Continuity at Infinity
A continuous map can behave badly far from every compact part of its domain. The one-point compactification makes this issue visible: points that escape all compact subsets approach a single added point, \(\infty\). For the extension to be continuous while sending \(\infty\) to \(\infty\), points far out in the domain must be sent far out in the target. The precise condition is properness.
Throughout, let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be noncompact locally compact Hausdorff spaces. A map \(f:X\to Y\) is proper if \(f^{-1}(K)\) is compact in \(X\) for every compact set \(K\subseteq Y\). This condition concerns preimages of all compact sets, not just bounded sets or individual points.
In a Hausdorff space, compact sets are closed. To recall why, if \(C\) is compact and \(x\notin C\), separate \(x\) from each \(c\in C\) by disjoint open sets. The open sets around points of \(C\) have a finite subcover; intersect the corresponding finitely many open neighborhoods of \(x\). This gives a neighborhood of \(x\) disjoint from \(C\), so \(X\setminus C\) is open. Thus the complements used in the definition of \(X^*\) are open in \(X\). Local compactness and Hausdorffness are the standard hypotheses under which this construction gives the one-point compactification.
The Properness Criterion
Proof. First consider a point \(x\in X\). If \(O\) is open in \(Y^*\) and \(f^*(x)=f(x)\in O\), then \(O\cap Y\) is open in \(Y\). Continuity of \(f\) gives the open set \(f^{-1}(O\cap Y)\) containing \(x\). It is also open in \(X^*\), since \(X\) is open in \(X^*\), and it is contained in \((f^*)^{-1}(O)\). Thus \(f^*\) is continuous at every point of \(X\).
It remains to check continuity at \(\infty_X\). For every compact \(K\subseteq Y\), the set \(Y^*\setminus K\) is an open neighborhood of \(\infty_Y\), and its inverse image is exactly $$ (f^*)^{-1}(Y^*\setminus K)=X^*\setminus f^{-1}(K). $$ By the definition of the topology on \(X^*\), this inverse image is open if and only if \(f^{-1}(K)\) is compact in \(X\). Therefore, if \(f\) is proper, the inverse image of every basic neighborhood \(Y^*\setminus K\) is an open neighborhood of \(\infty_X\). This proves continuity at \(\infty_X\), and hence continuity everywhere.
Conversely, suppose \(f^*\) is continuous. For each compact \(K\subseteq Y\), the set \(Y^*\setminus K\) is open, so its inverse image \(X^*\setminus f^{-1}(K)\) is open. Its complement in \(X^*\) is \(f^{-1}(K)\), which is a subset of \(X\). By the definition of the topology at infinity, that complement must be compact in \(X\). This holds for every compact \(K\), so \(f\) is proper. \(\square\)
The calculation in the proof is the key point: continuity at ordinary points comes from continuity of \(f\), while continuity at infinity is precisely the compact-preimage condition. No extra condition about a formula or a limit is needed.
Worked Example: The Cubic Map
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^3\), with both copies of \(\mathbb{R}\) carrying the standard topology. We check that \(f\) is proper, so it extends continuously by \(f^*(\infty_{\mathbb{R}})=\infty_{\mathbb{R}}\).
Let \(K\subseteq\mathbb{R}\) be compact. If \(K=\varnothing\), then \(f^{-1}(K)=\varnothing\), which is compact. Otherwise \(K\) is closed and bounded, so choose \(M\geq 0\) with \(|y|\leq M\) for every \(y\in K\). If \(x\in f^{-1}(K)\), then \(|x^3|\leq M\), hence \(|x|\leq M^{1/3}\). Also \(f^{-1}(K)\) is closed because \(f\) is continuous and \(K\) is closed. It is therefore closed and bounded in \(\mathbb{R}\), hence compact by the Heine–Borel theorem. This verifies properness for every compact \(K\).
Check: for \(K=[-8,1]\), the condition \(-8\leq x^3\leq 1\) is equivalent to \(-2\leq x\leq 1\), since cubing preserves order. Thus \(f^{-1}([-8,1])=[-2,1]\), which is compact. The endpoint values are \(f(-2)=(-2)^3=-8\) and \(f(1)=1^3=1\), as required.
Continuous Maps That Do Not Extend
Continuity of \(f\) alone does not control what happens at infinity. The next examples show how a compact target set can have a noncompact preimage, and how that failure appears directly in the topology of the compactifications.
