Compactness and Compactification

Compactness, for metric spaces, was the subject of 2B.3 Compactness: every sequence has a convergent subsequence, every open cover has a finite subcover, and the two are equivalent there. In general topological spaces the open-cover definition is the right one, and it combines with the Hausdorff condition to give the single most useful lemma for working with quotients: a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. With it, identifying a glued-together space with a familiar one becomes a matter of writing down a continuous map.

The chapter's second theme is adding a point at infinity. The plane plus one point is a sphere; three-dimensional space plus one point is the 3-sphere $S^3$, the space the Poincaré conjecture is about. This is the one-point compactification, and stereographic projection makes it concrete.

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

• use the open-cover definition of compactness, and prove that compact subsets of Hausdorff spaces are closed;
• prove and use the closed map lemma for continuous maps from compact to Hausdorff spaces;
• state Tychonoff's theorem and say where the axiom of choice enters;
• define local compactness and proper maps;
• construct the one-point compactification and prove $\mathbb{R}^n\cup\{\infty\} \cong S^n$ by stereographic projection.

2.1Mapping the poles

2.2Compactness in topological spaces

Several facts from 2B.3 Compactness hold with the same proofs: a closed subset of a compact space is compact, and a continuous image of a compact space is compact, so a continuous real function on a compact space attains its maximum and minimum (2A.9 Continuous Functions). What changes is the relation with sequences: in general spaces sequential compactness and compactness are different (each can hold without the other), which is why the open-cover definition is taken as basic. For metric spaces, and so for subsets of manifolds, the two agree.

Hausdorff spaces add one more fact, which is false without the separation condition.

In the cofinite topology on an infinite set, every subset is compact (Exercise 2.7) but only finite sets and the whole space are closed: the proposition genuinely needs the Hausdorff condition.

This lemma is the workhorse for quotients. To identify $X/{\sim}$ with a known space $Y$, it is enough to find a continuous surjection $f : X \to Y$, with $X$ compact and $Y$ Hausdorff, whose fibres are exactly the equivalence classes. For example $[0, 1] \to S^1$, $t \mapsto e^{2\pi it}$, shows $[0, 1]/(0 \sim 1) \cong S^1$ (7A.1 Topological Spaces and Quotients); and the map that sends the square to the doughnut surface by the usual angle coordinates shows that the glued square is the doughnut (Exercise 2.9).

2.3Products and Tychonoff's theorem

For finitely many factors the proof is elementary, by the tube lemma: if $X$ is compact and $W$ an open set in $X \times Y$ containing $X\times\{y\}$, then $W$ contains a whole "tube" $X\times V$ with $V$ a neighbourhood of $y$ (Exercise 2.8). From it, $X \times Y$ is compact when both factors are, and by induction any finite product. So closed bounded subsets of $\mathbb{R}^n$, being closed subsets of products of intervals, are compact, which is the Heine–Borel theorem.

For infinitely many factors the theorem needs the axiom of choice, and is in fact equivalent to it (John Kelley, 1950), in the sense of 2A.8 Infinite Sets. It is used in functional analysis, for example in the proof of the Banach–Alaoglu theorem (4A.6 Weak Convergence and the Direct Method), but not for manifolds, which are locally products of finitely many lines.

2.4Local compactness and proper maps

$\mathbb{R}^n$ is not compact, but every point has a compact neighbourhood (a closed ball). A Hausdorff space is locally compact if every point has a neighbourhood whose closure is compact. Every manifold is locally compact, since it is locally homeomorphic to $\mathbb{R}^n$ (7A.3 Manifolds and Surfaces). Infinite-dimensional normed spaces are not (4A.1 Banach Spaces and Bounded Operators).

A continuous map $f : X \to Y$ is proper if the preimage of every compact set is compact. Proper maps are the maps that "send infinity to infinity": as a point leaves every compact set of $X$, its image leaves every compact set of $Y$. A proper map into a locally compact Hausdorff space is closed. The polynomial map $\mathbb{R} \to \mathbb{R}$, $x \mapsto x^2$, is proper; $x \mapsto e^x$ is not (the preimage of $[0, 1]$ is $(-\infty, 0]$). Covering maps with finitely many sheets are proper (7A.6 Covering Spaces), and in Riemannian geometry a complete manifold is one whose distance functions are proper (9A.3 Geodesics and the Exponential Map).

