What the Remainder Measures
A compactification adds points so that a space sits densely inside a compact Hausdorff space. The Stone–Čech compactification adds enough points to extend every bounded continuous real-valued function. Its added part, the remainder, can therefore record many different ways in which points of the original space escape every compact region. For a locally compact noncompact space, we can prove that this remainder is itself compact, then see how much its size and structure vary in two examples.
Throughout, assume \(X\) is a locally compact Hausdorff space. We use \(\beta X\) for its Stone–Čech compactification and identify \(X\) with its embedded image in \(\beta X\). A compactification is a compact Hausdorff space containing \(X\) as a dense subspace. The Stone–Čech extension theorem, established in the earlier material on Stone–Čech compactifications, says that every bounded continuous function \(f:X\to\mathbb{R}\) has a continuous extension \(\beta f:\beta X\to\mathbb{R}\).
The General Remainder Theorem
Proof. Fix \(x\in X\). By local compactness, there is a compact set \(K\subseteq X\) and an open set \(U\) in \(X\) such that \(x\in U\subseteq K\). The inclusion of \(X\) in \(\beta X\) is an embedding, so there is an open set \(O\) in \(\beta X\) with \(O\cap X=U\). The set \(K\) is compact, and \(\beta X\) is Hausdorff, so \(K\) is closed in \(\beta X\). We claim \(O\subseteq K\). Otherwise \(O\setminus K\) would be a nonempty open subset of \(\beta X\). Since \(X\) is dense in \(\beta X\), it would meet \(X\), but \(O\cap X=U\subseteq K\), a contradiction. Thus \(x\in O\subseteq K\subseteq X\), and every \(x\in X\) has a \(\beta X\)-open neighborhood contained in \(X\). Hence \(X\) is open in \(\beta X\).
It follows that \(\beta X\setminus X\) is closed in the compact space \(\beta X\), so it is compact. If \(X\) is noncompact and \(\beta X\setminus X\) were empty, then \(\beta X=X\) as spaces, making \(X\) compact. This contradicts the assumption. Therefore the remainder is nonempty. \(\square\)
The theorem establishes compactness and nonemptiness, but does not say that the remainder is a single point, or that it has any particular shape. Those stronger conclusions depend on \(X\). The natural numbers give a description in terms of ultrafilters; the real line will show how bounded functions distinguish points of its remainder.
The Natural Numbers: Free Ultrafilters
For the discrete space \(\mathbb{N}\), the standard ultrafilter model of \(\beta\mathbb{N}\) identifies each \(n\in\mathbb{N}\) with its principal ultrafilter, consisting of the sets that contain \(n\). The other points are free ultrafilters. For \(A\subseteq\mathbb{N}\), write \(\widehat{A}\) for the set of ultrafilters containing \(A\). These sets form a clopen basis for \(\beta\mathbb{N}\), and \(\widehat{A}\cap\mathbb{N}=A\). In particular, writing \(A^*=\widehat{A}\cap(\beta\mathbb{N}\setminus\mathbb{N})\), we have a basic description of neighborhoods in the remainder.
Worked Example: A Basic Set in the Natural-Number Remainder
Let \(A=\{2,4,6,\ldots\}\), the set of positive even integers. The basic open set \(\widehat{A}\) consists of the ultrafilters that contain \(A\). Its principal ultrafilters are exactly the even integers, so \(\widehat{A}\cap\mathbb{N}=A\). Consequently \(A^*\) consists of the free ultrafilters containing the even integers.
The set \(A\) is infinite, and \(A^*\ne\varnothing\). To justify this existence, the cofinite filter on \(A\) extends to a maximal proper filter by Zorn’s lemma: the union of a chain of proper filters extending it is again a proper filter, so Zorn’s lemma applies. The resulting ultrafilter, viewed as an ultrafilter on \(\mathbb{N}\), contains \(A\) and contains no finite set. It is therefore a point of \(A^*\). By contrast, if \(A\) is finite, no free ultrafilter contains \(A\), so \(A^*=\varnothing\).
Check: for the chosen even-number set, \(A\) is infinite, so the construction gives at least one free ultrafilter in \(A^*\). This also illustrates why the remainder is not just a collection of the original integers: its points are nonprincipal ultrafilters, not elements of \(\mathbb{N}\).
Every free ultrafilter contains exactly one of \(E=\{2,4,6,\ldots\}\) and \(O=\{1,3,5,\ldots\}\). Indeed, an ultrafilter contains one of a set and its complement, and cannot contain both because their intersection is empty. Since \(E\) and \(O\) are both infinite, each of \(E^*\) and \(O^*\) is nonempty. These are disjoint clopen subsets of the remainder whose union is the entire remainder. Thus \(\beta\mathbb{N}\setminus\mathbb{N}\) is disconnected.
Worked Example: The Natural-Number Remainder Has No Isolated Points
Let \(p\) be any point of \(\beta\mathbb{N}\setminus\mathbb{N}\), so \(p\) is a free ultrafilter. A basic neighborhood of \(p\) in the remainder has the form \(A^*\) for some \(A\in p\). Because \(p\) is free, \(A\) is infinite. Split \(A\) into two disjoint infinite subsets \(B\) and \(C\). Such a split can be made by listing the elements of \(A\) in increasing order and putting alternating entries into \(B\) and \(C\).
