Show that \(\RR\) with the cofinite topology is compact.
Solution
Goal: Show that \(\RR\) with the cofinite topology is compact.
1. Let \(\mathcal{U} = \theset{U_\alpha}_{\alpha \in I}\) be an open cover of \(\RR\) (cofinite topology).
Arbitrary open cover.
2. Pick a nonempty member \(U_0\) of the cover.
The cover is nonempty because \(\RR \neq \emptyset\); and since the cover covers \(\RR\), some \(U_0\) contains a point, hence is nonempty.
(If \(U_0\) were empty it still exists; choose one that is nonempty since the union is all of \(\RR\).)
3. The complement \(\RR \setminus U_0\) is finite: \(\RR \setminus U_0 = \theset{x_1, \ldots, x_n}\).
Definition of the cofinite topology: nonempty open sets have finite complement.
4. For each \(j = 1, \ldots, n\), choose \(U_j \in \mathcal{U}\) with \(x_j \in U_j\).
\(\mathcal{U}\) covers \(\RR\).
5. \(\theset{U_0, U_1, \ldots, U_n}\) is a finite subcover of \(\RR\).
\(U_0\) covers everything except \(\theset{x_1, \ldots, x_n}\), and each \(x_j\) is covered by \(U_j\) (<1>4).
6. Q.E.D.
<1>1–<1>5 show every open cover of \(\RR\) (cofinite) has a finite subcover.