Mathematics Prelims

July 16, 2008

Some Compactness Properties

Filed under: Analysis,Functional Analysis — cjohnson @ 2:32 pm

In a metric space, a subset is called compact if every sequence in the space contains a convergent subsequence (In a more general setting this is called “sequentially compact,” and compactness refers to a set where every open cover has a finite subcover, but in the case of a metric space these two definitions are equivalent.)  As it turns out, every compact subset of a metric space will be both closed and bounded.

Suppose M is a compact subset of some metric space (X, d).  Suppose that M were not bounded, then for a fixed m \in M we could find a sequence (y_n)_{n \in \mathbb{N}} such that for each n \in \mathbb{N}, d(m, y_n) > n.  Note that such a sequence has no convergent subsequence.  Since unboundedness implies a set can’t be compact, every compact set must be bounded.  For closedness, let x \in \overline{M}.  There exists a sequence in M, (x_n)_{n \in \mathbb{N}} such that x_n \to x.  By compactness of M, this sequence has a convergent subsequence (x_{p_n}) which converges to an element of M.  However, every subsequence of a convergent sequence must have the same limit as the “original” sequence, so x_{p_n} \to x, and this implies x \in M, so \overline{M} = M and M is closed.

In a finite dimensional normed space, this becomes boundedness and closedness are no longer simply necessary conditions for compactness, but they are in fact sufficient conditions.  That is, in a finite dimensional normed space, a set is compact if and only if it is both closed and bounded.

We’ve already show that compactness implies closed and bounded, so we now show the converse.  Suppose (X, \|\cdot\|) is an n-dimensional normed space with basis vectors \{ e_1, e_2, ..., e_n \}.  Let M \subseteq X and suppose M is both closed and bounded.  Let x_m = \sum_{i=1}^n \alpha_i^{(m)} e_i for any sequence in M.  Since M is bounded, there exists a K > 0 such that for each m \in \mathbb{N} we have \| x_m \| \leq K.  We know (from an earlier lemma) there exists a c > 0 such that

\displaystyle K \geq \| x_m \| = \left\| \sum_{i=1}^n \alpha_i^{(m)} e_i \right\| \geq c \sum_{i=1}^n |\alpha_i^{(m)}|

This tells us for any fixed i, that (\alpha_i^{(m)}) is bounded, so it must have a convergent subsequence.  Call the limit of this subsequence \alpha_i.  We now have that (x_m) must have a convergent subsequence with limit x = \alpha_1 e_1 + ... + \alpha_n e_n.  Note this limit is in M, so M is compact.

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