A topological space \(X\) is called locally compact if every point in \(X\) has an open neighborhood whose closure is compact. Show that the Cartesian product of two locally compact spaces, with the product topology, is also locally compact.
Solution
Let \(X\) and \(Y\) be locally compact.
1. Fix \((x,y)\in X\times Y\) and choose open neighborhoods \(U\ni x\) and \(V\ni y\) such that \(\overline U\) and \(\overline V\) are compact.
This is the assumed local compactness of \(X\) at \(x\) and of \(Y\) at \(y\).
2. The set \(U\times V\) is an open neighborhood of \((x,y)\) in \(X\times Y\).
Products of open sets form a basis for the product topology.
3. One has\[ \overline{U\times V}=\overline U\times\overline V. \]
The set \(\overline U\times\overline V\) is closed in \(X\times Y\) and contains \(U\times V\), so \[ \overline{U\times V}\subseteq\overline U\times\overline V. \]
Conversely, let \((u,v)\in\overline U\times\overline V\) and let \(O\) be any open neighborhood of \((u,v)\). Choose a basic open set \(O_X\times O_Y\) with \[ (u,v)\in O_X\times O_Y\subseteq O. \] Since \(u\in\overline U\), the set \(O_X\) meets \(U\); since \(v\in\overline V\), the set \(O_Y\) meets \(V\). Hence \((O_X\times O_Y)\cap(U\times V)\) is nonempty. Thus every neighborhood of \((u,v)\) meets \(U\times V\), proving \((u,v)\in\overline{U\times V}\).
4. The closure \(\overline{U\times V}\) is compact.
By 1, both \(\overline U\) and \(\overline V\) are compact. We verify directly that their product is compact. Let \(\mathcal W\) be an open cover of \(\overline U\times\overline V\). Fix \(u\in\overline U\). For each \(v\in\overline V\), choose \(W_v\in\mathcal W\) containing \((u,v)\) and a basic open rectangle \[ N_v\times M_v\subseteq W_v \] with \(u\in N_v\) and \(v\in M_v\). Compactness of \(\overline V\) gives finitely many points \(v_1,\dots,v_r\) such that \[ \overline V\subseteq M_{v_1}\cup\cdots\cup M_{v_r}. \] Then \[ N_u:=N_{v_1}\cap\cdots\cap N_{v_r} \] is an open neighborhood of \(u\), and \(N_u\times\overline V\) is covered by the finitely many sets \(W_{v_1},\dots,W_{v_r}\).
As \(u\) varies, the sets \(N_u\) cover \(\overline U\). Compactness of \(\overline U\) yields \(u_1,\dots,u_s\) with \[ \overline U\subseteq N_{u_1}\cup\cdots\cup N_{u_s}. \] The finitely many members of \(\mathcal W\) chosen for these \(s\) neighborhoods cover all of \(\overline U\times\overline V\). Thus \(\overline U\times\overline V\) is compact, and 3 gives compactness of \[ \overline{U\times V}=\overline U\times\overline V. \]
5. Therefore \(X\times Y\) is locally compact.
4.This is precisely the stated definition of local compactness.