The Criterion and Its Hypotheses
A one-point compactification adds a single point, often denoted \(\infty\), to a space \(X\). The new space must be compact, and the original space must sit inside it with its original topology and be dense. The criterion identifies exactly when this can be done with a Hausdorff compact space. The assumption that \(X\) is noncompact matters: if \(X\) is already compact, adding a point cannot produce a compactification in which \(X\) is dense.
A space is locally compact here if every point has a compact neighborhood: for each \(x\in X\), some compact set \(K\subseteq X\) contains an open set \(U\) with \(x\in U\subseteq K\). A compactification of \(X\) is a compact Hausdorff space \(Y\) containing a homeomorphic copy of \(X\) as a dense subspace. A one-point compactification is such a compactification for which \(Y\setminus X\) consists of exactly one point.
The two separation requirements have different roles. The Hausdorff condition on \(X\) is inherited from any Hausdorff compactification. Local compactness is what allows neighborhoods of a point of \(X\) to be separated from the added point. We first construct the extension and prove these claims directly.
Constructing the Extension
Assume \(X\) is locally compact and Hausdorff. Let \(X^*=X\cup\{\infty\}\), where \(\infty\notin X\). Give \(X^*\) the following topology: every open subset of \(X\) is open in \(X^*\), and a set containing \(\infty\) is open precisely when its complement in \(X\) is compact. In the second clause, the complement is a subset of \(X\), so such a set has the form \(X^*\setminus K\) for a compact \(K\subseteq X\).
This is a topology. Arbitrary unions of open sets not containing \(\infty\) are open in \(X\). A finite intersection of sets containing \(\infty\) has complement a finite union of compact sets, which is compact. If an arbitrary union contains \(\infty\), at least one set in the union contains \(\infty\); the complement of the union is a closed subset of the compact complement of that set, and is therefore compact. Intersections with open sets in \(X\) are open in \(X\). These facts verify the topology axioms.
We use the standard fact that a compact subset of a Hausdorff space is closed. For completeness, if \(K\) is compact and \(x\notin K\), separate \(x\) from each \(y\in K\) by disjoint open sets \(U_y\ni x\) and \(V_y\ni y\). Finitely many \(V_y\) cover \(K\). The intersection of the corresponding finitely many \(U_y\) is a neighborhood of \(x\) disjoint from their union, hence disjoint from \(K\). Thus \(X\setminus K\) is open. In particular, the sets \(X^*\setminus K\) used above restrict to open sets of \(X\), and the subspace topology on \(X\) is exactly its given topology.
Proof. For compactness, take an open cover of \(X^*\) and choose a member \(G\) containing \(\infty\). Its complement \(K=X^*\setminus G\) is compact in \(X\). The other members of the cover cover \(K\), so finitely many of them cover \(K\). Together with \(G\), these finitely many sets cover \(X^*\).
For Hausdorffness, two distinct points \(x,y\in X\) have disjoint open neighborhoods in \(X\), since \(X\) is Hausdorff; these are also open in \(X^*\). To separate \(x\in X\) from \(\infty\), choose a compact neighborhood \(K\) of \(x\). By the definition of neighborhood, there is an open \(U\) with \(x\in U\subseteq K\). The sets \(U\) and \(X^*\setminus K\) are open in \(X^*\), contain \(x\) and \(\infty\) respectively, and are disjoint.
Finally, an open neighborhood of \(\infty\) has the form \(X^*\setminus K\) for a compact \(K\subseteq X\). Since \(X\) is noncompact, \(K\ne X\), so this neighborhood meets \(X\). Every point of \(X\) already belongs to \(X\), and every neighborhood of \(\infty\) meets \(X\); therefore \(\overline{X}=X^*\). \(\square\)
This proves the forward construction in the criterion: when \(X\) is noncompact, locally compact, and Hausdorff, \(X^*\) is a compact Hausdorff space containing \(X\) densely, with exactly one additional point.
