---
tag: G96C77E433
kind: proof
name: Proof that a compact closed unit ball implies finite dimension
msc: [46B50]
proves: 0VZVV8VME5
uses: [097C7KCAT5, 0K8SGQSV7G, 0CARW59WD2, 0ABP2F2ZP7, 0W41XG4W60, 018W3V3TCT, 0HR573CHNM, 0EPTJA5QV4, 06JYPJ4WHH]
sources:
  - url: https://en.wikipedia.org/wiki/Riesz%27s_lemma
    locator: Characterization of finite dimension
---

## Proof

We argue by contradiction: in infinite dimension, Riesz's lemma gives unit vectors that stay far apart.

1. Let $\bar{B} = \{x \in X : \|x\| \le 1\}$ be compact, and suppose that $X$ is not [[finite-dimensional|018W3V3TCT]]. Then $X$ has no finite spanning set. In particular $X \neq \{0\}$.
2. We build unit vectors $x_1, x_2, \ldots$ with $\|x_m - x_n\| \ge \tfrac{1}{2}$ for $m \neq n$. Choose any $x_1 \neq 0$ and replace it by $x_1 / \|x_1\|$.
3. Given $x_1, \ldots, x_n$, let $Y_n$ be their [[linear span|0HR573CHNM]]. It is a [[linear subspace|0EPTJA5QV4]], and $Y_n \neq X$ because $X$ has no finite spanning set.
4. $Y_n$ with the restricted norm is a finite-dimensional normed space. So it is complete, by [[finite-dimensional normed spaces are complete|0ABP2F2ZP7]].
5. Its metric is the restriction of the metric of $X$. So $Y_n$ is closed in $X$, by [[complete subspaces are closed|0W41XG4W60]].
6. By [[Riesz's lemma|097C7KCAT5]] with $\alpha = \tfrac{1}{2}$, there is $x_{n+1}$ with $\|x_{n+1}\| = 1$ and $\|x_{n+1} - y\| \ge \tfrac{1}{2}$ for all $y \in Y_n$.
7. For $m > n$, we have $x_n \in Y_{m-1}$. So $\|x_m - x_n\| \ge \tfrac{1}{2}$, by step 6 for $x_m$.
8. All $x_n$ lie in $\bar{B}$. As $\bar{B}$ is compact in the metric space $X$, it is [[sequentially compact|0CARW59WD2]], by [[this result|0K8SGQSV7G]].
9. So some subsequence $(x_{n_k})$ [[converges|06JYPJ4WHH]] to a point $z$. There is $k_0$ with $\|x_{n_k} - z\| < \tfrac{1}{4}$ for $k \ge k_0$.
10. Then $\|x_{n_{k_0}} - x_{n_{k_0 + 1}}\| \le \|x_{n_{k_0}} - z\| + \|z - x_{n_{k_0 + 1}}\| < \tfrac{1}{2}$, with $n_{k_0} \neq n_{k_0 + 1}$. This contradicts step 7.
11. So $X$ is finite-dimensional.

## Idea

In infinite dimension, each finite-dimensional subspace is a closed proper subspace. Riesz's lemma then gives a new unit vector far from all previous ones. The resulting sequence has no convergent subsequence, so the unit ball cannot be compact.