Compact closed unit ball implies finite dimension

Statement

Let $X$ be a normed space over $\mathbb{R}$ or $\mathbb{C}$. If the closed unit ball $\{ x \in X : \|x\| \le 1 \}$ is compact, then $X$ is finite-dimensional.

Remarks

Proof: if $X$ is infinite-dimensional, build unit vectors $x_1, x_2, \ldots$ by Riesz's lemma, with $x_{n+1}$ at distance at least $\tfrac{1}{2}$ from the span of $x_1, \ldots, x_n$. That span is finite-dimensional, so it is complete and closed. Then $\|x_m - x_n\| \ge \tfrac{1}{2}$ for $m \neq n$, so no subsequence converges, and the ball is not sequentially compact. With the Heine–Borel theorem in finite dimension, we get: a normed space is finite-dimensional if and only if its closed unit ball is compact. Equivalently, a normed space is locally compact if and only if it is finite-dimensional.

Explanations

L1 human

In infinite dimensions, the closed unit ball is never compact: there is always room for infinitely many unit vectors that stay far apart. This is one of the main differences between finite and infinite dimensions.