compactness of closed unit ball in normed spaces
The above result is false, in general, if one is considering other topologies in besides the norm topology (see, for example, the Banach-Alaoglu theorem). It follows that infinite dimensional (http://planetmath.org/Dimension2) normed spaces are not locally compact.
Suppose that is not finite dimensional. Pick an element such that and denote by the subspace (http://planetmath.org/VectorSubspace) generated by
Recall that, according to the Riesz Lemma (http://planetmath.org/RiezsLemma), there exists an element such that and , where denotes the distance between an element and a subspace
Now consider the subspace generated by . Since is infinite dimensional is a proper subspace and we can still apply the Riesz Lemma to find an element such that and
If we proceed inductively, we will find a sequence of norm 1 elements and a sequence of subspaces such that Under this setting it is easily seen that the sequence is in and satisfies for all . Therefore, is a sequence in that has no convergent (http://planetmath.org/ConvergentSequence) subsequence, i.e., is not compact.
Remarks on the proof - Note that the sequence 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 is finite dimensional if and only if every bounded sequence in has a Cauchy subsequence.
For Hilbert spaces the proof would be slightly simpler because one could just pick any orthonormal basis and it would for all with 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|
|Date of creation||2013-03-22 17:48:41|
|Last modified on||2013-03-22 17:48:41|
|Last modified by||asteroid (17536)|
|Synonym||closed unit ball in a normed space is compact iff the space is finite dimensional|