Compactness · Definition

Open cover

topologycompactnesstopological spaces

Before you start you need to know

It is the language in which compactness is stated: a way of covering a set by open pieces.

Definition

An open cover of a subset \(A\) of a topological space \(X\) is a family \(\mathcal{U}\) of open sets \(U\) of \(X\) whose union contains \(A\).

\[A\subseteq\bigcup_{U\in\mathcal{U}}U\]

Covering \(A\) means every point of \(A\) lies in at least one open set of the family; the open sets may stick out of \(A\) and the family may be infinite, even uncountable. A cover of the whole of \(X\) always exists, for instance \(\{X\}\), so the interesting question is not existence but how far a cover can be thinned.

A subcover is a subfamily that still covers. The cover \(\{(-n,n)\}\) of \(\mathbb{R}\) has no finite subcover, yet the same family does have one for \([-5,5]\), since \((-6,6)\) suffices. For the interval \((0,1)\), the family \(\{(1/n,1)\}\) covers it and no finite part does; this shows how missing endpoints spoil finiteness.

If \(\mathcal{U}\) covers \(A\), it also covers every \(B\subseteq A\), and unions of covers cover unions of sets. In the subspace \(A\) one can also speak of covers by open sets of \(A\), namely the traces \(U\cap A\); both formulations are equivalent for deciding what is compact.

All this prepares the next definition: a space is compact when each of its open covers contains a finite subcover.

Think about it

Why does \(\{(n,n+1):n\in\mathbb{Z}\}\) not cover \(\mathbb{R}\)?

Because the integers belong to no open interval of the family.

If \(\mathcal{U}\) covers \(A\) and \(B\subseteq A\), does \(\mathcal{U}\) cover \(B\)?

Yes: each point of \(B\) lies in \(A\), hence in some open set of \(\mathcal{U}\).

Exercises 10

References

  • James R. Munkres, Topology, 2nd edition, Pearson, 2000. §26
  • Stephen Willard, General Topology, Dover reprint, Dover Publications, 2004. cap. 5