Choosing a Compactification
A compactification is useful when it makes a noncompact space part of a compact Hausdorff space without changing the original topology. There are two common choices with different strengths. The one-point compactification adds a single point and is available under a local compactness condition. The Stone–Čech compactification can add many points, but it is designed to extend every continuous map from the original space into a compact Hausdorff space.
Throughout, a compactification of a space \(X\) means a compact Hausdorff space \(K\) together with an embedding \(i:X\to K\) whose image is dense in \(K\). An embedding is a map that is a homeomorphism from \(X\) to its image with the subspace topology. When convenient, we identify \(X\) with \(i(X)\). A space is locally compact if every point has a compact neighborhood. A Hausdorff space \(X\) is completely regular if, whenever \(x\in X\) and \(F\subseteq X\) is closed with \(x\notin F\), there is a continuous \(u:X\to[0,1]\) such that \(u(x)=0\) and \(u(y)=1\) for every \(y\in F\). A completely regular Hausdorff space is also called Tychonoff.
The Stone–Čech universal property is a standard result established in the earlier material on Stone–Čech compactifications. We will use its precise extension statement below. The one-point construction and the tests for when a map extends to its added point will be verified directly here.
Constructing and Checking the One-Point Compactification
Suppose \(X\) is noncompact and Hausdorff. Add a point \(\infty\notin X\). Keep every open subset of \(X\) open, and declare a subset containing \(\infty\) open when its complement in \(X\) is compact. In a Hausdorff space compact subsets are closed, so these rules define a topology. The resulting space is denoted \(X^*=X\cup\{\infty\}\).
Proof. Let an open cover of \(X^*\) be given. One member \(W\) contains \(\infty\), so \(C=X^*\setminus W\) is a compact subset of \(X\). The other members of the cover cover \(C\), and finitely many of them cover \(C\). Those finitely many sets together with \(W\) cover \(X^*\); hence \(X^*\) is compact.
Two points of \(X\) have disjoint open neighborhoods in \(X\), since \(X\) is Hausdorff. To separate \(x\in X\) from \(\infty\), local compactness gives a compact neighborhood \(K\) of \(x\); choose an open set \(U\) with \(x\in U\subseteq K\). Then \(U\) and \(\{\infty\}\cup(X\setminus K)\) are disjoint open neighborhoods of \(x\) and \(\infty\). Thus \(X^*\) is Hausdorff. Finally, every open neighborhood of \(\infty\) has compact complement in \(X\). That complement cannot equal \(X\), because \(X\) is noncompact, so the neighborhood meets \(X\). Open sets in \(X\) also meet \(X\). Thus every nonempty open set of \(X^*\) meets \(X\), which means \(X\) is dense. \(\square\)
The one-point compactification criterion from the earlier material gives the converse choice test: for a noncompact Hausdorff space, a Hausdorff one-point compactification is available exactly when the space is locally compact. It is important to check local compactness, rather than infer it just from the fact that \(X\) is a subspace of a familiar space.
Worked Example: Compactifying the Real Line
Give \(\mathbb{R}\) its standard topology. Every \(x\in\mathbb{R}\) has the compact neighborhood \([x-1,x+1]\), so \(\mathbb{R}\) is locally compact. It is noncompact, since the open cover \(\{(-n,n):n\in\mathbb{N}\}\) has no finite subcover. Its one-point compactification is therefore a compact Hausdorff space.
To identify it, map \(\mathbb{R}\) to the unit circle \(S^1\subseteq\mathbb{R}^2\) by \(t\mapsto (2t/(1+t^2),(t^2-1)/(1+t^2))\), and map \(\infty\) to \((0,1)\). The image of \(\mathbb{R}\) is \(S^1\setminus\{(0,1)\}\); the map is a homeomorphism onto that image. As \(t\to+\infty\) or \(t\to-\infty\), the first coordinate tends to \(0\) and the second tends to \(1\). The map therefore identifies the one-point compactification of \(\mathbb{R}\) with \(S^1\).