Applying the cofinite-filter extension argument from the previous example to \(B\) and to \(C\) gives free ultrafilters \(q\) and \(r\) containing \(B\) and \(C\), respectively. Since \(B,C\subseteq A\), both \(q\) and \(r\) contain \(A\), so \(q,r\in A^*\). They are distinct: \(q\) contains \(B\), whereas \(r\) cannot contain \(B\) because it already contains the disjoint set \(C\). Also neither is a principal ultrafilter. Thus every basic neighborhood of \(p\) contains a remainder point other than \(p\). Every open neighborhood contains a basic neighborhood, so \(p\) is not isolated. Since \(p\) was arbitrary, the remainder has no isolated points.
Check: the argument used only that \(A\in p\) is infinite, that it can be split into two infinite parts, and that each part lies in a free ultrafilter. It therefore applies to every point \(p\) of the remainder, not just to ultrafilters containing the even numbers.
The Real Line: Bounded Functions Separate Points at Infinity
Now take \(\mathbb{R}\) with its standard topology. It is locally compact Hausdorff and noncompact, so the general theorem says that \(\beta\mathbb{R}\setminus\mathbb{R}\) is compact and nonempty. We can prove more: the remainder contains at least two distinct points. The argument uses the continuous extension of the bounded function \(f(x)=\cos x\).
Worked Example: Two Distinct Points in the Real-Line Remainder
Consider the subsets \(S_+=\{2\pi n:n\in\mathbb{N}\}\) and \(S_-=\{(2n+1)\pi:n\in\mathbb{N}\}\) of \(\mathbb{R}\). Each is closed in \(\mathbb{R}\): each is an arithmetic progression with positive spacing, and every bounded interval meets it in only finitely many points. Each is noncompact. For example, the cover of \(S_+\) by its singleton subsets has no finite subcover; the same is true of \(S_-\).
The closures \(\overline{S_+}^{\,\beta\mathbb{R}}\) and \(\overline{S_-}^{\,\beta\mathbb{R}}\) are compact, since they are closed subsets of the compact space \(\beta\mathbb{R}\). Each closure meets the remainder. To check this for \(S_+\), suppose instead that \(\overline{S_+}^{\,\beta\mathbb{R}}\subseteq\mathbb{R}\). Its intersection with \(\mathbb{R}\) is the closure of \(S_+\) in \(\mathbb{R}\), which is \(S_+\) itself. The supposition would therefore give \(\overline{S_+}^{\,\beta\mathbb{R}}=S_+\), making \(S_+\) compact, a contradiction. The same reasoning applies to \(S_-\).
The function \(f(x)=\cos x\) is bounded and continuous, so it extends to \(\beta f:\beta\mathbb{R}\to\mathbb{R}\). On \(S_+\), \(\beta f\) has value \(1\), because \(\cos(2\pi n)=1\) for every \(n\in\mathbb{N}\). The level set \((\beta f)^{-1}(\{1\})\) is closed, so it contains the closure of \(S_+\). Choose \(p_+\) in that closure and in the remainder; then \(\beta f(p_+)=1\). On \(S_-\), \(\cos((2n+1)\pi)=-1\) for every \(n\in\mathbb{N}\). The closed level set \((\beta f)^{-1}(\{-1\})\) contains the closure of \(S_-\), so that closure contains a remainder point \(p_-\) with \(\beta f(p_-)=-1\). Since one real number cannot equal both \(1\) and \(-1\), \(p_+\ne p_-\).
Check: the values used are \(\cos(2\pi n)=1\) and \(\cos((2n+1)\pi)=-1\). The selected points lie in the remainder by the closure argument, and their different extended-function values verify that they are distinct.
This argument does not describe every point of \(\beta\mathbb{R}\setminus\mathbb{R}\), but it proves that the remainder is not a singleton. More generally, if two subsets escaping compact regions force a bounded continuous function to have different constant values on their closures, those closures must contribute different points of the remainder.
A Common Pitfall: Treating the Remainder as One Point
The one-point compactification of a locally compact noncompact Hausdorff space adds a single point, often interpreted as infinity. The Stone–Čech compactification is different: its remainder may have many points. The theorem above guarantees only that the remainder is compact and, in the noncompact case, nonempty; it does not identify it with the one-point remainder.
For \(\mathbb{R}\), the two points constructed above directly rule out a one-point remainder. There is also a useful function-based check. In the one-point compactification of \(\mathbb{R}\), the sequence \(2\pi n\) converges to the added point, as does \((2n+1)\pi\), because both sequences eventually leave every compact subset of \(\mathbb{R}\). A continuous extension of cosine to that compactification would have to take the same limiting value along both sequences. But the values are \(1\) along the first and \(-1\) along the second. Thus cosine has no continuous extension to the one-point compactification. It does extend over \(\beta\mathbb{R}\), whose remainder can keep the two limiting behaviors separate.
For \(\mathbb{N}\), the distinction is also substantial: the one-point compactification has one added point, whereas the Stone–Čech remainder contains free ultrafilters and has no isolated points. The phrase “point at infinity” is therefore safe only when the particular compactification really has a single added point.
Check Your Understanding
Use the definitions and arguments above to test what the general theorem does—and does not—tell you.
- Where in the proof that \(X\) is open in \(\beta X\) is density of \(X\) used?
- For a finite subset \(F\subseteq\mathbb{N}\), why is \(F^*=\varnothing\)? What changes when \(F\) is infinite?
- Explain why splitting an infinite member of a free ultrafilter into two infinite pieces rules out an isolated point.
- Which bounded continuous function and which two subsets of \(\mathbb{R}\) distinguish the two remainder points constructed above?
- Why does compactness and nonemptiness of \(\beta X\setminus X\) not imply that the remainder is a single point?