Recall that any two norms on a finite-dimensional vector space are equivalent. If \(\|\cdot\|_1\) and \(\|\cdot\|_2\) are norms on \(V\), then there exist constants \(c,C>0\) such that

$$ c\|x\|_1\le\|x\|_2\le C\|x\|_1 \qquad\text{for every }x\in V. $$

We will use this fact to study compactness of the closed unit ball

$$ B=\{x\in V:\|x\|\le 1\}. $$
Definition — Compact Set
Let \((X,d)\) be a metric space. A set \(E\subseteq X\) is compact if every open cover of \(E\) has a finite subcover. Explicitly, whenever \(E\subseteq\bigcup_{\alpha\in A}U_\alpha\) with each \(U_\alpha\) open in \(X\), there exist finitely many indices \(\alpha_1,\ldots,\alpha_m\) such that \(E\subseteq U_{\alpha_1}\cup\cdots\cup U_{\alpha_m}\).


Theorem
Every compact subset of a metric space is closed and bounded.


Heine–Borel Theorem
A subset of \(\mathbb{R}^n\), equipped with its usual Euclidean metric, is compact if and only if it is closed and bounded. The same statement holds in \(\mathbb{C}^n\), identified with \(\mathbb{R}^{2n}\).

More generally, suppose a metric space has the property that every bounded sequence has a convergent subsequence with limit in the space. Then every closed bounded subset is compact. Indeed, a sequence in such a subset has a convergent subsequence, and closedness ensures that its limit remains in the subset. In metric spaces, this sequential compactness is equivalent to compactness.

The Closed Unit Ball in Finite Dimensions

Lemma
Let \(V\) be a finite-dimensional normed space over \(\mathbb{R}\) or \(\mathbb{C}\). Its closed unit ball \(B=\{x\in V:\|x\|\le 1\}\) is compact.

Proof

If \(V=\{0\}\), its closed unit ball is the singleton \(\{0\}\), which is compact. Otherwise, choose a basis \(\{e_1,\ldots,e_n\}\) of \(V\). Every vector has a unique expression

$$ x=\sum_{i=1}^{n}x_i e_i. $$

1. Construct a countable dense subset.

For a complex vector space, define

$$ D=\left\{\sum_{i=1}^{n}q_i e_i:q_i\in\mathbb{Q}+i\mathbb{Q}\right\}. $$

For a real vector space, take \(q_i\in\mathbb{Q}\) instead. These sets are countable because a finite product of countable sets is countable.

To see density, fix \(x=\sum_{i=1}^{n}x_i e_i\) and \(\epsilon>0\). Approximate each coefficient \(x_i\) by a coefficient \(q_i\) from the appropriate countable set so that

$$ |x_i-q_i|\lt \frac{\epsilon}{n\|e_i\|} \qquad\text{for }i=1,\ldots,n. $$

Then \(q=\sum_{i=1}^{n}q_i e_i\) belongs to \(D\), and

$$ \|x-q\| \le\sum_{i=1}^{n}|x_i-q_i|\,\|e_i\| \lt\epsilon. $$

Thus \(D\) is dense in \(V\), so \(V\) is separable.

2. Show that bounded sequences have convergent subsequences.

Let \((v_j)\) be a bounded sequence in \(V\). Write

$$ v_j=\sum_{i=1}^{n}x_{j,i}e_i. $$

We first justify why each sequence of coordinates is bounded. The function

$$ \left\|\sum_{i=1}^{n}x_i e_i\right\|_{\mathrm{coord}} =\max_{1\le i\le n}|x_i| $$

is a norm on \(V\). By equivalence of norms, there exists \(K>0\) such that

$$ \max_{1\le i\le n}|x_{j,i}| \le K\|v_j\|. $$

Since \((v_j)\) is bounded, the right-hand side is bounded independently of \(j\). Hence every coordinate sequence \((x_{j,i})_{j=1}^{\infty}\) is bounded.

Apply the Bolzano–Weierstrass theorem to the first coordinate sequence to obtain a subsequence along which the first coordinate converges. From that subsequence, choose a further subsequence along which the second coordinate converges. Continue through all \(n\) coordinates.

