# A Normed Linear Space is Finite-Dimensional If and Only If the Closed Unit Ball is Compact

Recall from the Riesz's Lemma page that if $X$ is a normed linear space and $Y$ is a proper closed linear subspace of $X$ then for all $\epsilon$ such that $0 < \epsilon < 1$ there exists an $x_0 \in X$ with $\| x _0 \| = 1$ such that:

(1)We use Riesz' lemma to prove a very important result in classifying finite-dimensional normed linear spaces which says that a normed linear space is finite-dimensional if and only if the closed unit ball $\overline{B} (0, 1)$ in $X$ is compact.

Theorem 1: Let $X$ be a normed linear space. Then $X$ is finite-dimensional if and only if $\overline{B}(0, 1)$ is compact in $X$. |

*Here the notation $\overline{B}(0, 1)$ denotes the closed unit ball in $X$, that is:*

*Recall that a subset $S \subseteq X$ of a metric space $X$ is compact if and only if every open cover of $S$ has a finite open subcover. Another classification of compactness is that $S \subseteq X$ is compact if and only if every sequence in $S$ has a subsequence that converges in $S$. We use this second characterization in the proof below.*

**Proof:**$\Rightarrow$ Let $X$ be an infinite-dimensional normed linear space. We will construct a sequence in $\overline{B}(0, 1)$ that has no subsequence that converges in $\overline{B}(0, 1)$.

- Let $x_1 \in \overline{B}(0, 1)$ and let $X_1 = \mathrm{span} (x_1)$. Since $X$ is an infinite-dimensional normed linear space we have that $X_1 \neq X$. Since $X_1$ is a finite-dimensional subspace of a normed linear space $X$ we have that $X_1$ is closed. So $X_1$ is a proper closed linear subspace of $X$. By Riesz's lemma, for $\epsilon = \frac{1}{2}$ there exists an $x_2 \in \overline{B}(0, 1)$ (since $\| x_2 \| = 1$) such that:

- We continue on in this manner. For $x_1, x_2, ..., x_n \in \overline{B}(0, 1)$ we let $X_n = \mathrm{span} (x_1, x_2, ..., x_n)$. Once again, since $X$ is an infinite-dimensional normed linear space we have that $X_n \neq X$. Since $X_n$ is a finite-dimensional subspace of a normed linear space $X$ we have that $X_n$ is closed. So $X_n$ is a proper closed linear subspace of $X$. By Riesz's lemma, for $\epsilon = \frac{1}{2}$ there exists an $x_{n+1} \in \overline{B}(0, 1)$ (since $\| x_{n+1} \| = 1$) such that:

- Furthermore, since $X_1 \subseteq X_2 \subseteq ... \subset X_{n+1}$, for all $1 \leq j < k \leq n+1$ we have that:

- This shows that the sequence $(x_n)_{n=1}^{\infty}$ in $\overline{B}(0, 1)$ is not Cauchy. So every subsequence of $(x_n)_{n=1}^{\infty}$ is not Cauchy. Hence there is no subsequence of $(x_n)_{n=1}^{\infty}$ that converges in $\overline{B}(0, 1)$. So $\overline{B}(0, 1)$ is not compact.

- $\Leftarrow$ Let $X$ be a finite-dimensional normed linear space with $\mathrm{dim} (X) = n$. Let $T : X \to \mathbb{C}^n$ be an isomorphism from $X$ to $\mathbb{C}^n$.

- For all $x \in \overline{B}(0, 1)$ we have that $\| x \| \leq 1$ and so:

- This shows that $T(\overline{B}(0, 1))$ is bounded. Now since $T$ is an isomorphism, $T$ and $T^{-1}$ are bounded linear operator and hence $T$ and $T^{-1}$ are continuous. So $T(\overline{B}(0, 1))$ is closed. But every closed and bounded subset of $\mathbb{C}^n$ is compact. So $T(\overline{B}(0, 1))$ is compact. Since $T^{-1}$ is continuous, the inverse image of any compact set in $Y$ is compact in $X$. So $\overline{B}(0, 1) = T^{-1}(T(\overline{B}(0, 1)))$ is compact in $X$. $\blacksquare$