2.05 Compactness


Below the video you will find accompanying notes and some pre-class questions.


Definition of compactness

(0.29) A topological space \(X\) is compact if every open cover has a finite subcover. More precisely, if \(X=\bigcup_{i\in I}U_i\) for some collection \(\{U_i\ :\ i\in I\}\) of open sets indexed by a set \(I\) then there is a finite subset \(J\subset I\) such that \(X=\bigcup_{i\in J}U_i\).

(1.45) This definition is motivated by the Heine-Borel theorem, which says that, for metric spaces, this definition is equivalent to sequential compactness (every sequence has a convergent subsequence).

(2.12) This definition is extremely useful. For example, if \(X\) is a discrete space (every subset is open) then \(X\) is compact if and only if \(X\) is a finite set (if you had an infinite discrete space then the collection \(\{\{x\}\ :\ x\in X\}\) is an open cover with no finite subcover). This provides a way to show that some set is finite. For example, if you are trying to show that there are only finitely many solutions to some differential equation, you could try putting a topology on the space of solutions, and proving it's a compact, discrete space.

Results about compactness

(3.47) Let \(X,Y\) be topological spaces and let \(F\colon X\to Y\) be a continuous map. If \(A\subset X\) is compact (when given the subspace topology) then \(F(A)\subset Y\) is also compact (when given the subspace topology).

(5.12) Take an open cover \(\{U_i\ :\ i\in I\}\) of \(F(A)\) (in the subspace topology). By definition of the subspace topology, there exist open sets \(V_i\subset Y\), \(i\in I\), such that \(U_i=V_i\cap F(A)\). Since \(F\) is continuous, \(F^{-1}(V_i)\) are open sets in \(X\). Finally, \(\{A\cap F^{-1}(V_i)\ :\ i\in I\}\) is an open cover of \(A\) (in the subspace topology on \(A\)). Since \(A\) is compact, there is a finite subset \(J\subset I\) such that \(A=\bigcup_{i\in J}A\cap F^{-1}(V_i)\). Now \(F(A)=\bigcup_{i\in J}U_i\), so we get a finite subcover \(\{U_i\ :\ i\in J\}\) of the cover we started with.

Recall that a subset \(A\subset X\) is closed if \(X\setminus A\) is open.

(8.13) A closed subset \(A\subset X\) of a compact space \(X\) is compact (in the subspace topology on \(A\)).

Let \(\{U_i\ :\ i\in I\}\) be an open cover of \(A\) in the subspace topology. By definition of the subspace topology, there exist open sets \(V_i\subset X\) such that \(U_i=A\cap V_i\). The \(V_i\) might not cover the whole of \(X\), but \(X=(X\setminus A)\cup\bigcup_{i\in I}V_i\) (and \(X\setminus A\) is open because \(A\) is closed). Now there is a finite subcover \(\{X\setminus A\}\cup\{V_i\ :\ i\in J\}\) (for some finite \(J\subset I\)), so \(\{U_i\ :\ i\in J\}\) is a finite subcover of the cover we started with.


(11.52) Recall that one version of the Heine-Borel theorem tells us that any subspace of \(\mathbf{R}^n\) is compact if and only if it is closed and bounded. This implies that the following familiar spaces are all compact:

The circle in the plane; the sphere, torus or higher genus surfaces in \(\mathbf{R}^3\) are all compact.

(13.17) \(\mathbf{R}\) is not compact (for example, \(\{(n-\epsilon,n+1+\epsilon)\ :\ n\in\mathbf{Z}\}\) gives an open cover with no finite subcover); \(\mathbf{R}^n\) is not compact; if I cut out a point from any of the compact examples above, the result is not compact.

(15.00) Note that to prove that closed and bounded sets in \(\mathbf{R}^n\) are compact, it's sufficient to prove that the cube \([0,R]^n\) is compact: any bounded set will be contained in some cube, so by our lemma above, it will be a closed subset of a compact space, hence compact. Since a cube is a product of intervals, it suffices to prove that \([0,1]\) is compact and the products of compact spaces are compact. The fact that products of compact spaces are compact is Tychonoff's theorem. The fact that \([0,1]\) is compact is another way of stating the Heine-Borel theorem.

Pre-class questions

  1. Why is \(\mathbf{R}^n\) not compact? What about a punctured torus?
  2. Let \(X\) be a compact space. Using the first lemma from the video and the Heine-Borel theorem, prove that any continuous function \(F\colon X\to\mathbf{R}\) is bounded and attains its maximum somewhere in \(X\).


CC-BY-SA, Jonny Evans 2017