Compactness

Almost every existence argument in analysis follows the same three-step template:

Construct an approximate sequence of “almost-solutions” (minimizing sequences, Galerkin approximations, Euler polygonal approximations, maximizers of a Rayleigh quotient, etc.).
Extract a convergent subsequence. This is where compactness enters.
Pass to the limit. Show the limit actually solves the original problem.

In finite dimensions step 2 is free: Bolzano-Weierstrass guarantees that every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence. This is why eigenvalue decompositions, the SVD, the direct method of the calculus of variations, and Peano’s ODE existence theorem all “just work” in $\mathbb{R}^n$ . Each reduces to building an approximate sequence on a compact set and extracting a convergent subsequence. In infinite dimensions, closed bounded sets are no longer compact, and step 2 fails without additional structure.

Definition 1 (Compact, precompact, and totally bounded sets)

Let $(M, d)$ be a metric space and $A \subset M$ .

$A$ is compact if every open cover of $A$ has a finite subcover, or equivalently, if every sequence in $A$ has a subsequence converging to a point in $A$ .
$A$ is precompact (or relatively compact) if its closure $\overline{A}$ is compact, or equivalently, if every sequence in $A$ has a subsequence that is Cauchy.
$A$ is totally bounded if for every $\varepsilon > 0$ , $A$ can be covered by finitely many balls of radius $\varepsilon$ .

The key equivalence is:

$A$ is compact $\iff$ $A$ is complete and totally bounded.

In a complete metric space (such as a Banach space), precompactness and total boundedness coincide: $A$ is precompact if and only if it is totally bounded.

Why both conditions are needed.

Total boundedness without completeness fails. The open interval $(0,1)$ is totally bounded in $\mathbb{R}$ , but the sequence $1/n$ converges to $0 \notin (0,1)$ . The limit escapes through a “hole.”
Completeness without total boundedness fails. The unit ball of $\ell^2$ is complete, but the orthonormal sequence $(e_n)$ has all elements distance $\sqrt{2}$ apart, so no finite $\varepsilon$ -cover works for $\varepsilon < \sqrt{2}/2$ . There are “too many directions.”

Total boundedness is the condition that makes a set approximately finite-dimensional: at every resolution $\varepsilon$ , the set is indistinguishable from a finite set. Completeness then ensures that Cauchy subsequences (built by the pigeonhole principle at finer and finer scales) actually converge.

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.

Remark 2 (Compact sets in infinite dimensions)

Boundedness alone is not enough. The unit ball of $\ell^2$ is closed and bounded, yet the orthonormal sequence $(e_n)$ has every pair distance $\sqrt{2}$ apart, so no subsequence can be Cauchy. You need a constraint that suppresses this “too many directions” problem, i.e. you need total boundedness (Definition 1).

A concrete and important example: the unit ball of $H^1(\Omega)$ , viewed inside $L^2(\Omega)$ , is precompact (Rellich-Kondrachov). The reason is visible in Fourier space. An $H^1$ bound forces $|\hat{u}(k)|^2 \leq C/(1 + |k|^2)$ , so high-frequency modes are uniformly suppressed. For any $\varepsilon > 0$ , all the $L^2$ energy beyond a sufficiently large frequency $N$ is less than $\varepsilon$ , uniformly over the entire $H^1$ ball. The remaining low-frequency part lives in a finite-dimensional space. This is total boundedness: the derivative bound kills oscillations that would otherwise prevent clustering, and the bounded domain prevents mass from escaping to spatial infinity.

Definition 1 (Compact, precompact, and totally bounded sets)

Example 1 (Compactness of Unit Ball in Rn\mathbb{R}^nRn)

Example 2 (The Unit Ball in ℓ2\ell^2ℓ2 is Not Compact)

Remark 1 (Compactness in Infinite Dimensions)

Remark 2 (Compact sets in infinite dimensions)

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

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