Local Compactness In Subspaces And Products

Compact Neighborhoods as a Local Test

Local compactness is a pointwise condition: near each point, the space must contain a compact neighborhood. When the space is Hausdorff, compact neighborhoods can be chosen small enough to fit inside a prescribed open set. That ability to shrink neighborhoods is what makes the open-subspace result work. For closed subspaces, intersecting a compact neighborhood with the closed set is enough. For finite products, products of compact neighborhoods provide the required neighborhoods.

A neighborhood here means an open set containing the point, unless we explicitly say “compact neighborhood.” A subset \(K\) is a compact neighborhood of \(x\) if \(K\) is compact and contains an open set containing \(x\). We use the definition that a space is locally compact if every point has a compact neighborhood. All subspaces below have the subspace topology, and products have the product topology.

We establish the neighborhood-shrinking fact needed for the open-subspace argument. First, a compact subset \(K\) of a Hausdorff space \(X\) is closed: if \(p\in X\setminus K\), separate \(p\) from each \(k\in K\) by disjoint open sets. The open sets around the points \(k\) cover \(K\), so finitely many cover it. The intersection of the corresponding finitely many open sets around \(p\) is a neighborhood of \(p\) disjoint from \(K\). Thus \(X\setminus K\) is open.

Proof. The set \(C=K\setminus O\) is closed in the compact space \(K\), hence compact. For each \(y\in C\), the Hausdorff property gives disjoint open sets \(P_y,Q_y\) in \(K\) with \(x\in P_y\) and \(y\in Q_y\). Finitely many sets \(Q_{y_1},\ldots,Q_{y_m}\) cover \(C\). Set \(W=\bigcap_{i=1}^m P_{y_i}\) and \(Q=\bigcup_{i=1}^m Q_{y_i}\). Then \(W\) is open and contains \(x\), and \(W\cap Q=\varnothing\). Since \(Q\) is open and disjoint from \(W\), it is also disjoint from \(\overline{W}^{\,K}\). Therefore \(\overline{W}^{\,K}\subseteq K\setminus Q\). As \(C\subseteq Q\), this gives \(K\setminus Q\subseteq K\setminus C=O\), so \(\overline{W}^{\,K}\subseteq O\). If \(C=\varnothing\), take \(W=K\); the conclusion holds because \(O=K\). \(\square\)

This lemma also verifies the equivalence in the definition. If \(K\) is a compact neighborhood of \(x\) in a Hausdorff space \(X\), choose an open set \(G\) with \(x\in G\subseteq K\). The compact set \(K\) is closed in \(X\), so it is compact Hausdorff. Apply the lemma in \(K\) to the relative open set \(G\) to obtain \(W\) with \(x\in W\) and \(\overline{W}^{\,K}\subseteq G\). Write \(W=H\cap K\) for \(H\) open in \(X\), and set \(V=H\cap G\). Then \(V\) is open in \(X\), contains \(x\), and lies in \(K\). Since \(K\) is closed, \(\overline{V}^{\,X}\subseteq K\), and the subspace closure formula gives \(\overline{V}^{\,X}\subseteq\overline{W}^{\,K}\subseteq G\subseteq K\). Thus \(\overline{V}^{\,X}\) is a closed subset of the compact set \(K\), hence compact. Conversely, if \(x\) has an open neighborhood \(V\) with compact closure, that closure is a compact neighborhood of \(x\).

Open Subspaces: Shrink Before Restricting

Proof. The Hausdorff property passes to subspaces: two distinct points of \(U\) have disjoint open neighborhoods in \(X\), whose intersections with \(U\) are disjoint neighborhoods in \(U\). Let \(x\in U\). Choose a compact neighborhood \(K\) of \(x\) in \(X\), and an open set \(G\) with \(x\in G\subseteq K\). As proved above, \(K\) is closed in \(X\) and is a compact Hausdorff space. The set \(O=G\cap U\cap K\) is open in \(K\) and contains \(x\). By the lemma, choose \(W\) open in \(K\) with \(x\in W\) and \(\overline{W}^{\,K}\subseteq O\). Write \(W=H\cap K\) for some open \(H\subseteq X\), and set \(V=H\cap G\cap U\). This is open in \(U\), contains \(x\), and is contained in \(W\). Because \(K\) is closed, \(\overline{V}^{\,X}\subseteq K\), and therefore \(\overline{V}^{\,X}\subseteq\overline{W}^{\,K}\subseteq K\cap U\). The closure of \(V\) in \(U\) is \(U\cap\overline{V}^{\,X}\); it is closed in the compact set \(\overline{V}^{\,X}\), so it is compact. Thus \(V\) is a compact neighborhood of \(x\) in \(U\). Since \(x\) was arbitrary, \(U\) is locally compact. \(\square\)