Worked Example: A Constant Map
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=0\). This map is continuous, but it is not proper: the compact set \(K=\{0\}\) has preimage \(f^{-1}(K)=\mathbb{R}\), which is not compact. For example, the open cover \(\{(-n,n):n\in\mathbb{N}\}\) of \(\mathbb{R}\) has no finite subcover: a finite subcollection has a largest index \(N\), and its union is just \((-N,N)\), which does not cover \(\mathbb{R}\).
The failure also rules out continuity of the canonical extension. The set \(\mathbb{R}^*\setminus\{0\}\) is an open neighborhood of \(\infty_{\mathbb{R}}\). Its inverse image under \(f^*\) is exactly \(\{\infty_{\mathbb{R}}\}\), because every point of \(\mathbb{R}\) maps to \(0\). This singleton is not open in \(\mathbb{R}^*\): its complement is \(\mathbb{R}\), which is not compact. Thus \(f^*\) is not continuous at infinity.
Worked Example: The Inclusion of an Open Interval
Let \(i:(0,1)\to\mathbb{R}\) be the inclusion, where \((0,1)\) has the subspace topology and \(\mathbb{R}\) has its standard topology. Both spaces are locally compact and Hausdorff. Take the compact set \(K=[0,1]\subseteq\mathbb{R}\). Then \(i^{-1}(K)=(0,1)\), which is not compact.
To verify noncompactness, consider the open cover of \((0,1)\) by the sets \(U_n=(1/n,1)\), for \(n\geq 2\). These sets are open in the subspace topology and cover \((0,1)\): for any \(x\in(0,1)\), choose \(n\) with \(1/n<x\). A finite subcollection has a largest index \(N\), and its union is \(U_N=(1/N,1)\), which misses points in \((0,1/N]\). Hence the cover has no finite subcover.
Indeed, the open neighborhood \(\mathbb{R}^*\setminus[0,1]\) of \(\infty_{\mathbb{R}}\) has inverse image under \(i^*\) equal to \(\{\infty_{(0,1)}\}\). That singleton is not open in \((0,1)^*\), because its complement \((0,1)\) is not compact. The canonical extension is therefore discontinuous at infinity.
Composition and the Meaning of Properness
Proof. Let \(K\subseteq Z\) be compact. Properness of \(g\) implies that \(g^{-1}(K)\) is compact in \(Y\). Properness of \(f\) then implies that \(f^{-1}(g^{-1}(K))\) is compact in \(X\). Since \((g\circ f)^{-1}(K)=f^{-1}(g^{-1}(K))\), the composite is proper. For the equality of extensions, if \(x\in X\), both sides send \(x\) to \(g(f(x))\); at \(\infty_X\), both sides send \(\infty_X\) to \(\infty_Z\). The two maps agree at every point of \(X^*\). \(\square\)
Thus proper maps are exactly the continuous maps compatible with the added points at infinity, and this compatibility is preserved by composition. In practice, the test is not whether \(f\) is bounded or whether it has a limit along a particular sequence. The test is whether every compact target set has a compact preimage.
Common pitfall. One might argue that the constant map \(f(x)=0\) extends continuously because it is continuous everywhere on \(\mathbb{R}\). This checks only continuity at points of \(\mathbb{R}\). The actual inverse-image calculation for the neighborhood \(\mathbb{R}^*\setminus\{0\}\) gives \((f^*)^{-1}(\mathbb{R}^*\setminus\{0\})=\{\infty_{\mathbb{R}}\}\), which is not open. Continuity at infinity is a separate requirement, and properness is exactly what supplies it.
Check Your Understanding
Use the compact-preimage test and the topology of neighborhoods of infinity.
- For \(f:\mathbb{R}\to\mathbb{R}\) given by \(f(x)=x\), verify properness and describe its canonical extension.
- For \(f:\mathbb{R}\to\mathbb{R}\) given by \(f(x)=e^x\), test properness using the compact set \([0,1]\).
- Explain why the preimage of a compact set is the relevant test for continuity at infinity, rather than just the behavior of \(f(x)\) along one sequence.
- If \(f:X\to Y\) is proper and \(K\subseteq Y\) is compact, what does properness say about \(f^{-1}(K)\), including when \(K=\varnothing\)?
- Show directly from the definition that the composite of two proper maps has compact preimages of compact sets.