What “Unique” Means
A one-point compactification adds one point to a locally compact, noncompact Hausdorff space so that the resulting space is compact Hausdorff and the original space is dense. The added point is often called the point at infinity. A construction may present the original space and the new point using different names or sets; uniqueness means that these presentations have the same topology once we match the original points and match the added points.
Throughout, a compactification of \(X\) means a compact Hausdorff space \(Y\) containing a homeomorphic copy of \(X\) as a dense subspace. For clarity, we identify each copy with \(X\) itself, so \(X\subseteq Y\). A subset is dense if every nonempty open set meets it. Assume that \((X,\mathcal{T})\) is locally compact, Hausdorff, and noncompact, and that \(Y_1\) and \(Y_2\) are compact Hausdorff compactifications with \(Y_1\setminus X=\{p_1\}\) and \(Y_2\setminus X=\{p_2\}\).
The map is a bijection: it is the identity between the two copies of \(X\), and it matches the only point outside one copy with the only point outside the other. The substantive question is whether it preserves open sets in both directions. The key is that compact subsets of \(X\) control neighborhoods of the added point.
Compact Subsets and Neighborhoods at Infinity
We use a basic fact about compact Hausdorff spaces: every compact subset \(K\) of a Hausdorff space \(Y\) is closed. To see why, take \(y\in Y\setminus K\). For each \(k\in K\), the Hausdorff property gives disjoint open neighborhoods \(U_k\) of \(k\) and \(V_k\) of \(y\). The sets \(U_k\) cover \(K\), so compactness gives finitely many, \(U_{k_1},\ldots,U_{k_m}\), that cover \(K\). The open neighborhood \(V_{k_1}\cap\cdots\cap V_{k_m}\) of \(y\) misses all of \(K\). Thus every point outside \(K\) has a neighborhood disjoint from \(K\), which says \(Y\setminus K\) is open.
In particular, \(X\) is open in each \(Y_i\). Indeed, a singleton in a Hausdorff space is closed: for each other point, the Hausdorff property supplies a neighborhood missing that singleton, so its complement is open. Hence \(\{p_i\}\) is closed and its complement \(X\) is open.
Proof. If \(U\) is open and contains \(p_i\), then \(C=Y_i\setminus U\) is closed in the compact space \(Y_i\), hence compact. It misses \(p_i\), so \(C\subseteq X\). The subspace topology on \(X\) is the original topology, so \(C\) is compact as a subset of \(X\) as well. Conversely, let \(K\subseteq X\) be compact in \(X\). The inclusion of \(X\) into \(Y_i\) is continuous, so \(K\) is compact in \(Y_i\). Since \(Y_i\) is Hausdorff, the fact just proved makes \(K\) closed in \(Y_i\). Therefore \(Y_i\setminus K\) is open and contains \(p_i\). \(\square\)
Consequently, the neighborhoods of the added point have exactly the same description in either compactification: they are open sets whose complements are compact subsets of \(X\). We can now prove that the canonical map preserves the topology.
The Uniqueness Theorem
Proof. The map \(h\) is a bijection, as noted above. We prove continuity by taking an arbitrary open set \(U\subseteq Y_2\) and showing that \(h^{-1}(U)\) is open in \(Y_1\).
First suppose \(p_2\notin U\). Then \(U\subseteq X\), and \(U\) is open in the subspace \(X\) because it is open in \(Y_2\). The restriction of \(h\) to \(X\) is the identity, so \(h^{-1}(U)\) is the same relatively open subset of \(X\). Since \(X\) is open in \(Y_1\), every relatively open subset of \(X\) is open in \(Y_1\): it has the form \(X\cap W\) for some open \(W\subseteq Y_1\), and \(X\cap W\) is an intersection of open sets. Thus \(h^{-1}(U)\) is open.
Now suppose \(p_2\in U\). The proposition shows that \(K=Y_2\setminus U\) is a compact subset of \(X\). The same set \(K\), viewed in \(Y_1\), is compact because the topology induced on \(X\) is unchanged. Since \(Y_1\) is Hausdorff, \(K\) is closed in \(Y_1\). The map \(h\) fixes \(X\) and sends \(p_1\) to \(p_2\), so \(h^{-1}(U)=Y_1\setminus K\). This is open in \(Y_1\). Both cases prove continuity of \(h\).
For the inverse, interchange \(Y_1\) and \(Y_2\). The same two cases show that \(h^{-1}:Y_2\to Y_1\) is continuous. Therefore \(h\) is a homeomorphism. It is the only map that fixes every point of \(X\) and sends \(p_1\) to \(p_2\), since these requirements specify its value at every point of \(Y_1\). \(\square\)
Worked Examples: Applying the Theorem
Worked Example: Two Presentations for the Real Line
Let \(X=\mathbb{R}\) with its standard topology. Form two spaces \(Y_1=\mathbb{R}\cup\{p_1\}\) and \(Y_2=\mathbb{R}\cup\{p_2\}\), using the one-point compactification topology in each: open subsets not containing \(p_i\) are open in \(\mathbb{R}\), and neighborhoods of \(p_i\) have compact complement in \(\mathbb{R}\). The real line is locally compact, Hausdorff, and noncompact, so the one-point compactification criterion gives compact Hausdorff spaces \(Y_1,Y_2\).
