§ 29. Local Compactness

Definition 29.1 : Locally Compact

A space \(X\) is said to be locally compact at \(x\) if there is some compact subspace \(C\) of \(X\) that contains a neighborhood of \(x\). If \(X\) is locally compact at each of its points, \(X\) is said simply to be locally compact.

Note that a compact space is automatically locally compact.

Example

The real line \(\mathbb{R}\) is locally compact. The point \(x\) lies in some interval \((a, b)\), which in turn is contained in the compact subspace \([a, b]\).

Example

The space \(\mathbb{R}^{n}\) is locally compact; the point \(x\) lies in some basis element \(\left(a_{1}, b_{1}\right) \times \cdots \times\left(a_{n}, b_{n}\right)\), which in turn lies in the compact subspace \(\left[a_{1}, b_{1}\right] \times \cdots \times\left[a_{n}, b_{n}\right]\). The space \(\mathbb{R}^{\omega}\) is not locally compact; none of its basis elements are contained in compact subspaces. For if

\[ B=\left(a_{1}, b_{1}\right) \times \cdots \times\left(a_{n}, b_{n}\right) \times \mathbb{R} \times \cdots \times \mathbb{R} \times \cdots \]

were contained in a compact subspace, then its closure

\[ \bar{B}=\left[a_{1}, b_{1}\right] \times \cdots \times\left[a_{n}, b_{n}\right] \times \mathbb{R} \times \cdots \]

would be compact, which it is not.

Example

Every simply ordered set \(X\) having the least upper bound property is locally compact: Given a basis element for \(X\), it is contained in a closed interval in \(X\), which is compact.

Theorem 29.1

Let \(X\) be a space. Then \(X\) is locally compact Hausdorff if and only if there exists a space \(Y\) satisfying the following conditions:

1. \(X\) is a subspace of \(Y\).
2. The set \(Y-X\) consists of a single point.
3. \(Y\) is a compact Hausdorff space.

If \(Y\) and \(Y^{\prime}\) are two spaces satisfying these conditions, then there is a homeomorphism of \(Y\) with \(Y^{\prime}\) that equals the identity map on \(X\).

Proof

Step 1.
We first verify uniqueness. Let \(Y\) and \(Y^{\prime}\) be two spaces satisfying these conditions. Define \(h: Y \rightarrow Y^{\prime}\) by letting \(h\) map the single point \(p\) of \(Y-X\) to the point \(q\) of \(Y^{\prime}-X\), and letting \(h\) equal the identity on \(X\). We show that if \(U\) is open in \(Y\), then \(h(U)\) is open in \(Y^{\prime}\). Symmetry then implies that \(h\) is a homeomorphism.

First, consider the case where \(U\) does not contain \(p\). Then \(h(U)=U\). Since \(U\) is open in \(Y\) and is contained in \(X\), it is open in \(X\). Because \(X\) is open in \(Y^{\prime}\), the set \(U\) is also open in \(Y^{\prime}\), as desired.

Second, suppose that \(U\) contains \(p\). Since \(C=Y-U\) is closed in \(Y\), it is compact as a subspace of \(Y\). Because \(C\) is contained in \(X\), it is a compact subspace of \(X\). Then because \(X\) is a subspace of \(Y^{\prime}\), the space \(C\) is also a compact subspace of \(Y^{\prime}\). Because \(Y^{\prime}\) is Hausdorff, \(C\) is closed in \(Y^{\prime}\), so that \(h(U)=Y^{\prime}-C\) is open in \(Y^{\prime}\), as desired.

Step 2.
Now we suppose \(X\) is locally compact Hausdorff and construct the space \(Y\). Step 1 gives us an idea how to proceed. Let us take some object that is not a point of \(X\), denote it by the symbol \(\infty\) for convenience, and adjoin it to \(X\), forming the set \(Y=X \cup\{\infty\}\). Topologize \(Y\) by defining the collection of open sets of \(Y\) to consist of (1) all sets \(U\) that are open in \(X\), and (2) all sets of the form \(Y-C\), where \(C\) is a compact subspace of \(X\).

We need to check that this collection is, in fact, a topology on \(Y\). The empty set is a set of type (1), and the space \(Y\) is a set of type (2). Checking that the intersection of two open sets is open involves three cases:

\[ \begin{array}{rlrl} U_{1} \cap U_{2} & & \text { is of type (1). } \\ \left(Y-C_{1}\right) \cap\left(Y-C_{2}\right) & =Y-\left(C_{1} \cup C_{2}\right) & & \text { is of type (2). } \\ U_{1} \cap\left(Y-C_{1}\right) & =U_{1} \cap\left(X-C_{1}\right) & & \text { is of type (1), } \end{array} \]

because \(C_{1}\) is closed in \(X\). Similarly, one checks that the union of any collection of open sets is open:

\[ \begin{aligned} & \bigcup U_{\alpha}=U \quad \text { is of type (1). } \\ & \bigcup\left(Y-C_{\beta}\right)=Y-\left(\bigcap C_{\beta}\right)=Y-C \quad \text { is of type (2). } \\ & \left(\bigcup U_{\alpha}\right) \cup\left(\bigcup\left(Y-C_{\beta}\right)\right)=U \cup(Y-C)=Y-(C-U) \text {, } \end{aligned} \]

which is of type (2) because \(C-U\) is a closed subspace of \(C\) and therefore compact.

