The One-Point Compactification Topology

Adding a Point at Infinity

A noncompact space has no point that gathers all of its “directions of escape.” One way to add such a point is to require its neighborhoods to contain every point of the original space except those in some compact set. The compact set is the part of the space that a neighborhood of infinity is allowed to omit. The main task here is to check that this prescription really defines a topology.

Assume throughout that \((X,\mathcal{T})\) is a noncompact, locally compact Hausdorff space. Hausdorff means that any two distinct points have disjoint open neighborhoods. Locally compact means that each point has a compact neighborhood: a set containing an open neighborhood of the point that is itself compact. Write \(\infty\) for a new point not in \(X\), and set \(X^*=X\cup\{\infty\}\). The local compactness assumption is part of the usual setting for the construction; the topology-axiom verification below uses noncompactness and the Hausdorff property.

Definition: A subset \(V\subseteq X^*\) is declared open if either (i) \(\infty\notin V\) and \(V\in\mathcal{T}\), or (ii) \(\infty\in V\) and \(X\setminus V\) is a compact subset of \(X\). In case (ii), \(X\setminus V\) means the points of \(X\) omitted by \(V\); equivalently, \(V=\{\infty\}\cup(X\setminus K)\) for some compact \(K\subseteq X\).

Thus the open sets that contain \(\infty\) are precisely the sets \(N_K=\{\infty\}\cup(X\setminus K)\), where \(K\) is compact in \(X\). Compact here means that every cover by open sets has a finite subcover. The empty compact set gives \(N_{\varnothing}=X^*\). Since \(X\) is noncompact, \(X\) itself is not compact, so \(\{\infty\}=N_X\) is not among these open sets.

Compactness Facts Needed for the Axioms

Before checking arbitrary unions and intersections, we need a few facts about compact subsets of a Hausdorff space. These facts ensure that complements of the proposed open sets behave as required.

Lemma: In a Hausdorff space, every compact subset is closed. A closed subset of a compact space is compact, and a finite union of compact subsets is compact.

Proof. First let \(C\subseteq X\) be compact, and fix \(x\in X\setminus C\). For each \(c\in C\), the Hausdorff property gives disjoint open sets \(U_c\) and \(V_c\) with \(c\in U_c\) and \(x\in V_c\). The sets \(U_c\) cover \(C\), so compactness gives \(c_1,\ldots,c_n\in C\) such that \(C\subseteq U_{c_1}\cup\cdots\cup U_{c_n}\). The open set \(V_{c_1}\cap\cdots\cap V_{c_n}\) contains \(x\) and misses \(C\), because it is disjoint from every \(U_{c_i}\). Thus every point outside \(C\) has an open neighborhood outside \(C\), so \(X\setminus C\) is open and \(C\) is closed.

Next, let \(F\) be closed in a compact space \(C\), and take an open cover of \(F\) in the subspace topology. Each member is the intersection of \(F\) with an open set of \(C\). Those open sets, together with \(C\setminus F\), cover \(C\). A finite subcover of \(C\) therefore gives a finite subcover of \(F\). Finally, for compact subsets \(C_1,\ldots,C_n\) of \(X\), an open cover of their union restricts to an open cover of each \(C_i\). Each restriction has a finite subcover, and the union of these finitely many finite selections covers \(C_1\cup\cdots\cup C_n\). Hence the finite union is compact. \(\square\)

Verifying the Topology Axioms

Theorem: The collection of subsets of \(X^*\) declared open above is a topology.

Proof. The empty set is open by clause (i), and \(X^*\) is open because its complement in \(X\) is \(\varnothing\), which is compact.

Consider an arbitrary family \((V_i)_{i\in J}\) of declared open sets. If none contains \(\infty\), their union is a union of open sets of \(X\), hence is open in \(X\) and satisfies clause (i). If at least one contains \(\infty\), let \(J_\infty=\{i\in J:\infty\in V_i\}\), which is nonempty, and choose \(j\in J_\infty\). For each \(i\in J_\infty\), put \(K_i=X\setminus V_i\); each \(K_i\) is compact and therefore closed in \(X\) by the lemma. For \(i\notin J_\infty\), \(V_i\) is open in \(X\), so \(X\setminus V_i\) is closed in \(X\). The complement in \(X\) of the full union is \(\bigcap_{i\in J}(X\setminus V_i)\). This intersection is closed in \(X\), and it is contained in \(K_j\). It is therefore a closed subset of the compact set \(K_j\), so it is compact by the lemma. The union contains \(\infty\) and has compact complement in \(X\), so it is open by clause (ii).