The proof does more than establish local compactness: it gives, around each \(x\in U\), an open neighborhood in \(U\) whose closure in \(U\) is compact. The shrinking lemma is essential to ensuring that this closure stays inside \(U\).

Closed Subspaces: Intersect with a Compact Neighborhood

Proof. The Hausdorff property passes to subspaces by restricting disjoint neighborhoods. Fix \(x\in F\). Choose a compact neighborhood \(K\) of \(x\) in \(X\) and an open set \(G\) in \(X\) with \(x\in G\subseteq K\). The set \(G\cap F\) is open in \(F\) and contains \(x\). The set \(F\cap K\) is compact: \(F\cap K\) is closed in \(K\), because \(F\) is closed in \(X\), and a closed subset of a compact space is compact. Also \(G\cap F\subseteq F\cap K\). Hence \(F\cap K\) is a compact neighborhood of \(x\) in \(F\). This holds for every \(x\in F\), so \(F\) is locally compact. \(\square\)

Unlike the open-subspace proof, this argument needs no shrinking: closedness makes the intersection of \(F\) with the compact set \(K\) compact. In fact, for this inheritance of local compactness, Hausdorffness is not needed; it is included in the theorem because the conclusion is locally compact Hausdorff.

Finite Products

For two spaces, the product topology has a basis consisting of sets \(U\times V\), where \(U\) and \(V\) are open in the respective factors. We use the finite case of Tychonoff’s theorem, established in the earlier material on product compactness: a finite product of compact spaces is compact. The Hausdorff property also passes to finite products, as the next result verifies.

Proof. First consider two spaces \(X\) and \(Y\). If \((x_1,y_1)\ne(x_2,y_2)\), then either \(x_1\ne x_2\) or \(y_1\ne y_2\). In the first case, choose disjoint open neighborhoods \(U_1,U_2\) of \(x_1,x_2\); then \(U_1\times Y\) and \(U_2\times Y\) are disjoint open neighborhoods of the two product points. In the second case, use disjoint open neighborhoods in \(Y\) and take their products with \(X\). Thus \(X\times Y\) is Hausdorff.

Now let \((x,y)\in X\times Y\). Choose compact neighborhoods \(K\) of \(x\) and \(L\) of \(y\), and open sets \(U,V\) with \(x\in U\subseteq K\) and \(y\in V\subseteq L\). The basis property makes \(U\times V\) an open neighborhood of \((x,y)\) contained in \(K\times L\). By finite-product compactness, \(K\times L\) is compact. Hence it is a compact neighborhood of \((x,y)\), proving local compactness. Applying this two-factor argument repeatedly proves the result for any positive finite number of factors. A product with no factors is a one-point space, which is compact and Hausdorff, so it also satisfies the claim. \(\square\)

A Common Pitfall

Intersecting an ambient compact neighborhood with an open subspace does not necessarily give a compact neighborhood. In \(U=(0,1)\subset\mathbb{R}\), take \(x=1/2\) and the compact neighborhood \(K=[0,1]\) in \(\mathbb{R}\). The tempting calculation is \(K\cap U=(0,1)\), followed by the incorrect claim that this is compact in \(U\). It is not compact: the open cover of \((0,1)\) by the sets \((1/n,1)\), for \(n\ge2\), has no finite subcover. Indeed, any finite subcollection has a smallest index \(N\), and its union is \((1/N,1)\), which misses points of \((0,1/N]\).

The correction is to shrink around the point before restricting: \(V=(1/4,3/4)\) is an open neighborhood of \(1/2\) in \(U\), and its closure in \(U\) is \([1/4,3/4]\), a compact interval contained in \(U\). The distinction is that the closure of an unshrunk set may reach the boundary of the open subspace, while a smaller neighborhood can have compact closure entirely inside it.