Why The One-Point Extension Is Compact

The Compactness Argument

The extra point in a one-point extension is designed so that its neighborhoods contain all but a compact part of the original space. This gives a direct strategy for proving compactness: from an open cover, choose one set containing the extra point; the points not yet covered lie in a compact subset of the original space, so finitely many more cover sets suffice.

Let \((X,\mathcal{T})\) be a noncompact, locally compact Hausdorff space. Here Hausdorff means that any two distinct points have disjoint open neighborhoods, and locally compact means that each point has a compact neighborhood. Let \(X^*=X\cup\{\infty\}\), where \(\infty\notin X\). We use the one-point extension topology: open sets not containing \(\infty\) are open in \(X\), and the open sets containing \(\infty\) are \(X^*\setminus K\), where \(K\) is a compact subset of \(X\). Compact subsets of a Hausdorff space are closed, so these sets are open in the extension topology.

The central point is that an open set containing \(\infty\) has a compact complement in \(X\). The complement is not merely bounded or small in some informal sense: it is compact, so it satisfies the open-cover definition of compactness. That definition applies to any open cover of the complement, including the cover obtained by restricting the other members of a cover of \(X^*\).

Proof. Let \(\mathcal{U}\) be an arbitrary open cover of \(X^*\). Since \(\mathcal{U}\) covers \(\infty\), there is some \(U_\infty\in\mathcal{U}\) with \(\infty\in U_\infty\). By the definition of the topology, \(U_\infty=X^*\setminus K\) for a compact subset \(K\subseteq X\). Thus \(U_\infty\) already covers \(\infty\) and every point of \(X\setminus K\). For each \(U\in\mathcal{U}\), the set \(U\cap X\) is open in \(X\); the family \(\{U\cap X:U\in\mathcal{U}\}\) therefore covers \(K\). By compactness of \(K\), finitely many members \(U_1,\ldots,U_m\) of \(\mathcal{U}\) have intersections with \(X\) covering \(K\). Then \(U_\infty,U_1,\ldots,U_m\) cover \(X^*\): \(U_\infty\) covers the point at infinity and \(X\setminus K\), and the other finitely many sets cover \(K\). Hence every open cover of \(X^*\) has a finite subcover, so \(X^*\) is compact. \(\square\)

Why the Restriction Is a Cover

In the proof, restricting the cover to \(X\) does not require the cover members to be subsets of \(X\). For any open \(U\subseteq X^*\), \(U\cap X\) is open in \(X\): if \(U\) does not contain \(\infty\), it is open in \(X\) by the construction; if it contains \(\infty\), then \(U=X^*\setminus L\) for compact \(L\subseteq X\), and \(U\cap X=X\setminus L\), which is open because \(L\) is closed in the Hausdorff space \(X\). This verifies that the restrictions used in the proof are legitimate open sets for applying compactness.

Also, the restricted family really covers \(K\). Given any \(x\in K\), the original family covers \(X^*\), so some \(U\in\mathcal{U}\) contains \(x\). Since \(x\in X\), it belongs to \(U\cap X\). Thus the finite-subcover property is applied to the whole compact remainder, not just to some of its points.

A Common Endpoint Mistake

Consider the cover of \(\mathbb{R}^*\) with \(U_\infty=\mathbb{R}^*\setminus[-1,1]\) and \(V_n=(-n,n)\) for \(n\in\mathbb{N}\). It is an open cover: \(U_\infty\) contains \(\infty\) and all real numbers outside \([-1,1]\), while the intervals \(V_n\) cover \([-1,1]\). A tempting but incorrect claim is that \(U_\infty\) and \(V_1=(-1,1)\) suffice. They do not: \(U_\infty\) omits both \(-1\) and \(1\), and \(V_1\) also omits both endpoints because its inequalities are strict. Thus \(-1,1\notin U_\infty\cup V_1\).

The correction is to choose \(V_2=(-2,2)\), which does contain both endpoints, since \(-2<-1<1<2\). Then \(U_\infty\cup V_2=\mathbb{R}^*\). The general compactness proof does not identify a particular finite collection in advance; compactness of the remainder guarantees that some finite collection works. But once a collection is chosen in an example, its endpoint membership must be checked using the actual strict or non-strict inequalities.

What the Argument Does—and Does Not—Use

The proof is a cover argument, not an assertion that every open set containing \(\infty\) has a finite complement. Its complement must be compact, and compact subsets can be infinite. For example, \([-3,3]\subset\mathbb{R}\) is infinite but compact, so \(\mathbb{R}^*\setminus[-3,3]\) is a valid open set containing \(\infty\).

Nor does the proof say that one member of every cover suffices. The chosen set containing \(\infty\) leaves its compact complement to be covered. The conclusion comes in two separate steps: compactness of that complement supplies finitely many more sets, and the chosen infinity-neighborhood covers everything outside the complement. Keeping these roles distinct is the main practical check when applying the argument.

The hypotheses have different roles in the construction and in the proof. The topology is defined using compact subsets of \(X\), and the local compactness and Hausdorff assumptions are part of the standard setting in which this construction gives the one-point extension topology. Once that topology is in place, the compactness proof itself uses the defining form of neighborhoods of \(\infty\) and compactness of their complements. The argument therefore explains compactness directly, without needing to establish separation properties of the extension.