Subset of Finite Dimensional Normed Vector Space is Compact iff Closed and Bounded/Sufficient Condition

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

Let $K \subset X$ be a compact subset.

Then $K$ is closed and bounded.

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

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.