2.5The one-point compactification

Neighbourhoods of $\infty$ are the complements of compact sets: a sequence converges to $\infty$ exactly when it eventually leaves every compact subset of $X$. Then $X^*$ is a compact Hausdorff space containing $X$ as an open dense subspace (Exercise 2.10). Compactness: any open cover contains a set around $\infty$, which misses only a compact $K$, and finitely many other sets cover $K$. Hausdorff: two points of $X$ are separated in $X$, and a point $x \in X$ is separated from $\infty$ by a neighbourhood of $x$ with compact closure $\overline U$, and $(X\setminus\overline U)\cup\{\infty\}$. This is where local compactness is used.

2.6History

The open-cover formulation of compactness was introduced by Pavel Alexandroff and Pavel Urysohn in the 1920s; Andrey Tychonoff proved his product theorem in 1930 (for products of intervals) and 1935 in general, and John Kelley showed in 1950 that it implies the axiom of choice. Alexandroff introduced the one-point compactification in 1924. Stereographic projection is ancient, used by Hipparchus and Ptolemy for star charts and astrolabes; its conformality was proved in the seventeenth century, by Thomas Harriot and later Edmond Halley.

2.7Exercises

Solution

Given an open cover of a subset $A$, pick one non-empty member $U$; it misses only finitely many points of the whole space, and each of those in $A$ is covered by one more member. So finitely many members cover $A$. Any infinite proper subset (such as the even numbers in $\mathbb{N}$) is compact but not closed, since closed sets are finite or everything.

Solution

(b) $(0, 1) \cong S^1\setminus\{1\}$ via $t \mapsto e^{2\pi it}$, and the extension sending $\infty \mapsto 1$ is a continuous bijection from a compact space to $S^1$. $\mathbb{N}^*$: send $n \mapsto \frac1n$ and $\infty \mapsto 0$; neighbourhoods of $\infty$ are cofinite sets plus $\infty$, which correspond to neighbourhoods of $0$. (c) The open disc gives $S^2$. The open annulus has two "ends", the inner and outer boundaries, and both are squeezed to the single point $\infty$: the result is a sphere with two points identified, and near $\infty$ it looks like two discs touching at their centres, not like a single disc.

Solution

(a) $|2y|^2 + (|y|^2 - 1)^2 = (|y|^2 + 1)^2$. With $x = \sigma^{-1}(y)$, $1 - x_{n+1} = \frac{2}{|y|^2 + 1}$, so $\sigma(x) = \frac{2y/(|y|^2 + 1)}{2/(|y|^2 + 1)} = y$. (b) $|\sigma(x)|^2 = \frac{1 - x_{n+1}^2}{(1 - x_{n+1})^2} = \frac{1 + x_{n+1}}{1 - x_{n+1}}$, using $\sum_{i\leq n}x_i^2 = 1 - x_{n+1}^2$. (c) $x_3 = 0$ gives $|\sigma| = 1$, and $x_3 < 0$ gives $|\sigma| < 1$.

Solution

(a) On $A\cap B$, $|z| = |w|$ and $|z|^2 + |w|^2 = 1$ give $|z| = |w| = \frac{1}{\sqrt2}$, so the set is $\{\frac{1}{\sqrt2}e^{i\alpha}\}\times\{\frac{1}{\sqrt2}e^{i\beta}\}$. (b) On $A$, $|z|^2 \leq \frac12$, so $|\sqrt2z| \leq 1$, and $w \neq 0$. The map is continuous on the compact set $A$; it is injective because $|w| = \sqrt{1 - |z|^2}$ is determined by $z$, so $w$ is recovered from $\frac{w}{|w|}$ and $z$; and it is onto. The closed map lemma makes it a homeomorphism. (c) Similarly for $B$ with the roles of $z$ and $w$ exchanged. On the common torus, the circle along which $z$ varies and $w$ is fixed bounds a disc in $A$ (a meridian of $A$) but runs parallel to the core of $B$ (a longitude of $B$), and the circle along which $w$ varies does the opposite.