Now take two declared open sets \(V\) and \(W\). If neither contains \(\infty\), their intersection is open in \(X\). If exactly one contains \(\infty\), write that set as \(N_K\), and call the other set \(U\), where \(U\) is open in \(X\). Then \(V\cap W=U\setminus K\). The compact set \(K\) is closed in \(X\), so \(U\setminus K=U\cap(X\setminus K)\) is open in \(X\), satisfying clause (i). If both contain \(\infty\), write them as \(N_{K_1}\) and \(N_{K_2}\). Their intersection is \(N_{K_1\cup K_2}\), since a point of \(X\) is in both exactly when it is outside both \(K_1\) and \(K_2\). The set \(K_1\cup K_2\) is compact by the lemma, so this intersection is open by clause (ii). The collection contains the empty set and \(X^*\), and is closed under arbitrary unions and finite intersections. It is a topology. \(\square\)

The proof uses Hausdorffness for a specific reason: it makes compact subsets closed. That closedness is needed when an infinity-neighborhood is intersected with an open subset of \(X\), and when checking complements of arbitrary unions. The argument does not claim that compactness of \(X^*\) follows from these topology axioms.

The Original Space Inside the Extension

Proposition: The inclusion \(i:X\to X^*\), defined by \(i(x)=x\), identifies \(X\) with an open dense subspace of \(X^*\), and the subspace topology on \(X\) is \(\mathcal{T}\).

Proof. The set \(X\subseteq X^*\) is open by clause (i), since \(X\) is open in itself. If \(V\) is open in \(X^*\) and does not contain \(\infty\), then \(V\in\mathcal{T}\). If \(V=N_K\) contains \(\infty\), then \(V\cap X=X\setminus K\), which is open in \(X\) because \(K\) is compact and hence closed. Thus every intersection of an open set of \(X^*\) with \(X\) belongs to \(\mathcal{T}\). Conversely every \(U\in\mathcal{T}\) is open in \(X^*\) by clause (i), so the induced subspace topology is exactly \(\mathcal{T}\).

To prove density, it suffices to show that every open neighborhood of every point of \(X^*\) meets \(X\). If the point is \(x\in X\), every neighborhood of \(x\) contains \(x\), so it meets \(X\). If the point is \(\infty\), an open neighborhood has the form \(N_K\) for compact \(K\subseteq X\). Because \(X\) is noncompact, \(K\ne X\), so \(X\setminus K\ne\varnothing\) and \(N_K\cap X\ne\varnothing\). Hence every open neighborhood of every point of \(X^*\) meets \(X\), which is the neighborhood criterion for \(X\) to be dense in \(X^*\). \(\square\)

Worked Examples

Worked Example: Neighborhoods of Infinity in the Real Line

Let \(X=\mathbb{R}\) with its standard topology. This is a noncompact, locally compact Hausdorff space. For compactness and local compactness of closed bounded intervals, we use the standard Heine–Borel theorem: a subset of \(\mathbb{R}\) is compact if and only if it is closed and bounded. Noncompactness follows, for example, from the open cover \(\{(-n,n):n\in\mathbb{N}\}\): any finite subfamily is contained in \((-N,N)\) for its largest index \(N\), and so does not cover \(\mathbb{R}\).

Take \(K=[-2,3]\). It is closed and bounded, hence compact by Heine–Borel. Its corresponding neighborhood is \(N_K=\{\infty\}\cup((-\infty,-2)\cup(3,\infty))\). The endpoints \(-2\) and \(3\) are excluded because they belong to \(K\). For example, \(-3\in N_K\), \(3\notin N_K\), and \(\infty\in N_K\).

Taking \(K=\{0\}\), which is compact because a singleton has a finite subcover from any open cover, gives \(N_{\{0\}}=\{\infty\}\cup(\mathbb{R}\setminus\{0\})\). Here \(0\notin N_{\{0\}}\), while \(1\in N_{\{0\}}\). Both sets are open in \(X^*\) because their omitted subsets of \(\mathbb{R}\) are compact.

