Let \(X\) be a set with two elements \(\{a,b\}\). Give \(X\) the indiscrete topology. Give \(X\times\mathbb R\) the product topology. Let \(A\subset X\times\mathbb R\) be \((\{a\}\times[0,1])\cup(\{b\}\times(0,1))\). Prove that \(A\) is compact.
Solution
1. Every nonempty open subset of \(X\times\mathbb R\) is of the form \(X\times V\) for some open \(V\subseteq\mathbb R\).
The only nonempty open subset of the indiscrete space \(X\) is \(X\) itself. Hence the basic open sets of the product topology are precisely the sets \(X\times V\) with \(V\) open in \(\mathbb R\). Arbitrary unions of such sets again have this form.
2. Every open subset of the subspace \(A\) is therefore of the form\[ A\cap(X\times V) \]for some open \(V\subseteq\mathbb R\).
This is the definition of the subspace topology together with 1.
3. Let \(\{A\cap(X\times V_i)\}_{i\in I}\) be an arbitrary open cover of \(A\).Then \(\{V_i\}_{i\in I}\) covers \([0,1]\).
For every \(t\in[0,1]\), the point \((a,t)\) belongs to \(A\). Since the displayed family covers \(A\), there is an \(i\) with \((a,t)\in X\times V_i\), hence \(t\in V_i\).
4. The interval \([0,1]\) admits a finite subcover from \(\{V_i\}_{i\in I}\).
We prove this directly. Let \[ S=\{x\in[0,1]:[0,x]\text{ is covered by finitely many }V_i\}. \] Some \(V_i\) contains \(0\), so \(S\) is nonempty. Let \(s=\sup S\). Choose \(V_j\) containing \(s\) and an open interval \((s-\varepsilon,s+\varepsilon)\subseteq V_j\).
If \(s<1\), choose \(x\in S\) with \(x>s-\varepsilon/2\). A finite family covers \([0,x]\), and adjoining \(V_j\) covers \[ [0,s+\varepsilon/2], \] contradicting the definition of \(s\) as an upper bound for \(S\). Thus \(s=1\). Now choose \(x\in S\) with \(x>1-\varepsilon/2\); a finite cover of \([0,x]\) together with \(V_j\) covers all of \([0,1]\).
5. The corresponding finitely many members of the original cover cover \(A\).
Choose indices \(i_1,\dots,i_n\) such that \[ [0,1]\subseteq V_{i_1}\cup\cdots\cup V_{i_n}. \] Every point of \(A\) has second coordinate in \([0,1]\), so it lies in \[ A\cap(X\times V_{i_1}),\ldots,A\cap(X\times V_{i_n}) \] for at least one index.
6. Therefore \(A\) is compact.
The open cover in 3 was arbitrary, and 5 produced a finite subcover.