A space \((X,\mathcal{T})\) is called locally compact if, for every \(x \in X\) and \(x \in U \in \mathcal{T}\), there is a compact set \(C \subseteq U\) containing an open neighborhood of \(x\). Show that the Cartesian product of two locally compact spaces is locally compact.
Solution
Let \(X\) and \(Y\) be locally compact. Fix \[ (x,y)\in X\times Y \] and an open set \(W\subseteq X\times Y\) with \((x,y)\in W\).
1. There are open sets \(U\subseteq X\) and \(V\subseteq Y\) such that\[ (x,y)\in U\times V\subseteq W. \]
The sets \(U\times V\), with \(U\) open in \(X\) and \(V\) open in \(Y\), form a basis for the product topology. Since \(W\) is an open neighborhood of \((x,y)\), one such basic open set lies between \((x,y)\) and \(W\).
2. There are compact sets \(K\subseteq U\) and \(L\subseteq V\) and open sets \(U'\subseteq X\), \(V'\subseteq Y\) such that\[ x\in U'\subseteq K\subseteq U \qquad\text{and}\qquad y\in V'\subseteq L\subseteq V. \]
Apply the definition of local compactness in \(X\) to the point \(x\) and the open neighborhood \(U\). It gives a compact set \(K\subseteq U\) containing an open neighborhood \(U'\) of \(x\).
Apply the same definition in \(Y\) to \(y\in V\) to obtain a compact set \(L\subseteq V\) containing an open neighborhood \(V'\) of \(y\).
3. The set\[ C=K\times L \]is compact and satisfies\[ (x,y)\in U'\times V'\subseteq C\subseteq W, \]where \(U'\times V'\) is open in \(X\times Y\).
The finite product of compact spaces is compact, so \(K\times L\) is compact. The product of the open sets \(U'\) and \(V'\) is open, and by 2, \[ (x,y) \in U'\times V' \subseteq K\times L \subseteq U\times V. \] By 1, \(U\times V\subseteq W\). Thus \(C\) is a compact subset of \(W\) containing an open neighborhood of \((x,y)\).
4. Therefore \(X\times Y\) is locally compact.
The point \((x,y)\) and its open neighborhood \(W\) were arbitrary. By 3, every such pair admits a compact subset of \(W\) containing an open neighborhood of \((x,y)\), which is exactly the stated definition of local compactness.