_1\in\Bbb N$ such that for $k>N_1$, we have $\Vert x_{n_k}-x_0\Vert<\varepsilon/2.$ Also, since $\{x_n\}$ is Cauchy, for every $\varepsilon>0$, there is an $N_2$ such that $\Vert x_m-x_n\Vert<\varepsilon/2\Vert$ for $m,n>N_2$.

Thus, fix $\varepsilon>0$ and define $N=\max\{N_1,N_2\}$ from above. Now, let $m,n>N$ so that we have $$\Vert x_n-x_0\Vert\leq\Vert x_m-x_n\Vert+\Vert x_0-x_m\Vert<\varepsilon.$$ Since $\varepsilon>0$ was arbitrary, this concludes the proof.

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Your saying that if I fix a cauchy sequence in S, then for all sequences in S that converge in V have to converge in S since it is closed? did i understand what u said correctly? –  bobdylan Feb 13 '13 at 3:25
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I'm not sure you did understand. Fix a Cauchy sequence, say $\{x_n\}_{n=1}^\infty$. Then we know there is a convergent subsequence, say $\lim_{k\to\infty}x_{n_k}=x_0$ (we know this since we're in a compact subset of a normed space). Since $x_n$ is Cauchy, you can show that $\lim_{n\to\infty}x_n=x_0$. This shows that every Cauchy sequence is convergent, hence the compact subset is complete. –  Clayton Feb 13 '13 at 3:27

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