Riesz's lemma

Statement

Let $X$ be a normed space over $\mathbb{R}$ or $\mathbb{C}$, and let $Y \neq X$ be a closed linear subspace of $X$. Let $\alpha$ be a real number with $0 < \alpha < 1$. Then there is $x \in X$ with $\|x\| = 1$ and $\|x - y\| \ge \alpha$ for all $y \in Y$.

Remarks

Proof idea: take $x_0 \notin Y$ and let $r = \inf_{y \in Y} \|x_0 - y\| > 0$, which is positive because $Y$ is closed. Choose $y_0 \in Y$ with $\|x_0 - y_0\| \le r / \alpha$, and let $x = (x_0 - y_0) / \|x_0 - y_0\|$. If $Y$ is finite-dimensional, one can take $\alpha = 1$. The lemma is a substitute for orthogonal vectors in normed spaces without an inner product. Its main application is that a compact closed unit ball forces finite dimension.

Explanations

L1 human

For each closed proper subspace, there is a unit vector that is almost as far from the subspace as possible. In an inner product space, a unit vector orthogonal to the subspace works with distance $1$.