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Open covers · Topology | Math Infinitum

Intuition

To cover a space is to choose open sets that between them leave nothing out. Compactness asks whether any such choice, however large, can be cut down to finitely many that still cover.

A hall lit by a great many lamps. The question is not whether few lamps could have been installed, but whether some finite handful of the ones present already lights the whole hall.

To cover a space is to choose open sets that between them leave nothing out; three of the chosen sets are drawn, and the rest of the cover is not. Compactness asks whether every such choice, however lavish, already holds finitely many that cover — a demand about all covers, which is why proving it is so much harder than denying it.

Covers, subcovers and compactness

A family of open sets whose union is the whole of $X$ is an open cover. A subcover is a subfamily of that same family which still covers. The space is compact when every open cover has a finite subcover. Read the order of the words carefully: the demand is not that some finite cover exists, but that any cover at all can be thinned to a finite one.

\begin{aligned}&X=\textstyle\bigcup_{i\in I}U_i\\ \implies\ &X=\textstyle\bigcup_{k\le n}U_{i_k}\end{aligned}

What the definition does and does not ask

Every space has a finite open cover: the single set $X$ is one. That is why the definition speaks about thinning an arbitrary cover rather than about small covers existing.
The finite subfamily has to come out of the cover you were handed. Producing some other finite family of open sets proves nothing.
The index set $I$ may be anything at all, finite or infinite. Only the subcover is required to be finite.
Compactness is stated in open sets alone, so a subset is called compact when it is compact in the topology it inherits.

A finite space is compact

X\ \text{finite}\implies X\ \text{compact}

Take any open cover of a space with finitely many points. Covering means every point lies in at least one member, so go through the points one at a time and pick, for each, a single member containing it. That is one choice per point and there are finitely many points, so finitely many members have been picked. They already cover, because every point is in the one picked for it. The cover was arbitrary, so every cover thins, which is the definition.

Proof steps

X=\{x_1,\ldots,x_n\}

The space has finitely many points; list them.

x_k\in U_{i_k}

A cover reaches every point, so for each point one member containing it can be chosen.

X=\textstyle\bigcup_{k\le n}U_{i_k}

Those members already cover, since every point lies in the one chosen for it.

X\ \text{is compact}

An arbitrary cover has been thinned to finitely many members, which is what compactness demands.

Applications

Practice

Thinning, not existing

X=\bigcup_{i\in I}U_i\implies X=\bigcup_{k\le n}U_{i_k}

Arbitrary on the left, finite on the right, and drawn from the same family.

Try it

What does compactness demand?