# Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact/Sufficient Condition

## Theorem

Let $X$ be a normed vector space.

Let $\Bbb S = \map {\Bbb S_1} 0$ be the unit sphere centred at $0$ in $X$.

Then $X$ is finite dimensional if and only if $\Bbb S$ is compact.

## Proof

Suppose that $X$ is not finite dimensional.

Then any finite dimensional subspace of $X$ is proper.

From Compact Subspace of Metric Space is Sequentially Compact in Itself, it suffices to show that $\Bbb S$ is not sequentially compact.

For this, it suffices to show that there exists an $\epsilon > 0$ and sequence $\sequence {x_n}_{n \mathop \in \N}$ with $\norm {x_n} = 1$ for each $n$ and:

- $\norm {x_i - x_j} \ge \epsilon$ for each $i \ne j$.

Then no subsequence of $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy.

We construct $\sequence {x_n}_{n \mathop \in \N}$ iteratively.

From Convergent Sequence is Cauchy Sequence, we will have that no subsequence of $\sequence {x_n}_{n \mathop \in \N}$ can converge.

Take $x_1 \in X$ with $\norm {x_1} = 1$.

Let:

- $Y_1 = \span \set {x_1}$

From Finite Dimensional Subspace of Normed Vector Space is Closed, $Y_1$ is closed.

From Riesz's Lemma, there exists $x_2 \in X$ with $\norm {x_2} = 1$ such that:

- $\ds \norm {x_2 - x_1} > \frac 1 2$

Now suppose that we have constructed $x_1, x_2, \ldots, x_n$ such that:

- $\ds \norm {x_i - x_j} > \frac 1 2$

for all $i \ne j$.

Let:

- $Y_n = \span \set {x_1, x_2, \ldots, x_n}$

From Finite Dimensional Subspace of Normed Vector Space is Closed, $Y_n$ is closed.

From Riesz's Lemma, there exists $x_{n + 1} \in X$ with $\norm {x_{n + 1} } = 1$ such that:

- $\ds \norm {x_{n + 1} - y} > \frac 1 2$

for all $y \in Y_n$.

In particular:

- $\ds \norm {x_{n + 1} - x_k} > \frac 1 2$

for all $1 \le k \le n$.

From Norm Axiom $\text N 2$: Positive Homogeneity, we have:

- $\ds \norm {x_i - x_j} > \frac 1 2$

for each $1 \le i, j \le {n + 1}$ with $i \ne j$.

Continuing this way, we have a sequence $\sequence {x_n}_{n \mathop \in \N}$ such that $\norm {x_n} = 1$ for each $n \in \N$ and:

- $\ds \norm {x_i - x_j} > \frac 1 2$

for each $i \ne j$.

So for any subsequence $\sequence {x_{n_j} }_{j \mathop \in \N}$ of $\sequence {x_n}_{n \mathop \in \N}$ we have:

- $\ds \norm {x_{n_j} - x_{n_k} } > \frac 1 2$

for each $j \ne k$.

So no subsequence $\sequence {x_{n_j} }_{j \mathop \in \N}$ can be Cauchy, and hence none can converge.

$\blacksquare$