Separating a Point from Infinity
In a one-point extension, a new point \(\infty\) is added to a space \(X\). Its neighborhoods are designed to contain all but a compact part of \(X\). This makes compactness central to the question of whether \(\infty\) can be separated from each point of \(X\). The exact answer is that the extension is Hausdorff precisely when \(X\) is both Hausdorff and locally compact.
Let \((X,\mathcal{T})\) be a topological space, and let \(\infty\) be a point not in \(X\). We use the one-point extension topology on \(X^*=X\cup\{\infty\}\): its open sets are the open subsets of \(X\), together with the sets containing \(\infty\) whose complements in \(X\) are closed and compact. Requiring the complements to be closed makes this definition a topology even when \(X\) is not Hausdorff.
Here compactness means that every open cover has a finite subcover. The definition of local compactness above does not require \(K\) itself to be open. In a Hausdorff space, however, compact subsets are closed, so a compact neighborhood can be used to construct an open neighborhood of \(\infty\).
Compact Subsets in Hausdorff Spaces
Proof. Let \(X\) be Hausdorff, let \(K\subseteq X\) be compact, and fix \(x\in X\setminus K\). For each \(y\in K\), the Hausdorff property gives disjoint open sets \(U_y,V_y\) with \(x\in U_y\) and \(y\in V_y\). The sets \(V_y\), as \(y\) ranges over \(K\), cover \(K\). Compactness gives a finite subcover \(V_{y_1},\ldots,V_{y_n}\). Set \(U=\bigcap_{i=1}^n U_{y_i}\) and \(V=\bigcup_{i=1}^n V_{y_i}\). Then \(U\) is an open neighborhood of \(x\), and \(V\) is an open set containing \(K\). Since each \(U_{y_i}\) is disjoint from \(V_{y_i}\), the set \(U\) is disjoint from \(V\), and hence from \(K\). Thus every point outside \(K\) has an open neighborhood disjoint from \(K\), so \(X\setminus K\) is open and \(K\) is closed. If \(K\) is empty, it is closed directly. \(\square\)
We will also use that continuous images of compact spaces are compact. Indeed, for an open cover of the image, taking preimages gives an open cover of the domain; a finite subcover of the domain then gives a finite subcover of the image. In particular, if \(K\subseteq X\), the inclusion map \(K\to X\) is continuous, so compactness of \(K\) as a subspace makes it compact as a subset of \(X\).
The Hausdorffness Criterion
Proof. First suppose \(X\) is Hausdorff and locally compact. Consider two distinct points of \(X^*\). If both lie in \(X\), the Hausdorff property of \(X\) gives disjoint open neighborhoods in \(X\); these are also open in \(X^*\). Now let the two points be \(x\in X\) and \(\infty\). By local compactness, there is a compact neighborhood \(K\) of \(x\) and an open set \(U\) with \(x\in U\subseteq K\). The lemma makes \(K\) closed in \(X\). Therefore \(U\) is open in \(X^*\), and \(V=\{\infty\}\cup(X\setminus K)\) is open in \(X^*\) by its definition. The sets \(U\) and \(V\) are disjoint because \(U\subseteq K\). Thus every distinct pair in \(X^*\) has disjoint open neighborhoods.
Conversely, suppose \(X^*\) is Hausdorff. Its subspace \(X\) is Hausdorff: for two points of \(X\), take disjoint open neighborhoods in \(X^*\) and intersect them with \(X\). Now fix \(x\in X\). Hausdorffness gives disjoint open neighborhoods \(O\) of \(x\) and \(V\) of \(\infty\) in \(X^*\). Since \(O\) does not contain \(\infty\), it is an open subset of \(X\). Since \(V\) contains \(\infty\), its complement \(C=X\setminus V\) is closed and compact in \(X\). Disjointness gives \(O\subseteq C\). Thus \(C\) is a compact neighborhood of \(x\), and \(X\) is locally compact. Both required properties of \(X\) follow. \(\square\)
The two directions use different information. To separate \(x\) from \(\infty\), local compactness supplies a compact neighborhood and Hausdorffness ensures that this compact set is closed. In the reverse direction, an open neighborhood of \(x\) disjoint from an open neighborhood of \(\infty\) must fit inside the compact complement of the latter.
Examples of Separation at Infinity
Worked Example: The Real Line
Give \(\mathbb{R}\) its standard topology. We show explicitly how to separate \(2\) from \(\infty\) in its one-point extension. The interval \(K=[1,3]\) is compact by the Heine–Borel theorem, and it is closed in \(\mathbb{R}\). The open interval \(U=(1,3)\) contains \(2\) and is contained in \(K\). The corresponding neighborhood of infinity is \(V=\{\infty\}\cup((-\infty,1)\cup(3,\infty))\). It is open because its complement in \(\mathbb{R}\) is the closed compact set \([1,3]\). Since \(U\subseteq[1,3]\), we have \(U\cap V=\varnothing\). The real line is Hausdorff, and this construction works at every real point, so the theorem gives that \(\mathbb{R}^*\) is Hausdorff.
