Compactness

Completeness (2B.2 Completeness and Contraction) produces solutions as limits of sequences that we construct, step by step, to be Cauchy. Compactness produces solutions in a different way. It takes any sequence, possibly one that wanders without settling, and extracts from it a subsequence that converges. We don't need to know in advance where the limit is, or build the sequence carefully. We only need the sequence to live in a compact space.

On the real line, compactness was the Bolzano–Weierstrass theorem (2A.6 Sequences), and it gave the maximum principle and uniform continuity (2A.9 Continuous Functions). This chapter makes it a property of metric spaces, proves that three quite different definitions agree, and puts it to work. It then states in full the proof pattern that the guidebook follows all the way to Perelman, the contradiction–compactness template of 2A.5 Quantifiers and the Shape of a Proof, and works through it on a model case where every step is visible. Finally it lists the three ways in which compactness can fail. Those three failures, in increasingly sophisticated forms, are the ways a Ricci flow can develop a singularity.

By the end of this chapter you will be able to:

• prove that a set is compact or not, using sequences, total boundedness or open covers, whichever is easiest;
• use compactness to show that a maximum exists, that a continuous function is uniformly continuous, and that a positive quantity has a positive lower bound;
• run the contradiction–compactness template: negate, normalise, extract a limit, contradict;
• prove that all norms on $\mathbb{R}^n$ are equivalent, and say why the proof fails in infinite dimensions;
• recognise the three ways a sequence can fail to have a convergent subsequence.

3.1A soap film that snaps

A finer point, which a real experiment shows: for separations between about $1.055$ and $1.3255$, two flat discs have less total area than the fat catenoid, yet the film stays a catenoid until $1.3255$. The catenoid there is a local minimum of area but not the global one. Which local minimum a physical system sits in depends on its history, a theme that returns with the calculus of variations (6A.9 Calculus of Variations and Gradient Flows) and with min–max methods (10A.8 Min–Max and Width).

3.2Sequential compactness

Tao takes this sequential definition as primary, as we do. Its first consequences are quick.

In $\mathbb{R}^n$ the converse holds.

In general metric spaces "closed and bounded" is not enough.

Both examples have something in common: points that stay a fixed distance apart, with no way of being grouped into finitely many small clusters. That observation is the content of the next characterisation.

3.3Complete and totally bounded

Totally bounded means "finite at every resolution". However fine a resolution $\varepsilon$ you choose, finitely many probes suffice to see every point to within $\varepsilon$. In $\mathbb{R}^n$, a bounded set is totally bounded: put it in a cube and use the points of a fine grid. In the discrete example, it fails at $\varepsilon = \tfrac12$, since each ball of radius $\tfrac12$ contains a single point.

The diagonal argument in the last step, "refine infinitely often, then take the diagonal", is a technique to remember. It reappears in the proof of the Arzelà–Ascoli theorem (2B.5 Uniform Convergence and Arzelà–Ascoli) and in every compactness theorem built on it.

3.4Open covers

There is a third characterisation, which looks nothing like the other two. It is the one that generalises to spaces without a metric (7A.2 Compactness and Compactification), and it is often the most convenient for proofs that patch local information together.

The proof uses a lemma that is worth knowing on its own.

3.5Continuous functions on compact spaces

Continuity between metric spaces is defined as on $\mathbb{R}$: $f : X \to Y$ is continuous if $x_n \to x$ implies $f(x_n) \to f(x)$, or equivalently if for every $x$ and $\varepsilon > 0$ there is $\delta > 0$ with $d_Y(f(x), f(x')) < \varepsilon$ whenever $d_X(x, x') < \delta$. Equivalently again, preimages of open sets are open (Exercise 3.14).

Part 3 is Heine–Cantor (2A.9 Continuous Functions) in general, with the same proof. Part 2 is the maximum principle of 2A.9 Continuous Functions, now valid for continuous functions on any compact space: on a closed square, on a sphere, on the space of positions of a mechanical linkage, or on a compact manifold, which is the setting of Hamilton's maximum principle (11A.4 Maximum Principles under Ricci Flow).

A fourth consequence is that a continuous bijection from a compact space has a continuous inverse (Exercise 3.15). This is the reason compact manifolds can be recognised by building a continuous bijection, without checking the inverse separately.

