compactness of closed unit ball in normed spaces
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finite dimensional \PMlinkescapephraseinfinitedimensional \PMlinkescapephrasesubspace^{} \PMlinkescapephraseconvergent
Theorem  Let $X$ be a normed space^{} and $\overline{{B}_{1}(0)}\subset X$ the closed ball^{} (http://planetmath.org/Ball). Then $\overline{{B}_{1}(0)}$ is compact^{} if and only if $X$ is finite dimensional (http://planetmath.org/Dimension2).
The above result is false, in general, if one is considering other topologies^{} in $X$ besides the norm topology (see, for example, the BanachAlaoglu theorem). It follows that infinite dimensional (http://planetmath.org/Dimension2) normed spaces are not locally compact.
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Proof:

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$(\u27f8)$ This is the easy part. Since $X$ is finite dimensional it is isomorphic to some ${\mathbb{R}}^{n}$ (with the standard topology). The result then follows from the HeineBorel theorem.
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$(\u27f9)$ Suppose that $X$ is not finite dimensional. Pick an element ${x}_{1}\in X$ such that $\parallel {x}_{1}\parallel =1$ and denote by ${S}_{1}$ the subspace (http://planetmath.org/VectorSubspace) generated by ${x}_{1}.$
Recall that, according to the Riesz Lemma (http://planetmath.org/RiezsLemma), there exists an element ${x}_{2}\in X$ such that $\parallel {x}_{2}\parallel =1$ and $d({x}_{2},{S}_{1})\ge \frac{1}{2}$, where $d(y,M):=inf\{\parallel yz\parallel :z\in M\}$ denotes the distance between an element $y\in X$ and a subspace $M\subset X.$
Now consider the subspace ${S}_{2}$ generated by ${x}_{1},{x}_{2}$. Since $X$ is infinite dimensional ${S}_{2}$ is a proper subspace and we can still apply the Riesz Lemma to find an element ${x}_{3}$ such that $\parallel {x}_{3}\parallel =1$ and $d({x}_{3},{S}_{2})\ge \frac{1}{2}.$
If we proceed inductively, we will find a sequence $\{{x}_{n}\}$ of norm 1 elements and a sequence of subspaces ${S}_{n}:=\u27e8{x}_{1},\mathrm{\dots},{x}_{n}\u27e9$ such that $d({x}_{n+1},{S}_{n})\ge \frac{1}{2}.$ Under this setting it is easily seen that the sequence $\{{x}_{n}\}$ is in $\overline{{B}_{1}(0)}$ and satisfies $\parallel {x}_{n}{x}_{m}\parallel \ge \frac{1}{2}$ for all $m\ne n\in \mathbb{N}$. Therefore, $\{{x}_{n}\}$ is a sequence in $\overline{{B}_{1}(0)}$ that has no convergent (http://planetmath.org/ConvergentSequence) subsequence, i.e., $\overline{{B}_{1}(0)}$ is not compact. $\mathrm{\square}$
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Remarks on the proof  Note that the sequence $\{{x}_{n}\}$ constructed in the proof does not have a Cauchy subsequence (http://planetmath.org/CauchySequence). Thus we have in fact proven the slightly stronger result that $X$ is finite dimensional if and only if every bounded^{} sequence in $X$ has a Cauchy subsequence.
For Hilbert spaces^{} the proof would be slightly simpler because one could just pick any orthonormal basis^{} $\{{e}_{n}\},$ and it would $\parallel {e}_{n}{e}_{m}\parallel =\sqrt{2}$ for all $m,n\in \mathbb{N}$ with $m\ne n,$ therefore having no convergent subsequence. For general normed spaces we cannot just pick orthonormal elements, since this notion does not exist. Thus, we have to use Riesz Lemma to assure the existence of elements with some .
Title  compactness of closed unit ball in normed spaces 

Canonical name  CompactnessOfClosedUnitBallInNormedSpaces 
Date of creation  20130322 17:48:41 
Last modified on  20130322 17:48:41 
Owner  asteroid (17536) 
Last modified by  asteroid (17536) 
Numerical id  13 
Author  asteroid (17536) 
Entry type  Theorem 
Classification  msc 46B50 
Synonym  closed unit ball in a normed space is compact iff the space is finite dimensional 