State the definition of compactness for topological spaces. Prove from your definition that the closed unit interval \([0,1]\) is compact.
Solution
A topological space \(X\) is compact if every open cover of \(X\) has a finite subcover. Explicitly, whenever \(\{U_\alpha\}_{\alpha\in I}\) is a family of open subsets of \(X\) satisfying \[ X=\bigcup_{\alpha\in I}U_\alpha, \] there are \(\alpha_1,\ldots,\alpha_n\in I\) such that \[ X=U_{\alpha_1}\cup\cdots\cup U_{\alpha_n}. \]
1. Fix an arbitrary open cover \(\mathcal U\) of \([0,1]\) and define\[ S=\left\{x\in[0,1]:[0,x]\text{ is covered by finitely many members of }\mathcal U\right\}. \]Then \(S\) is nonempty and bounded above.
Since \(\mathcal U\) covers \([0,1]\), some \(U_0\in\mathcal U\) contains \(0\). Thus the singleton interval \[ [0,0]=\{0\} \] is covered by the one member \(U_0\), so \(0\in S\). Hence \(S\ne\varnothing\). Also \(S\subseteq[0,1]\), so \(1\) is an upper bound for \(S\).
2. Let\[ s=\sup S. \]Then \(s=1\).
The supremum exists by 1 and the least-upper-bound property of \(\RR\). Suppose for contradiction that \(s<1\). Because \(\mathcal U\) covers \([0,1]\), choose \(U\in\mathcal U\) with \[ s\in U. \] Since \(U\) is open in the subspace \([0,1]\), there is \(\varepsilon>0\) such that \[ (s-\varepsilon,s+\varepsilon)\cap[0,1]\subseteq U. \] Set \[ \delta=\min\left\{\frac{\varepsilon}{2},\frac{1-s}{2}\right\}>0. \] By the definition of \(s=\sup S\), there is \(x\in S\) with \[ x>s-\delta. \] Indeed, if no such \(x\) existed, then \(s-\delta<s\) would be an upper bound for \(S\).
Because \(x\in S\), finitely many members of \(\mathcal U\) cover \([0,x]\). Moreover, \[ x>s-\delta\ge s-\frac{\varepsilon}{2}>s-\varepsilon \] and \[ s+\delta\le s+\frac{\varepsilon}{2}<s+\varepsilon. \] Therefore \(U\) contains the whole interval \[ [x,s+\delta]. \] Adding \(U\) to the finite cover of \([0,x]\) gives a finite cover of \[ [0,s+\delta]. \] Thus \(s+\delta\in S\). But \(s+\delta>s\), contradicting that \(s\) is an upper bound for \(S\). Hence \(s=1\).
3. The whole interval \([0,1]\) has a finite subcover from \(\mathcal U\).
Choose \(U_1\in\mathcal U\) with \(1\in U_1\). Since \(U_1\) is open in \([0,1]\), there is \(\varepsilon>0\) such that \[ (1-\varepsilon,1]\subseteq U_1. \] By 2, \[ \sup S=1. \] Hence there exists \(x\in S\) with \[ x>1-\varepsilon; \] otherwise \(1-\varepsilon\) would be an upper bound for \(S\) smaller than its supremum. Because \(x\in S\), there are finitely many members of \(\mathcal U\) covering \([0,x]\). Together with \(U_1\), they cover \[ [0,x]\cup(1-\varepsilon,1]=[0,1]. \] Thus \(\mathcal U\) has a finite subcover of \([0,1]\).
4. Therefore \([0,1]\) is compact.
The open cover \(\mathcal U\) in 1 was arbitrary, and 3 produced a finite subcover. This is exactly the definition of compactness. Hence \[ \boxed{[0,1]\text{ is compact}.} \]