Show that \([0, 1]\) is compact.
Solution
Let \(\mathcal U\) be an open cover of \([0,1]\) and define \[ S=\{x\in[0,1]:[0,x]\text{ is covered by finitely many members of }\mathcal U\}. \]
1. \(S\) is nonempty.
Choose \(U_0\in\mathcal U\) with \(0\in U_0\). Since \(U_0\) is open in \(\RR\), there is \(\delta>0\) such that \[ [0,\delta)\subseteq U_0 \] after decreasing \(\delta\) if necessary. Hence every \[ x\in[0,\min\{\delta/2,1\}] \] belongs to \(S\).
2. If \(s=\sup S\), then \(s=1\).
The least-upper-bound property of \(\RR\) applies because \(S\) is nonempty by 1 and is bounded above by \(1\).
Suppose \(s<1\). Choose \(U\in\mathcal U\) with \(s\in U\). Since \(U\) is open, there is \(\varepsilon>0\) such that \[ (s-\varepsilon,s+\varepsilon)\subseteq U. \] By the defining property of the supremum, choose \(x\in S\) with \[ s-\varepsilon/2<x\le s. \] A finite subfamily of \(\mathcal U\) covers \([0,x]\). Adjoining \(U\) therefore gives a finite cover of \[ [0,s+\varepsilon/2], \] so \(s+\varepsilon/2\in S\), contradicting that \(s\) is an upper bound of \(S\). Thus \(s=1\).
3. The cover \(\mathcal U\) has a finite subcover.
Choose \(U_1\in\mathcal U\) with \(1\in U_1\). For some \(\varepsilon>0\), \[ (1-\varepsilon,1]\subseteq U_1. \] By 2, \(\sup S=1\), so choose \(x\in S\) with \[ 1-\varepsilon<x\le1. \] By the definition of \(S\), finitely many members of \(\mathcal U\) cover \([0,x]\). Together with \(U_1\), they cover all of \([0,1]\).
4. \([0,1]\) is compact.
The open cover \(\mathcal U\) was arbitrary, and 3 gives it a finite subcover. This is exactly compactness.