Worked Example: The Open Interval
Take \(X=(0,1)\) with the subspace topology inherited from the standard topology on \(\mathbb{R}\). First, \(X\) is Hausdorff because distinct real numbers have disjoint open interval neighborhoods, which can be intersected with \(X\).
For any \(x\in(0,1)\), let \(\delta=\min(x,1-x)/2\). Both \(x\) and \(1-x\) are positive, so \(\delta>0\). The compact interval \(K=[x-\delta,x+\delta]\) lies in \((0,1)\), and the open neighborhood \((x-\delta,x+\delta)\) of \(x\) lies in \(K\). Compactness of \(K\) follows from the Heine–Borel theorem. Thus \(X\) is locally compact.
To check noncompactness, consider the open cover \(U_n=(1/n,1)\), for \(n\ge 2\). Each \(U_n\) is open in \(X\), and their union is \(X\): given \(x\in(0,1)\), choose \(n>1/x\), so \(1/n<x<1\). A finite subcollection has a largest index \(N\), and its union is \(U_N=(1/N,1)\), which misses, for example, \(1/(2N)\in X\). There is no finite subcover. The criterion therefore gives a compact Hausdorff one-point compactification of \((0,1)\).
For instance, \(K=[1/4,3/4]\) is compact in \(X\), so \(X^*\setminus K=\{\infty\}\cup((0,1)\setminus[1/4,3/4])\) is an open neighborhood of \(\infty\). It is disjoint from the open neighborhood \((1/3,1/2)\) of \(5/12\), which lies inside \(K\). This illustrates how compact neighborhoods separate finite points from infinity.
Why the Criterion Is Necessary
Now suppose a compact Hausdorff space \(Y\) is a one-point compactification of \(X\). Identify \(X\) with its embedded copy in \(Y\), and write \(Y\setminus X=\{\infty\}\). Since Hausdorff spaces have closed singletons, \(X=Y\setminus\{\infty\}\) is open in \(Y\). The subspace \(X\) is Hausdorff because it is a subspace of a Hausdorff space.
To obtain local compactness, we need to find, near each \(x\in X\), an open set whose closure in \(Y\) avoids \(\infty\). The needed separation fact follows from compactness and Hausdorffness:
Proof. The closed subset \(F\) of the compact space \(Y\) is compact. For each \(y\in F\), the Hausdorff condition gives disjoint open sets \(U_y\ni x\) and \(V_y\ni y\). Finitely many of the sets \(V_y\) cover \(F\), say \(V_{y_1},\ldots,V_{y_m}\). Let \(W=U_{y_1}\cap\cdots\cap U_{y_m}\), an open neighborhood of \(x\). For each \(i\), \(W\subseteq U_{y_i}\), and \(U_{y_i}\cap V_{y_i}=\varnothing\). Since \(V_{y_i}\) is open and disjoint from \(W\), it is also disjoint from \(\overline{W}\). The union of these finitely many \(V_{y_i}\) contains \(F\), so \(\overline{W}\cap F=\varnothing\). \(\square\)
Proof of necessity in the theorem. Fix \(x\in X\) and apply the lemma with \(F=\{\infty\}\), which is closed in \(Y\). We obtain an open \(W\ni x\) with \(\overline{W}\cap\{\infty\}=\varnothing\). Hence \(W\subseteq X\), and \(\overline{W}\) is a closed subset of the compact space \(Y\), so it is compact. Its subspace topology as a subset of \(X\) agrees with its topology as a subset of \(Y\); thus it is compact in \(X\) as well. Since \(W\) is open in \(Y\), it is open in \(X\), and \(x\in W\subseteq\overline{W}\). Therefore \(x\) has a compact neighborhood in \(X\). This holds for every \(x\), proving that \(X\) is locally compact. Together with the Hausdorff conclusion above, this proves necessity. \(\square\)
Worked Example: The Rationals Do Not Qualify
Let \(X=\mathbb{Q}\) with the subspace topology inherited from \(\mathbb{R}\). It is Hausdorff, but it is not locally compact. Suppose, to the contrary, that some \(q\in\mathbb{Q}\) has a compact neighborhood \(K\subseteq\mathbb{Q}\). Then \(K\) contains a relatively open neighborhood of \(q\), and so contains \(\mathbb{Q}\cap(a,b)\) for some real numbers \(a<b\).