There are only finitely many coordinates, so this process produces a subsequence \((v_{j_k})\) and scalars \(x_1,\ldots,x_n\) such that

$$ x_{j_k,i}\to x_i \qquad\text{for every }i=1,\ldots,n. $$

Set \(v=\sum_{i=1}^{n}x_i e_i\). By the triangle inequality,

$$ \begin{aligned} \|v_{j_k}-v\| &=\left\|\sum_{i=1}^{n}(x_{j_k,i}-x_i)e_i\right\|\\ &\le\sum_{i=1}^{n}|x_{j_k,i}-x_i|\,\|e_i\| \longrightarrow 0. \end{aligned} $$

Thus every bounded sequence in \(V\) has a convergent subsequence.

3. Apply this to the closed unit ball.

Let \((v_j)\) be any sequence in \(B\). Since \(\|v_j\|\le 1\), the sequence is bounded. By the preceding argument, some subsequence \((v_{j_k})\) converges to a vector \(v\in V\).

The reverse triangle inequality gives

$$ \bigl|\|v_{j_k}\|-\|v\|\bigr| \le\|v_{j_k}-v\|\to 0. $$

Therefore \(\|v\|=\lim_{k\to\infty}\|v_{j_k}\|\le 1\), so \(v\in B\). Every sequence in \(B\) has a subsequence converging to an element of \(B\). Hence \(B\) is sequentially compact, and therefore compact.

We have proved that finite-dimensionality implies compactness of the closed unit ball. The next theorem establishes the converse.

Theorem
Let \((V,\|\cdot\|)\) be a Banach space. Then \(V\) is finite-dimensional if and only if its closed unit ball \(B=\{v\in V:\|v\|\le 1\}\) is compact.

Example 1: The Discrete Metric

Consider \(\mathbb{R}\) equipped with the discrete metric

$$ d(x,y)= \begin{cases} 0, & x=y,\\ 1, & x\ne y. \end{cases} $$

The whole space \(\mathbb{R}\) is closed and bounded in this metric. Indeed, every pair of points has distance at most \(1\). For every \(x\in\mathbb{R}\), the singleton \(\{x\}\) is open because \(B_d(x,1/2)=\{x\}\). Thus

$$ \mathbb{R}=\bigcup_{x\in\mathbb{R}}\{x\} $$

is an open cover. Any finite collection of these singletons contains only finitely many points, so it cannot cover \(\mathbb{R}\). Therefore this open cover has no finite subcover, and the space is not compact.

Example 2: The Closed Unit Ball of \(\ell^1(\mathbb{N})\)

Recall that

$$ \ell^1(\mathbb{N}) =\left\{(x_n)_{n=1}^{\infty}: x_n\in\mathbb{C},\ \sum_{n=1}^{\infty}|x_n|\lt\infty\right\}, \qquad \|(x_n)\|_1=\sum_{n=1}^{\infty}|x_n|. $$

Its closed unit ball is

$$ B=\left\{(x_n)\in\ell^1(\mathbb{N}): \sum_{n=1}^{\infty}|x_n|\le 1\right\}. $$

For each positive integer \(n\), let \(e_n\) be the sequence with a \(1\) in the \(n\)th coordinate and zeros elsewhere. Then \(\|e_n\|_1=1\), so every \(e_n\) belongs to \(B\).

For distinct indices \(n\) and \(m\), the sequence \(e_n-e_m\) has one coordinate equal to \(1\) and another equal to \(-1\). Hence

$$ \|e_n-e_m\|_1=|1|+|-1|=2. $$

Cover \(B\) by open balls of radius \(1/100\) centered at its points:

$$ B\subseteq\bigcup_{x\in B}B_{\ell^1}\left(x,\frac{1}{100}\right). $$

Suppose \(B\) were compact. Then finitely many of these balls would cover it, so there would exist \(x_1,\ldots,x_M\in B\) such that

$$ B\subseteq\bigcup_{i=1}^{M} B_{\ell^1}\left(x_i,\frac{1}{100}\right). $$

Every \(e_n\) must lie in one of these finitely many balls. By the pigeonhole principle, two distinct vectors \(e_n\) and \(e_m\) lie in the same ball, say the ball centered at \(x_i\). The triangle inequality then gives

$$ \begin{aligned} 2 &=\|e_n-e_m\|_1\\ &\le\|e_n-x_i\|_1+\|x_i-e_m\|_1\\ &\lt\frac{1}{100}+\frac{1}{100} =\frac{2}{100}. \end{aligned} $$

This is a contradiction. Thus the closed unit ball of \(\ell^1(\mathbb{N})\) is not compact.

