# 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.