3.6The contradiction–compactness template, in full

In 2A.5 Quantifiers and the Shape of a Proof the contradiction–compactness template was stated in words, and 2A.9 Continuous Functions used it once, to show that a continuous positive function on $[a, b]$ has a positive lower bound. Here is the full template, and a case with all its steps, including the one that the earlier example did not need: normalisation.

To prove a uniform estimate:

1. Suppose it fails. Negate the quantifiers to get, for each $n$, a counterexample that is "$n$ times worse".
2. Normalise. Rescale each counterexample so that it lies in a fixed compact set, without changing the failure.
3. Extract a limit. By compactness, a subsequence converges.
4. Pass to the limit. Use continuity to find what the limit satisfies.
5. Contradict. Show that no such limit can exist.

A norm on $\mathbb{R}^n$ is a function $N : \mathbb{R}^n \to [0, \infty)$ with $N(x) = 0$ only for $x = 0$, $N(\lambda x) = |\lambda|\,N(x)$, and $N(x + y) \leq N(x) + N(y)$. Each norm gives a metric, $d(x, y) = N(x - y)$, and the three metrics of 2B.1 Metric Spaces come from the norms $\|x\|_1$, $\|x\|_2$ and $\|x\|_\infty$. We saw there that those three are uniformly equivalent. In fact all norms are.

Every step was needed. Without normalisation, the $x_k$ could run off to infinity or shrink to $0$, and no compactness theorem would apply. Without compactness of the sphere, there would be no limit. Without continuity, the limit wouldn't inherit the smallness of $N(u_k)$. The pattern of the proof is drawn in Figure 3.4.

Why it fails in infinite dimensions. On $C([0, 1])$, the sup norm $\|f\|_\infty$ and the area norm $\|f\|_1 = \int_0^1|f|$ are both norms, and they are not equivalent (2B.1 Metric Spaces). Run the template and see where it breaks. Counterexamples are the tents $f_n$, with $\|f_n\|_1 = \tfrac1n$ and $\|f_n\|_\infty = 1$, already normalised. Step 3 needs a convergent subsequence on the unit sphere $\{\|f\|_\infty = 1\}$, and there isn't one: the sup-norm unit sphere of $C([0, 1])$ is not compact. That is the whole difference between finite and infinite dimensions. A large part of functional analysis, from weak compactness to the Rellich theorem (4A.6 Weak Convergence and the Direct Method, 4A.10 Sobolev Embeddings and Critical Exponents), is about finding substitutes for step 3 when the unit sphere is not compact.

3.7Three ways to lose compactness

When a sequence has no convergent subsequence, it is worth asking how it escapes. In practice there are three ways, and it pays to recognise each on sight.

1. Escape to infinity. In $\mathbb{R}$, the sequence $x_n = n$ leaves every bounded set. The space is complete, but not bounded.
2. Approaching a missing point. In $(0, 1]$, the sequence $\tfrac1n$ is Cauchy and heads for $0$, which isn't there. The space is bounded, but not complete.
3. Oscillation in infinitely many directions. In $C([0, 1])$, take bumps $g_n$ of height $1$ with disjoint supports, say $g_n$ supported on $[2^{-n}, 2^{1-n}]$. Then $d_\infty(g_m, g_n) = 1$ for $m \neq n$: the sequence is bounded and lives in a complete space, but its terms point in infinitely many independent directions, and no two are close. The space is complete and bounded, but not totally bounded.

The three modes match the three conditions of Theorem 3.6 (bounded, complete, totally bounded), and in later books they reappear in disguise:

• In 2A.11 The Riemann Integral and 3A.3 The Lebesgue Integral, area under a sequence of functions escapes by moving to infinity, by concentrating at a point, or by spreading thin. Concentration is a "missing point" phenomenon in a function space.
• In Sobolev spaces (4A.10 Sobolev Embeddings and Critical Exponents), compactness of embeddings fails exactly through translation (escape) and through concentration under rescaling.
• In Riemannian geometry (9B.3 Collapsing and Noncollapsing), a sequence of manifolds with bounded curvature can fail to converge because points run off to infinity, because the curvature blows up somewhere (a singularity), or because the manifolds collapse: they become thinner and thinner in some direction, like a cylinder whose circumference shrinks to zero. Perelman's κ-noncollapsing theorem (12A.4 κ-Noncollapsing) rules out the third mode for Ricci flow, and it is the step that makes his compactness arguments work.