Consider the continuous function \(f:\mathbb{R}\to[0,1]\) given by \(f(t)=1/(1+t^2)\). Define its proposed value at infinity to be \(0\). For every \(\varepsilon>0\), if \(\varepsilon\geq1\) then \(f(t)<\varepsilon\) for all \(t\), since \(f(t)\leq1\) and, when \(\varepsilon=1\), equality occurs only at \(t=0\); the neighborhood \(\mathbb{R}^* \setminus\{0\}\) of infinity removes that point. If \(0<\varepsilon<1\), then \(f(t)<\varepsilon\) whenever \(t^2>1/\varepsilon-1\). In either case, the values of \(f\) lie within \(\varepsilon\) of \(0\) outside a compact subset of \(\mathbb{R}\). This proves continuity at infinity of the extended map.
Check: for \(0<\varepsilon<1\), choose \(R>\sqrt{1/\varepsilon-1}\). If \(|t|>R\), then \(1+t^2>1/\varepsilon\), so \(f(t)<\varepsilon\). The exceptional interval \([-R,R]\) is compact, as required by the neighborhood definition at infinity.
When Does a Map Extend to Infinity?
There is an exact test for extending a map over the added point. Let \(X\) have its one-point compactification \(X^*\), let \(Y\) be a topological space, let \(f:X\to Y\) be continuous, and choose \(y_\infty\in Y\). Define \(\widetilde f:X^*\to Y\) by \(\widetilde f|_X=f\) and \(\widetilde f(\infty)=y_\infty\).
Proof. Continuity of \(\widetilde f\) at points of \(X\) holds because \(f\) is continuous and \(X\) has its original subspace topology. At \(\infty\), continuity means that \(\widetilde f^{-1}(V)\) is an open neighborhood of \(\infty\) for every open neighborhood \(V\) of \(y_\infty\). Its complement in \(X\) is exactly \(X\setminus f^{-1}(V)\). By the definition of the topology at \(\infty\), this preimage is an open neighborhood of \(\infty\) exactly when that complement is compact. This proves both implications. \(\square\)
When \(Y\) is compact Hausdorff, this condition says that the map eventually stays in every neighborhood of the proposed value: the part of \(X\) whose images lie outside the neighborhood must fit inside a compact subset. It does not say that every continuous map extends to \(X^*\). The Stone–Čech compactification has a stronger extension property.
Worked Example: Why One Point Is Not Enough for the Integers
Give \(\mathbb{N}\) the discrete topology. It is locally compact because every singleton is an open compact neighborhood, and it is noncompact because the cover by singletons has no finite subcover. In \(\mathbb{N}^*\), a neighborhood of \(\infty\) has finite complement in \(\mathbb{N}\), so it contains all but finitely many integers.
Define \(g:\mathbb{N}\to\{0,1\}\) by \(g(n)=0\) for even \(n\) and \(g(n)=1\) for odd \(n\), where \(\{0,1\}\) has the discrete topology. This map is continuous because every subset of \(\mathbb{N}\) is open. It cannot extend continuously over \(\infty\). Indeed, if the proposed value were \(0\), continuity would require a neighborhood of \(\infty\) mapped into the open set \(\{0\}\); that would require all but finitely many integers to be even. There are infinitely many odd integers. If the proposed value were \(1\), the same reasoning would require all but finitely many integers to be odd, but there are infinitely many even integers.
By contrast, \(\beta\mathbb{N}\) extends \(g\). The extension is unique, and it takes values in \(\{0,1\}\): it is a continuous extension into \([0,1]\) by the Stone–Čech extension property, and its image lies in the closed set \(\{0,1\}\) because it does so on the dense subset \(\mathbb{N}\). This example shows why the number of added points matters: a single point cannot record both persistent values of \(g\) at infinity.
