[Munkres Topology] 29. Local Compactness

 This article is one of the posts in the Textbook Commentary Project.

Definition. 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.

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 $Y'$ are two spaces satisfying these conditions. then there is a homeomorphism of $Y$ with $Y'$ that equals the identity map on $X$.

Proof. 

Step 1. We first verify uniqueness. Let $Y$ and $Y'$ be two spaces satisfying these conditions. Define $h:Y\rightarrow Y'$ by letting $h$ map the single point $p$ of $Y-X$ to the point $q$ of $Y'-X$, and letting $h$ equal the identity on $X$ (In $X$, it's identity, but in $Y$ and points, it's different only for point). We show that if $U$ is open in $Y$, then $h(U)$ is open in $Y'$. Symmetry then implies that $h$ is a homeomorphism.

First, consider the case where $Y$ 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 $Y'$ ($X\in \mathcal{T}_X$), the set $U$ is also open in $Y'$, 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 $Y'$, the space $C$ is also a compact subspace of $Y'$. Because $Y'$ is Hausdorff, $C$ is closed in $Y'$, so that $h(U)=Y'-C$ is open in $Y'$, 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 (by Step 1, it's unique.), and adjoint 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 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 the intersection of two open sets is open involves three cases: $$U_1 \cap U_2 \quad \mbox{is of type (1).}$$ $$(Y_C_1)\cap (Y_C_2)=Y_(C_1\cup C_2)\quad \mbox{is of type (2).}$$ $$U_1\cap (Y-C_1)=U_1 \cap (X-C_1)\quad \mbox{is of type (1),$$ because $C_1$ is closed in $X$. Similarly, one checks that the union of any collection of open sets is open: $$\bigcup U_\alpha=U \quad \mbox{is of type (1).}$$ $$\bigcup (Y-C_\beta)=Y-(\bigcap C_\beta)=Y-C \quad \mbox{is of type (2).$$ $$(\bigcup U_\alpha) \cup (\bigcup (Y_C_\beta))=U\cup (Y-C)=Y-(C-U),$$ 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 $X$. It $U$ is of type (1), then $U\cap X=U$; it $Y-C$ 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.

Th 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) exsits. Then $X$ is Hausdorff because it is a subspace of 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 $X$ is not compact, then the point $\infty$ is a limit point of $X$, so that $\bar{X}=Y$.

Definition. 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 $X$.

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 $\bar{x}$ is compact and $\bar{V}\subset \bar{U}$.

Proof. ($\Leftarrow$) Clearly this new formulation implies local compactness; the set $C=\bar{V}$ is the desired compact set containing a neighborhood of $x$. ($\Rightarrow$) 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 use the Hausdorff condition here.)

Suppose now that $A$ is open in $X$. Given $x\in A$, we apply the preceding theorem to choose a neighborhood $V$ of $x$ in $X$ such that $\bar{V}$ is compact an $\bar{V}\subset A$. 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. Theorem 29.1. and Corollary 29.3.