3.8History

Bernard Bolzano (1817) and Karl Weierstrass (in lectures from the 1860s) used the principle that bounded sequences of reals have convergent subsequences. Eduard Heine's 1872 proof of uniform continuity contained the covering idea implicitly; Émile Borel proved in 1895 that a countable cover of a closed bounded interval by open intervals has a finite subcover, and Henri Lebesgue extended it to arbitrary covers. The word "compact" is Maurice Fréchet's, from his 1906 thesis, the same work that introduced metric spaces (2B.1 Metric Spaces). The contradiction–compactness method has no single inventor. It is the standard way analysts turn qualitative compactness into quantitative estimates, and it runs through geometric analysis from Gromov's work in the 1980s to Perelman's in the 2000s.

3.9Exercises

Solution

(a) Yes: closed (a preimage of $(-\infty, 1]$ under a continuous function) and bounded ($|x|, |y| \leq 1$). (b) Yes: every sequence either takes some value infinitely often or has terms $\tfrac1{n_k}$ with $n_k \to \infty$, converging to $0$. (c) No: $\tfrac1n \to 0$, which is missing. (d) No: not closed in $\mathbb{R}$ (a sequence of rationals can converge to $1/\sqrt2$). (e) Yes: the great-circle metric is equivalent to the chordal metric from $\mathbb{R}^3$ (Exercise 3.13), and the sphere is closed and bounded in $\mathbb{R}^3$. (f) No: Example 3.4.

Solution

$\|x - y\|^2 = 2 - 2x\cdot y = 2 - 2\cos\theta = 4\sin^2(\theta/2)$. For $0 \leq s \leq \pi/2$, $\frac{2}{\pi}s \leq \sin s \leq s$ (the sine is concave there), and apply with $s = \theta/2$.

Solution

A closed $F \subseteq X$ is compact; $f(F)$ is compact, hence closed in $Y$. So $(f^{-1})^{-1}(F) = f(F)$ is closed for every closed $F$, which (taking complements) is continuity of $f^{-1}$ by Exercise 3.14. For the circle: points $(\cos t, \sin t)$ with $t$ slightly less than $2\pi$ are close to $(1, 0)$, but their preimages are near $2\pi$, far from $0 = f^{-1}(1, 0)$. Here $[0, 2\pi)$ is not compact.

Solution

If not, there are $x_n$ with $g(x_n) - f(x_n) < \tfrac1n$. A subsequence converges to some $x \in K$, and by continuity $g(x) - f(x) \leq 0$, a contradiction. (No normalisation is needed here.) Counterexample: $f(x) = 0$ and $g(x) = x$ on $(0, 1]$.

Solution

For fixed $m$ and $n_k \geq m$, $f_m(x_{n_k}) \geq f_{n_k}(x_{n_k}) \geq \varepsilon$. Let $k \to \infty$: by continuity of $f_m$, $f_m(x) \geq \varepsilon$. This holds for all $m$, contradicting $f_m(x) \to 0$. On $[0, 1)$, $x^n$ decreases to $0$ pointwise but $\sup_{[0,1)} x^n = 1$ for every $n$.

Hint

Upper bound: a square of side $\varepsilon\sqrt2$ lies in the ball of radius $\varepsilon$ around its centre. Lower bound: compare areas.

Solution

If not, there are $x_k \in K$ and $v_k \neq 0$ with $v_k^{\mathsf T}A(x_k)v_k < \tfrac1k\|v_k\|^2$. Normalise $u_k = v_k/\|v_k\|$: $u_k^{\mathsf T}A(x_k)u_k < \tfrac1k$. The product $K \times S^{n-1}$ is compact (a sequence has a subsequence converging in the first coordinate, and a further subsequence converging in the second), so $(x_{k_j}, u_{k_j}) \to (x, u)$ with $\|u\| = 1$. The function $(x, u) \mapsto u^{\mathsf T}A(x)u$ is continuous, so $u^{\mathsf T}A(x)u \leq 0$, contradicting positive definiteness of $A(x)$.