Riesz’s Lemma

To extend the preceding argument to an arbitrary infinite-dimensional Banach space, we need a way to construct unit vectors that stay a fixed distance apart.

Riesz’s Lemma
Let \(V\) be a normed space, and let \(Y\subsetneq Z\subseteq V\) be linear subspaces with \(Y\) closed in \(Z\). For every \(\theta\in(0,1)\), there exists \(v\in Z\) such that \(\|v\|=1\) and \(\displaystyle\operatorname{dist}(v,Y) =\inf_{y\in Y}\|v-y\|\ge\theta\).

The distance condition means that \(\|v-y\|\ge\theta\) for every \(y\in Y\). Thus the lemma produces a unit vector that stays at least \(\theta\) away from the entire subspace.

Remark — The Hilbert Space Case

If \(V\) is a Hilbert space and \(Y\) is a closed proper subspace of \(Z\), we can take \(\theta=1\). Choose \(z\in Z\setminus Y\) and let \(P_Yz\) be its orthogonal projection onto \(Y\). Then \(w=z-P_Yz\) is a nonzero vector in \(Z\) orthogonal to \(Y\). Normalize it by setting \(v=w/\|w\|\). For every \(y\in Y\),

$$ \|v-y\|^2=\|v\|^2+\|y\|^2=1+\|y\|^2\ge 1. $$

Taking \(y=0\) shows that \(\operatorname{dist}(v,Y)=1\).

Proof That Compactness Forces Finite Dimension

Assume Riesz’s lemma. Suppose \(V\) is infinite-dimensional. We will construct a sequence of unit vectors satisfying

$$ \|e_n-e_m\|\ge\frac12 \qquad\text{whenever }n\ne m. $$

Choose any unit vector \(e_1\in V\) and let \(Y_1=\operatorname{span}\{e_1\}\). This subspace is finite-dimensional, so it is closed. Since \(V\) is infinite-dimensional, \(Y_1\) is a proper subspace of \(V\).

Apply Riesz’s lemma with \(Y=Y_1\), \(Z=V\), and \(\theta=1/2\). It gives a unit vector \(e_2\) such that

$$ \inf_{y\in Y_1}\|e_2-y\|\ge\frac12. $$

In particular, \(e_1\in Y_1\), so \(\|e_2-e_1\|\ge 1/2\).

Now let \(Y_2=\operatorname{span}\{e_1,e_2\}\). This is again a closed proper subspace of \(V\). Applying Riesz’s lemma gives a unit vector \(e_3\) whose distance from every vector in \(Y_2\) is at least \(1/2\). Since \(e_1,e_2\in Y_2\), we obtain

$$ \|e_3-e_1\|\ge\frac12, \qquad \|e_3-e_2\|\ge\frac12. $$

Continue inductively. Once \(e_1,\ldots,e_n\) have been chosen, the subspace

$$ Y_n=\operatorname{span}\{e_1,\ldots,e_n\} $$

is finite-dimensional, closed, and proper. Riesz’s lemma supplies a unit vector \(e_{n+1}\) such that

$$ \inf_{y\in Y_n}\|e_{n+1}-y\|\ge\frac12. $$

Every earlier vector belongs to \(Y_n\), so the resulting sequence satisfies \(\|e_n-e_m\|\ge 1/2\) for all distinct indices.

Suppose the closed unit ball \(B\) were compact. Cover it by all open balls of radius \(1/100\) centered at points of \(B\). Compactness gives a finite subcover

$$ B\subseteq\bigcup_{i=1}^{M}B_V\left(x_i,\frac{1}{100}\right). $$

Every \(e_n\) lies in \(B\). By the pigeonhole principle, two distinct vectors \(e_n\) and \(e_m\) lie in the same ball. Thus

$$ \frac12 \le\|e_n-e_m\| \le\|e_n-x_i\|+\|x_i-e_m\| \lt\frac{2}{100}, $$

which is impossible. Hence the closed unit ball of an infinite-dimensional Banach space is not compact.

Equivalently, if the closed unit ball is compact, then the space is finite-dimensional.

References

  • SFSU Fourier Analysis by Professor Chun-Kit Lai
  • Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi