notes/docs/mathematics/functional-analysis/normed-spaces/compactness.md

Compactness

Definition 1: a metric space X is compact if every sequence in X has a convergent subsequence. A subset M of X is compact if every sequence in M has a convergent subsequence whose limit is an element of M.

A general property of compact sets is expressed in the following proposition.

Proposition 1: a compact subset M of a metric space (X,d) is closed and bounded.

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The converse of this proposition is generally false.

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However, for a finite dimensional normed space we have the following proposition.

Proposition 2: in a finite dimensional normed space (X, \|\cdot\|) a subset M \subset X is compact if and only if M is closed and bounded.

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A source of interesting results is the following lemma.

Lemma 1: let Y and Z be subspaces of a normed space (X, \|\cdot\|), suppose that Y is closed and that Y is a strict subset of Z. Then for every \alpha \in (0,1) there exists a z \in Z, such that

1. \|z\| = 1,
2. \forall y \in Y: \|z - y\| \geq \alpha.

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Lemma 1 gives the following remarkable proposition.

Proposition 3: if a normed space (X, \|\cdot\|) has the property that the closed unit ball M = \{x \in X | \|x\| \leq 1\} is compact, then X is finite dimensional.

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Compact sets have several basic properties similar to those of finite sets and not shared by non-compact sets. Such as the following.

Proposition 4: let (X,d_X) and (Y,d_Y) be metric spaces and let T: X \to Y be a continuous mapping. Let M be a compact subset of (X,d_X), then T(M) is a compact subset of (Y,d_Y).

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From this proposition we conclude that the following property carries over to metric spaces.

Corollary 1: let M \subset X be a compact subset of a metric space (X,d) over a field F, a continuous mapping T: M \to F attains a maximum and minimum value.

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