Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact

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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

Necessary Condition

Let $X$ be a finite dimensional vector space $\R^d$.

We have that all norms on finite-dimensional vector space are equivalent.

Choose Euclidean norm $\norm {\, \cdot \,}_2$.

Let $\struct {\R^d, \norm {\, \cdot \,}_2}$ be the normed finite-dimensional real vector space with Euclidean norm.

Let $\map {\Bbb S^{d - 1}_1} 0$ be a unit sphere in $\struct {\R^d, \norm {\, \cdot \,}_2}$.

By definition, $\Bbb S^{d - 1}$ is bounded.

Let $\mathbf L := \tuple {L_1, \dots, L_d} \in \R^d$.

Let $\mathbf x_n := \tuple {x_n^{\paren 1}, \dots, x_n^{\paren d}}$.

Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ be a sequence in $\Bbb S^{d - 1}$.

Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converge to $\mathbf L$:

$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \norm {\mathbf x_n - \mathbf L}_2 < \epsilon$

Furthermore, for all $k \in \N : k \le d$ we have that:

\(\ds \size {x_n^{\paren k} - L_k}\) \(=\) \(\ds \sqrt {\size {x_n^{\paren k} - L_k}^2 }\)
\(\ds \) \(\le\) \(\ds \sqrt{\sum_{j \mathop = 1}^d \size {x_n^{\paren j} - L_k}^2}\)
\(\ds \) \(=\) \(\ds \norm {\mathbf x_n - \mathbf L}_2\) Definition of Euclidean Norm

Thus:

$\forall k \in \N : k \le d : \forall \epsilon \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \size {x_n^{\paren k} - L_k} < \epsilon$

In other words, $\ds \lim_{n \mathop \to \infty} x_n^{\paren k} = L_k$.

We have that $\mathbf x_n \in \Bbb S^{d - 1}$.

Therefore:

$\ds \norm {\mathbf x_n}_2^2 = \sum_{j \mathop = 1}^d \paren {x_n^{\paren j}}^2 = 1$

$\map f x = x^2$ is a continuous function in its whole range.

By Limit of Composite Function, taking the limit $n \to \infty$ results in:

$\ds \norm {\mathbf L}_2^2 = \sum_{j \mathop = 1}^d \paren {L_j}^2 = 1$

Therefore, $\mathbf L \in \Bbb S^{d - 1}$.

Hence, $\Bbb S^{d - 1}$ is closed.

By Heine-Borel theorem, $\Bbb S^{d - 1}$ is compact.

$\Box$

Sufficient Condition

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.

$\blacksquare$