Closed Unit Ball Criterion for Finite Dimensional Normed Linear Spaces

Recall from the Riesz's Lemma page that if $(X, \| \cdot \|)$ is a normed linear space and $Y \subset X$ is a proper and closed subspace of $X$ then for all $0 < \epsilon < 1$ there exists an $x_0 \in X$ such that for every $y \in Y$:

(1)

\begin{align} \quad \| x_0 - y \| \geq 1 - \epsilon \end{align}

We will use Riesz's lemma to show that if $(X, \| \cdot \|)$ is a normed linear space then the closed unit ball in $X$ is compact if and only if $X$ is finite dimensional.

Theorem 1 (Compact Unit Ball Criterion for Finite Dimensional Normed Linear Spaces): Let $(X, \| \cdot \|)$ be a normed linear space. Then the closed unit ball in $X$ is compact if and only if $X$ is finite-dimensional.

• Proof: $\Rightarrow$ We show that contrapositive of this direction - that is, if $X$ is infinite-dimensional then the closed unit ball in $X$ is not compact.

• Suppose that $X$ is infinite-dimensional normed linear space. We will find a sequence $(x_n)$ in $X$ on the unit ball in $X$ such that $\| x_m - x_n \| \geq \frac{1}{2}$ for all $m, n \in \mathbb{N}$ with $m \neq n$.

• Let $x_1 \in \bar{B}(0, 1)$. Let $Y_1 = \mathrm{span} (x_1)$. Then $Y_1$ is a a proper subspace of $X$ since $X$ is infinite-dimensional. Also, $Y_1$ is closed since every finite-dimensional subspace of a vector space is closed. By Riesz' lemma for $\epsilon = \frac{1}{2}$ there exists an $x_2 \in X$ with $\| x_2 \| = 1$ such that $\| x_2 - x_1 \| \geq \frac{1}{2}$. Since $\| x_2 \| = 1$ we have that $x_2 \in \bar{B}(0, 1)$.

• We continue this process. Suppose that $x_1, x_2, ..., x_n \in \bar{B}(0, 1)$ are obtained as above. Let $Y_n = \mathrm{span} (x_1, x_2, ..., x_n)$. Then $Y_n$ is a proper subspace of $X$ since $X$ is infinite-dimensional. Also, $Y_n$ is closed since every finite-dimensional subspace of a vector space is closed. So by Riesz' lemma for $\epsilon = \frac{1}{2}$ there exists an $x_{n+1} \in X$ with $\| x_{n+1} \| = 1$ such that $\| x_{n+1} - x_k \| \geq \frac{1}{2}$ for all $k \in \{ 1, 2, ..., n \}$. Since $\| x_{n+1} \| = 1 $] ]we have that [[$ x_{n+1} \in \bar{B}(0, 1)$.

• So $(x_n)$ is a sequence in $\bar{B}(0, 1)$ such that $\| x_m - x_n \| \geq \frac{1}{2}$ for all $m, n \in \mathbb{N}$ with $m \neq n$. But then the sequence $(x_n)$ has no Cauchy subsequence. This implies that $\bar{B}(0, 1)$ is not compact.

• $\Leftarrow$ Suppose that $X$ is finite-dimensional. Then $X$ is isomorphic to $\mathbb{R}^n$. Since the closed unit ball in $\mathbb{R}^n$ is compact, so is the closed unit ball in $X$. $\blacksquare$