Now we show that \(X\) is a subspace of \(Y\). Given any open set of \(Y\), we show its intersection with \(X\) is open in \(X\). If \(U\) is of type (1), then \(U \cap X=U\); if \(Y-C\) is of type (2), then \((Y-C) \cap X=X-C\); both of these sets are open in \(X\). Conversely, any set open in \(X\) is a set of type (1) and therefore open in \(Y\) by definition.

To show that \(Y\) is compact, let \(\mathcal{A}\) be an open covering of \(Y\). The collection \(\mathcal{A}\) must contain an open set of type (2), say \(Y-C\), since none of the open sets of type (1) contain the point \(\infty\). Take all the members of \(\mathcal{A}\) different from \(Y-C\) and intersect them with \(X\); they form a collection of open sets of \(X\) covering \(C\). Because \(C\) is compact, finitely many of them cover \(C\); the corresponding finite collection of elements of \(\mathcal{A}\) will, along with the element \(Y-C\), cover all of \(Y\).

To show that \(Y\) is Hausdorff, let \(x\) and \(y\) be two points of \(Y\). If both of them lie in \(X\), there are disjoint sets \(U\) and \(V\) open in \(X\) containing them, respectively. On the other hand, if \(x \in X\) and \(y=\infty\), we can choose a compact set \(C\) in \(X\) containing a neighborhood \(U\) of \(x\). Then \(U\) and \(Y-C\) are disjoint neighborhoods of \(x\) and \(\infty\), respectively, in \(Y\).

Step 3.
Finally, we prove the converse. Suppose a space \(Y\) satisfying conditions (1)-(3) exists. Then \(X\) is Hausdorff because it is a subspace of the Hausdorff space \(Y\). Given \(x \in X\), we show \(X\) is locally compact at \(x\). Choose disjoint open sets \(U\) and \(V\) of \(Y\) containing \(x\) and the single point of \(Y-X\), respectively. Then the set \(C=Y-V\) is closed in \(Y\), so it is a compact subspace of \(Y\). Since \(C\) lies in \(X\), it is also compact as a subspace of \(X\); it contains the neighborhood \(U\) of \(x\).

If \(X\) itself should happen to be compact, then the space \(Y\) is not very interesting, for it is obtained from \(X\) by adjoining a single isolated point. However, if \(X\) is not compact, then the point of \(Y-X\) is a limit point of \(X\), so that \(\bar{X}=Y\).

Definition 29.2 : Compactification / One-Point Compactification

If \(Y\) is a compact Hausdorff space and \(X\) is a proper subspace of \(Y\) whose closure equals \(Y\), then \(Y\) is said to be a compactification of \(X\). If \(Y-X\) equals a single point, then \(Y\) is called the one-point compactification of \(X\).

By Theorem 29.1, we have shown that \(X\) has a one-point compactification \(Y\) if and only if \(X\) is a locally compact Hausdorff space that is not itself compact. We speak of \(Y\) as "the" one-point compactification because \(Y\) is uniquely determined up to a homeomorphism.

Theorem 29.2

Let \(X\) be a Hausdorff space. Then \(X\) is locally compact if and only if given \(x\) in \(X\), and given a neighborhood \(U\) of \(x\), there is a neighborhood \(V\) of \(x\) such that \(\bar{V}\) is compact and \(\bar{V} \subset U\).

Proof

Clearly this new formulation implies local compactness; the set \(C=\bar{V}\) is the desired compact set containing a neighborhood of \(x\).

To prove the converse, suppose \(X\) is locally compact; let \(x\) be a point of \(X\) and let \(U\) be a neighborhood of \(x\). Take the one-point compactification \(Y\) of \(X\), and let \(C\) be the set \(Y-U\). Then \(C\) is closed in \(Y\), so that \(C\) is a compact subspace of \(Y\). Apply Lemma 26.4 to choose disjoint open sets \(V\) and \(W\) containing \(x\) and \(C\), respectively. Then the closure \(\bar{V}\) of \(V\) in \(Y\) is compact; furthermore, \(\bar{V}\) is disjoint from \(C\), so that \(\bar{V} \subset U\), as desired.

Corollary 29.3

Let \(X\) be locally compact Hausdorff; let \(A\) be a subspace of \(X\). If \(A\) is closed in \(X\) or open in \(X\), then \(A\) is locally compact.

Proof

Suppose that \(A\) is closed in \(X\). Given \(x \in A\), let \(C\) be a compact subspace of \(X\) containing the neighborhood \(U\) of \(x\) in \(X\). Then \(C \cap A\) is closed in \(C\) and thus compact, and it contains the neighborhood \(U \cap A\) of \(x\) in \(A\). (We have not used the Hausdorff condition here.)

Suppose now that \(A\) is open in \(X\). Given \(x \in A\), we apply the Theorem 29.2 to choose a neighborhood \(V\) of \(x\) in \(X\) such that \(\bar{V}\) is compact and \(\bar{V} \subset A\). Then \(C=\bar{V}\) is a compact subspace of \(A\) containing the neighborhood \(V\) of \(x\) in \(A\).

Corollary 29.4

A space \(X\) is homeomorphic to an open subspace of a compact Hausdorff space if and only if \(X\) is locally compact Hausdorff.

Proof

This follows from Theorem 29.1 and Corollary 29.3.