Compactness and Topological Equivalence

We now turn to two fundamental topological properties of Banach spaces that distinguish finite-dimensional spaces from infinite-dimensional ones: compactness of bounded sets and equivalence of norms.

Compactness¶

Definition 1 (Compact Set)

A subset $E \subset X$ is compact if either:

each open cover contains a finite subcover.
each sequence $\{x_n\} \subset E$ contains a convergent subsequence.

In finite dimensions, compactness is characterized by the Heine-Borel theorem: a set is compact if and only if it is closed and bounded.

Example 1 (Compactness of Unit Ball in $\mathbb{R}^n$ )

The closed unit ball $B_1(0) = \{x \in \mathbb{R}^n : \|x\|_p \leq 1\}$

is compact by Heine-Borel. Equivalently, we can take any sequence $\{x_n\} \subset B_1(0)$ . Since that sequence is bounded Bolzano-Weierstrass implies that it has a convergent subsequence.

This picture changes dramatically in infinite dimensions.

Example 2 (The Unit Ball in $\ell^2$ is Not Compact)

Consider the Hilbert space $\ell^2$ of square-summable sequences with the standard orthonormal basis $\{e_n\}_{n=1}^\infty$ where $e_n = (0, \ldots, 0, 1, 0, \ldots)$ has a 1 in position $n$ and zeros elsewhere.

Each $e_n$ lies in the closed unit ball since $\|e_n\|_2 = 1$ . However, for any $n \neq m$ :

\|e_n - e_m\|_2 = \sqrt{\|e_n\|_2^2 + \|e_m\|_2^2} = \sqrt{2}

(1)

Every pair of distinct elements is at distance $\sqrt{2}$ apart. Therefore no subsequence of $\{e_n\}$ can be Cauchy, and hence no subsequence converges. The closed unit ball in $\ell^2$ is not compact.

Remark 1 (Compactness in Infinite Dimensions)

The failure of the Heine-Borel theorem in infinite dimensions is one of the most important distinctions between finite- and infinite-dimensional analysis. Closed and bounded sets need not be compact, and establishing compactness requires additional structure (e.g. equicontinuity in the Arzelà-Ascoli theorem, or tightness in probability theory). For this reason, compactness arguments play a crucial and often delicate role throughout functional analysis.

Equivalence of Norms¶

Definition 2 (Equivalent Norms)

Two norms $\|x\|^1$ and $\|x\|^2$ are equivalent if there exists $a, b > 0$ such that $a \|x\|^1 \leq \|x\|^2 \leq b \|x\|^1\quad \forall x \in X$

Note that the previous definition calls two norms equivalent when they induce the same underlying topology on $X$ . In other words, they generate the same open sets or intuitively induce the same notion of two elements being close.

Geometric intuition. The condition $a\|x\|^1 \leq \|x\|^2 \leq b\|x\|^1$ says that the unit ball of one norm can be scaled to fit inside the unit ball of the other, and vice versa. In $\mathbb{R}^2$ , for example, the $\ell^1$ ball (a diamond), the $\ell^2$ ball (a circle), and the $\ell^\infty$ ball (a square) all have different shapes — but each one can be scaled up or down to contain, or be contained in, any of the others. This mutual containment of balls is exactly what equivalent topologies means: the same sequences converge, the same sets are open, and the same functions are continuous.

Theorem 1 (Equivalence of Norms in $\mathbb{R}^n$ )

In $\mathbb{R}^n$ all norms are equivalent.

Proof 1

This is merely a sketch of a proof. The key idea is that in $\mathbb{R}^n$ all the unit balls can be scaled so that they fit into each other. Allowing us to show that the open sets in $(\mathbb{R}^n, \|x\|_p)$ are the same as in $(\mathbb{R}^n, \|x\|_q)$ .

Remark 2 (Failure in Infinite Dimensions)

This theorem fails in infinite dimensions. For instance, on $\mathcal{C}^0([0,1])$ the norms $\|\cdot\|_\infty$ and $\|\cdot\|_1$ are not equivalent: a sequence of functions can converge in $L^1$ without converging uniformly. The choice of norm — and hence the topology — matters fundamentally in infinite-dimensional analysis and determines which operators are bounded, which functionals are continuous, and which sequences converge.