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