Constructing the Stone–Čech Compactification
For a Tychonoff space \(X\), let \(C(X,[0,1])\) be the set of continuous functions from \(X\) to \([0,1]\). The evaluation map puts all these functions into one product at once:
The product has the product topology. Complete regularity ensures that these coordinates separate points from closed sets, which makes \(e\) an embedding; in particular, it separates any two distinct points. The Stone–Čech construction takes the closure of the image in this product:
This construction uses the Tychonoff theorem that an arbitrary product of compact spaces is compact; its proof uses the axiom of choice. Thus \(\beta X\) is compact, and it is Hausdorff as a subspace of a product of Hausdorff spaces. It contains the dense embedded copy \(e(X)\). The standard Stone–Čech universal property says that, for every compact Hausdorff space \(K\) and continuous \(f:X\to K\), there is a unique continuous \(\overline f:\beta X\to K\) with \(\overline f\circ e=f\). For each \(u\in C(X,[0,1])\), the corresponding product projection gives the extension of \(u\) directly. The full assertion for compact Hausdorff targets is the Stone–Čech universal property.
Worked Example: A Non-Locally-Compact Space
Give \(\mathbb{Q}\) the subspace topology inherited from \(\mathbb{R}\). It is Tychonoff because it is a metric space, so \(\beta\mathbb{Q}\) exists. But \(\mathbb{Q}\) is not locally compact, so its Hausdorff one-point compactification criterion fails.
To check the local compactness failure at \(0\), suppose there were a compact neighborhood \(K\) of \(0\) in \(\mathbb{Q}\). Then \(K\) would contain a relative open neighborhood \(\mathbb{Q}\cap(-r,r)\) for some \(r>0\). The inclusion \(\mathbb{Q}\to\mathbb{R}\) is continuous, so \(K\), compact in \(\mathbb{Q}\), is compact in \(\mathbb{R}\). Compact subsets of the Hausdorff space \(\mathbb{R}\) are closed. Since \(K\) contains \(\mathbb{Q}\cap(-r,r)\), it must contain its closure \([-r,r]\) in \(\mathbb{R}\). This is impossible because \(K\subseteq\mathbb{Q}\) while \([-r,r]\) contains irrational numbers. Thus no such compact neighborhood exists.
For a concrete extension test, the bounded function \(u(q)=q^2/(1+q^2)\) is continuous from \(\mathbb{Q}\) to \([0,1]\). It extends over \(\beta\mathbb{Q}\) by the Stone–Čech property. The formula alone does not prescribe values on the remainder \(\beta\mathbb{Q}\setminus\mathbb{Q}\); existence and uniqueness of those values are supplied by that property.
A Practical Selection Rule and a Common Pitfall
When choosing a compactification, first identify the required behavior of maps, then check the hypotheses.
- If adding exactly one point is the goal, check that \(X\) is noncompact, Hausdorff, and locally compact. Then use \(X^*\).
- If every continuous map into every compact Hausdorff target must extend, check that \(X\) is Tychonoff. Then use \(\beta X\).
- If a particular map must extend to \(X^*\), specify its proposed value at infinity and apply the compact-complement test to every neighborhood of that value.
- If a particular map into a compact Hausdorff space must extend and \(X\) is Tychonoff, the Stone–Čech universal property gives a unique extension.
Common pitfall. It is incorrect to assume that every bounded continuous real-valued function extends over the one-point compactification. For the alternating function \(g:\mathbb{N}\to\{0,1\}\) above, boundedness and continuity both hold, but continuity at a proposed added point would force one of the two infinite fibers to be finite. The correction is to match the compactification to the extension requirement: the one-point construction only records one limiting value, whereas Stone–Čech extends every continuous map into a compact Hausdorff target.
Check Your Understanding
For each question, state the topology and check the relevant hypothesis or extension condition.
- Why does the one-point compactification criterion apply to \(\mathbb{R}\) with its standard topology?
- For a map \(f:X\to Y\) and proposed value \(y_\infty\), write the exact compactness condition that tests continuity at infinity.
- Why does the alternating map from discrete \(\mathbb{N}\) to \(\{0,1\}\) fail to extend over \(\mathbb{N}^*\)?
- Which condition fails for \(\mathbb{Q}\) with its usual subspace topology, and which condition allows its Stone–Čech compactification?
- State the target-space and domain hypotheses in the Stone–Čech extension property.