# Heine-Borel Theorem/Normed Vector Space

## Theorem

Let $\struct {X, \norm {\,\cdot\,}}$ be a finite-dimensional normed vector space.

A subset $K \subseteq X$ is compact if and only if $K$ is closed and bounded.

## Proof

### Necessary Condition

Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $K$.

$\sequence {x_n}_{n \mathop \in \N}$ is bounded.

We have that bounded sequence in finite-dimensional space has a convergent subsequence.

Hence, $\sequence {x_n}_{n \mathop \in \N}$ has a convergent subsequence $\sequence {x_{n_k} }_{k \mathop \in \N}$.

Denote the limit $\ds \lim_{k \mathop \to \infty} \sequence {x_{n_k} } = L$.

By assumption, $K$ is closed.

By equivalence of definitions, $L \in K$.

By definition, $K$ is compact.

$\Box$

### Sufficient Condition

### Closedness

Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $K$.

Suppose, $\sequence {x_n}_{n \mathop \in \N}$ converges to $L \in K$.

Then there is a subsequence $\sequence {x_{n_k}}_{k \mathop \in \N}$ convergent to $L' \in K$.

But $\sequence {x_{n_k}}_{k \mathop \in \N}$ is a subsequence of $\sequence {x_n}_{n \mathop \in \N}$.

Hence, $\sequence {x_{n_k}}_{k \mathop \in \N}$ converges to $L$.

By uniqueness of limits, $L = L'$.

By definition, $K$ is closed.

### Boundedness

Suppose, $K$ is not bounded.

Then:

- $\forall n \in \N : \exists x_n \in K: \norm {x_n} > n$

Therefore, no subsequence of $\sequence {x_n}_{n \mathop \in \N}$ is convergent.

Hence, $K$ is not compact.

This is in contradiction with the assumption.

Thus, $K$ is bounded.

$\blacksquare$

## Sources

- 2017: Amol Sasane:
*A Friendly Approach to Functional Analysis*... (previous) ... (next): Chapter $\S 1.5$: Normed and Banach spaces. Compact sets