The inclusion map from \(\mathbb{Q}\) to \(\mathbb{R}\) is continuous, so \(K\), compact in \(\mathbb{Q}\), is compact as a subset of \(\mathbb{R}\). The compact-subset lemma above implies that \(K\) is closed in \(\mathbb{R}\). Since the rationals are dense in \(\mathbb{R}\), closedness and \(\mathbb{Q}\cap(a,b)\subseteq K\) imply \([a,b]\subseteq K\). But every nonempty real interval contains an irrational number, contradicting \(K\subseteq\mathbb{Q}\). Thus no point has a compact neighborhood.
Also, \(\mathbb{Q}\) is noncompact: the open sets \(U_n=(-n,n)\cap\mathbb{Q}\), \(n\in\mathbb{N}\), cover \(\mathbb{Q}\), but any finite subcollection is contained in \(U_N\) for its largest index \(N\), and misses the rational number \(N+1\). By the criterion, \(\mathbb{Q}\) has no compact Hausdorff one-point compactification.
Why Noncompactness Cannot Be Dropped
If \(X\) is compact Hausdorff, it is already compact and locally compact. Nevertheless, adding a point does not give a compactification with \(X\) dense. In the topology constructed above, \(X\) itself is compact, so \(X^*\setminus X=\{\infty\}\) is open. The point \(\infty\) therefore has a neighborhood disjoint from \(X\), and \(X\) is not dense in \(X^*\).
Worked Example: A Finite Space and the Density Failure
Let \(X=\{a,b\}\) with the discrete topology. It is Hausdorff and compact: every open cover has a finite subcover because there are only two points. Every subset is compact, so in the one-point extension \(X^*=\{a,b,\infty\}\), the set \(\{\infty\}=X^*\setminus X\) is open. It is a neighborhood of \(\infty\) that misses \(X\), proving directly that \(X\) is not dense in \(X^*\).
More generally, no compact Hausdorff compactification can add a point to a compact space while keeping the original space dense. The embedded copy of a compact space in a Hausdorff space is compact, hence closed. If it is also dense, it must equal the whole space. Therefore a compact Hausdorff space cannot have a one-point compactification in the dense-embedding sense.
Common pitfall. “Locally compact Hausdorff” by itself is not the full criterion. The finite example satisfies both conditions, but the one-point extension is not a compactification because it is not dense. In the construction, density is exactly where noncompactness is used: every compact \(K\subseteq X\) must be a proper subset, so every neighborhood \(X^*\setminus K\) of \(\infty\) meets \(X\). Conversely, one must not infer local compactness merely from the existence of an arbitrary added point; the necessity proof uses compactness and Hausdorffness of the whole space to find neighborhoods with compact closure inside \(X\).
| Condition on \(X\) | Role in the criterion |
|---|---|
| Hausdorff | Inherited from the Hausdorff compactification; also separates points of \(X\) from one another. |
| Locally compact | Provides compact neighborhoods that separate points of \(X\) from \(\infty\). |
| Noncompact | Ensures every neighborhood of \(\infty\) meets \(X\), making the extension dense. |
Check Your Understanding
For each question, identify which part of the criterion or construction is being tested.
- In the constructed topology on \(X^*\), why does compactness of \(K\subseteq X\) make \(X^*\setminus K\) open?
- Which step in the proof that \(X^*\) is Hausdorff uses local compactness?
- Why does noncompactness imply that every open neighborhood of \(\infty\) intersects \(X\)?
- In the necessity proof, why is \(W\subseteq X\) when \(\overline{W}\) avoids \(\infty\)?
- Explain why a compact Hausdorff space cannot have a one-point compactification in which it is dense.