The canonical map is \(h(x)=x\) for \(x\in\mathbb{R}\) and \(h(p_1)=p_2\). Consider the open neighborhood \(U=\{p_2\}\cup(\mathbb{R}\setminus[-2,3])\). Its complement is \([-2,3]\), a compact subset of \(\mathbb{R}\) by the Heine–Borel theorem, so \(U\) is open by the specified topology. The preimage is \(h^{-1}(U)=\{p_1\}\cup(\mathbb{R}\setminus[-2,3])\), also open because its complement is the same compact interval.
For an open set that does not contain the added point, take \(V=(0,1)\subseteq Y_2\). Its preimage is \(h^{-1}(V)=(0,1)\), which is open in \(Y_1\) because \(\mathbb{R}\) is open in \(Y_1\) and \((0,1)\) is open in the subspace \(\mathbb{R}\). These calculations illustrate the two cases in the proof. In particular, the map does not depend on the names \(p_1\) and \(p_2\).
Worked Example: Testing a Neighborhood at Infinity
Continue with the two compactifications of \(\mathbb{R}\). Let \(K=[0,1]\) and \(W=Y_2\setminus K\). The interval \([0,1]\) is compact by Heine–Borel, so \(W\) is an open neighborhood of \(p_2\). Its preimage is \(h^{-1}(W)=Y_1\setminus[0,1]\). The complement is compact in \(Y_1\), hence closed there, so this preimage is open and contains \(p_1\).
In contrast, consider \(V=(0,1)\cup\{p_2\}\). Its complement in \(Y_2\) is \((-\infty,0]\cup[1,\infty)\), which is not compact in \(\mathbb{R}\). For example, the open sets \((-n,n)\), for \(n\in\mathbb{N}\), cover \((-\infty,0]\cup[1,\infty)\), but no finite subcollection covers it: a finite subcollection has a largest index \(N\), and misses every point greater than \(N\). Thus \(V\) is not a neighborhood of \(p_2\) in the one-point compactification topology. The compact-complement test distinguishes genuine neighborhoods of infinity from sets that merely contain infinity.
Worked Example: Why Hausdorffness Cannot Be Dropped
Here is a finite example showing that compactness alone does not imply uniqueness. Let \(X=\{a,b\}\) with the discrete topology, and let \(Y=\{a,b,p\}\). Give \(Y\) either of these topologies:
\(\mathcal{T}_1=\mathcal{P}(X)\cup\{Y\}\), or \(\mathcal{T}_2=\mathcal{P}(X)\cup\{U\subseteq Y:p\in U\text{ and }a\in U\}\). Each collection is a topology: it contains \(\varnothing,Y\), and unions and finite intersections of its open sets remain in the collection. In both spaces \(X\) is an open subspace with the discrete topology. It is dense because every nonempty open set containing \(p\) also meets \(X\). Both spaces are compact, since a finite space has a finite subcover from every open cover.
Neither topology is Hausdorff. In \((Y,\mathcal{T}_1)\), every open neighborhood of \(p\) is \(Y\), so it cannot be separated from \(a\) by disjoint open neighborhoods. In \((Y,\mathcal{T}_2)\), every open neighborhood of \(p\) contains \(a\), so again \(p\) and \(a\) have no disjoint open neighborhoods. The identity bijection on \(Y\), which fixes \(X\) and \(p\), is not a homeomorphism: \(\{a,p\}\in\mathcal{T}_2\), but \(\{a,p\}\notin\mathcal{T}_1\). Thus the Hausdorff hypothesis in the uniqueness theorem is essential.
A Common Pitfall
When checking continuity, it is not enough to observe that the part of a set inside \(X\) is open in \(X\). For instance, in the one-point compactification of \(\mathbb{R}\), the set \(V=(0,1)\cup\{p\}\) has \(V\cap\mathbb{R}=(0,1)\), which is open in \(\mathbb{R}\). A mistaken argument might conclude that \(V\) is open in the compactification. But its complement \((-\infty,0]\cup[1,\infty)\) is not compact, as the open cover by \((-n,n)\), \(n\in\mathbb{N}\), has no finite subcover. Therefore \(V\) is not open. The correct continuity proof separates open sets according to whether they contain the added point; in the latter case it checks the compactness of the complement.
The uniqueness result matters because it shows that the one-point compactification is determined by \(X\), not by the choice of a particular construction or label for infinity. The compact-complement description is the mechanism: both spaces have exactly the same neighborhoods at infinity when those neighborhoods are read through the shared copy of \(X\).
Check Your Understanding
Use the compact-complement description and the two cases in the continuity proof to answer the following.
- Why is \(X\) open in each compact Hausdorff compactification with a one-point remainder?
- Let \(K\) be compact in \(X\). Explain why \(Y_i\setminus K\) is an open neighborhood of \(p_i\).
- In the uniqueness proof, why must open sets containing \(p_2\) be handled separately from open sets contained in \(X\)?
- In the real-line example, decide whether \(\{p_2\}\cup(\mathbb{R}\setminus[1,4])\) is open in the one-point compactification, and justify your answer.
- In the finite example, identify an open set in \(\mathcal{T}_2\) that is not open in \(\mathcal{T}_1\), and explain why this prevents the identity bijection from being a homeomorphism.