Worked Example: The Discrete Natural Numbers

Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with the discrete topology, in which every subset is open. A subset \(K\subseteq\mathbb{N}\) is compact if and only if it is finite. Indeed, if \(K\) is finite, any open cover has a finite subcover by choosing one cover member for each point. If \(K\) is infinite, the cover by its open singletons \(\{\{k\}:k\in K\}\) has no finite subcover. Thus \(X\) is noncompact. It is Hausdorff, and each singleton is an open compact neighborhood, so it is locally compact.

For \(K=\{2,5\}\), the set \(N_K=\{\infty\}\cup(\mathbb{N}\setminus\{2,5\})\) is open. Membership is determined exactly by exclusion of those two points: \(2\notin N_K\), \(5\notin N_K\), and \(1,3,\infty\in N_K\). In contrast, \(N_{\{2,5,8\}}\) omits exactly \(2,5,8\); the strictness is exclusion, not inclusion, because each of those points belongs to the compact complement.

The set \(\{1,3\}\), which does not contain \(\infty\), is also open in \(X^*\) because it is open in the discrete topology on \(X\). Its complement in \(X^*\) need not be compact for clause (i) to apply: the compact-complement condition is required only for open sets containing \(\infty\).

Worked Example: Intersections and a Non-Open Candidate

Again let \(X=\mathbb{R}\) with the standard topology. Set \(K_1=[0,1]\) and \(K_2=[2,4]\). Both are compact by Heine–Borel. The intersection of their infinity-neighborhoods is \(N_{K_1}\cap N_{K_2}=N_{K_1\cup K_2}\), where \(K_1\cup K_2=[0,1]\cup[2,4]\) is compact. For example, \(1\notin N_{K_1}\cap N_{K_2}\), \(2\notin N_{K_1}\cap N_{K_2}\), and \(5\in N_{K_1}\cap N_{K_2}\).

The sets \(K_1\) and \(K_2\) are disjoint, so \(K_1\cap K_2=\varnothing\). Therefore \(N_{K_1}\cup N_{K_2}=N_{\varnothing}=X^*\): every real number is outside at least one of the two compact sets, and \(\infty\) belongs to both neighborhoods. This computes the union directly and agrees with the arbitrary-union axiom.

For a mixed intersection, take \(U=(0,3)\), open in \(\mathbb{R}\), and \(N_{[1,2]}\). Their intersection is \(U\cap N_{[1,2]}=(0,1)\cup(2,3)\). It is open in \(\mathbb{R}\), as required by clause (i). In particular, \(1\) and \(2\) are excluded, while \(1/2\) and \(5/2\) are included.

A Common Pitfall

Common mistake: treating every bounded set as compact and therefore declaring \(\{\infty\}\cup(\mathbb{R}\setminus(0,1))\) open. The omitted set \((0,1)\) is bounded, but it is not compact: in the Hausdorff space \(\mathbb{R}\), every compact set is closed, whereas \((0,1)\) is not closed. For instance, every neighborhood of \(0\) meets \((0,1)\), but \(0\notin(0,1)\). The proposed set is therefore not open under the definition. The correction is to use a compact omitted set, such as \([0,1]\); then \(\{\infty\}\cup(\mathbb{R}\setminus[0,1])\) is open. Boundedness alone does not suffice: in the real line, Heine–Borel requires both closedness and boundedness.

Takeaway: The construction declares old open sets open and gives \(\infty\) precisely the neighborhoods whose omitted parts are compact. The Hausdorff hypothesis ensures those compact parts are closed, which is essential to verifying the topology axioms. The original topology is recovered on the open subspace \(X\), and noncompactness ensures that every neighborhood of \(\infty\) still meets \(X\).

Check Your Understanding

For each question, state which part of the definition or proof justifies your answer.

In the one-point extension of \(\mathbb{R}\), is \(\{\infty\}\cup(\mathbb{R}\setminus[-1,2])\) open? Which real endpoints does it omit?
Why does the compact-complement condition apply to an open set containing \(\infty\), but not to an open set \(U\subseteq X\) that omits \(\infty\)?
In the discrete topology on \(\mathbb{N}\), determine whether the complement of \(\{\infty\}\cup(\mathbb{N}\setminus\{3,7\})\) in \(X^*\) is compact.
In the topology-axiom proof, where is Hausdorffness used, and what fact about compact subsets does it provide?
Explain why every open neighborhood of \(\infty\) meets \(X\) when \(X\) is noncompact.