Check: the point \(2\) belongs to \((1,3)\), while every real number in \(V\) is either less than \(1\) or greater than \(3\). Neither set contains \(\infty\) in \(U\), so the neighborhoods are indeed disjoint.
Worked Example: The Rationals
Let \(X=\mathbb{Q}\) with the subspace topology inherited from \(\mathbb{R}\). This space is Hausdorff, but it is not locally compact. To verify the latter directly, fix \(q\in\mathbb{Q}\) and suppose a compact subset \(C\subseteq\mathbb{Q}\) were a neighborhood of \(q\). Then \(C\) would contain a relative open neighborhood \((q-\varepsilon,q+\varepsilon)\cap\mathbb{Q}\) for some \(\varepsilon>0\). The inclusion \(\mathbb{Q}\to\mathbb{R}\) is continuous, so \(C\), compact in \(\mathbb{Q}\), is compact as a subset of \(\mathbb{R}\). Since \(\mathbb{R}\) is Hausdorff, the lemma implies that \(C\) is closed in \(\mathbb{R}\).
Rational numbers are dense in \(\mathbb{R}\), so the closure in \(\mathbb{R}\) of \((q-\varepsilon,q+\varepsilon)\cap\mathbb{Q}\) contains the whole interval \((q-\varepsilon,q+\varepsilon)\). Because \(C\) is closed and contains that set of rationals, it must contain the interval, including irrational numbers. That contradicts \(C\subseteq\mathbb{Q}\). Thus \(q\) has no compact neighborhood in \(\mathbb{Q}\); the argument applies to every \(q\).
Consequently, \(\mathbb{Q}^*\) is not Hausdorff. More explicitly, if disjoint open neighborhoods \(O\) of \(q\) and \(V\) of \(\infty\) existed, then \(O\) would be a relative open neighborhood of \(q\), and the closed compact set \(\mathbb{Q}\setminus V\) would contain \(O\). This would give the compact neighborhood just shown to be impossible.
Check: the contradiction relies on closedness in \(\mathbb{R}\), not merely closedness in \(\mathbb{Q}\). Compactness gives closedness in \(\mathbb{R}\) because the inclusion into the Hausdorff space \(\mathbb{R}\) is continuous.
Worked Example: A Finite Non-Hausdorff Space
Let \(X=\{a,b\}\) have the topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). Every subset of a finite space is compact: from any open cover, choose one member for each point, giving a finite subcover. In particular \(X\) is a compact neighborhood of each point, so \(X\) is locally compact.
However, \(X\) is not Hausdorff. Every open neighborhood of \(b\) is \(X\), so no open neighborhood of \(a\) can be disjoint from an open neighborhood of \(b\). The extension \(X^*\) is also not Hausdorff: its subspace \(X\) already has this failure. For example, any open sets in \(X^*\) separating \(a\) and \(b\) would restrict to disjoint open neighborhoods of \(a\) and \(b\) in \(X\), which do not exist.
Check: local compactness alone does not imply that the extension is Hausdorff. The theorem requires both local compactness and Hausdorffness of \(X\).
A Common Mistake
A tempting but incorrect argument is: “If \(X\) is Hausdorff, then \(X^*\) is Hausdorff, because \(\infty\) is a new point.” The missing issue is whether a neighborhood of \(x\) can be made disjoint from a neighborhood of \(\infty\). Such a neighborhood of infinity is the complement of a closed compact subset of \(X\), so its complement must contain an open neighborhood of \(x\). Hausdorffness of \(X\) alone does not provide that compact neighborhood.
For instance, in \(\mathbb{Q}\), the singleton \(\{q\}\) is compact and closed, so \(\{\infty\}\cup(\mathbb{Q}\setminus\{q\})\) is an open neighborhood of \(\infty\). But it does not separate \(\infty\) from \(q\): every open neighborhood of \(q\) in \(\mathbb{Q}\) contains rational points other than \(q\), and hence intersects this set. Replacing \(\{q\}\) by a larger compact complement cannot fix the problem, since \(\mathbb{Q}\) has no compact neighborhood of \(q\). The theorem identifies precisely the additional condition that the mistaken argument omits.
| Pair of points to separate | What provides the separation | Required condition |
|---|---|---|
| Two points \(x,y\in X\) | Disjoint open neighborhoods in \(X\) | \(X\) is Hausdorff |
| A point \(x\in X\) and \(\infty\) | An open neighborhood inside a closed compact neighborhood of \(x\) | \(X\) is locally compact and Hausdorff |
Check Your Understanding
For each question, identify which part of the Hausdorffness criterion is being used.
- In the one-point extension of \(\mathbb{R}\), give disjoint open neighborhoods of \(0\) and \(\infty\).
- Why does a compact neighborhood \(K\) in a Hausdorff space provide a closed compact set suitable for defining a neighborhood of infinity?
- Explain why the one-point extension of \(\mathbb{Q}\) is not Hausdorff, despite \(\mathbb{Q}\) itself being Hausdorff.
- Can a locally compact space have a one-point extension that is not Hausdorff? Give the finite example from the tutorial and identify the failed condition.
- If \(X\) is compact and Hausdorff, what is the neighborhood \(\{\infty\}\) in \(X^*\), and why is it open?