# Why does $\mathbb{N}$ have only two topological compactifications?

Definition: A compactiﬁcation $$\gamma X$$ of a space $$X$$ is said to be a topological compactiﬁcation if all autohomeomorphisms of the space $$X$$ can be continuously extended to a mapping of $$\gamma X$$ into $$\gamma X$$.

The space $$\mathbb{N}$$ with its usual discrete topology only admits two topological compactifications - that is the one-point compactification $$\alpha \mathbb{N}$$ and the Stone-Čech compactification $$\beta \mathbb{N}$$.

I couldn´t find a proof of this. Why is this true? Intuitively, it does make sense to me, as $$\mathbb{N}$$ is a discrete space and also spaces with more automorphisms tend to have less compactifications extending them. But I am not able to prove that.

Thank you for providing advice or source fot this.

UPDATE: Apart from consulting this at my university, I have found some sources supporting the statement so far. However, I dont see it clearly from the sources and my lecturer told me that it should be easily seen.

Source 1: Van Douwen - Characterizations of $$\beta \mathbb{Q}$$ and $$\beta \mathbb{R}$$ (unfortunaetly I couldnt find it online): If $$D$$ is a discrete spaee with $$\mid D \mid = \aleph_\psi$$, then D has precisely $$\psi$$ + 2 topological compaetifications.

Source 2: Vejnar - Topological Compactifications: Corollary 11. :The only topological compactiﬁcations of $$\omega$$ are $$\alpha \omega$$ and $$\beta \omega$$. (I suppose this is another notation for natural numbers?)

I dont understand much any of these though, maybe this is the best way how to show that.

• Do you have a reference for the statement? Sep 4, 2022 at 14:41
• @LeeMosher I added an update. However, the first source states that as an exercise and for a general case and the second does it quite complicately in my opinion. Sep 4, 2022 at 15:28
• I will try to find answer to this. Sep 4, 2022 at 15:38
• By the way, $\omega$ is the usual notation for the first infinite ordinal number, and is indeed order isomorphic to $\mathbb N$. Furthermore, in set theory $\mathbb N$ is defined to be $\omega$ with the additional structures of addition and multiplication that are themselves defined using Peano's Axioms. Sep 4, 2022 at 15:45
• So are you asking how Corollary 11 of that paper by Vejnar is derived from the earlier results in the paper? Sep 4, 2022 at 15:54

Nota bene: We are working in the context in which compact includes Hausdorff as part of the definition. Not everyone uses that definition. Otherwise, we could have $$\mathbb{N}\cup\{-\infty, +\infty\}$$ with the topology $$\{A:\ A\subset\mathbb{N}\}\cup\{A\cup\{-\infty,+\infty\}:\ A\subset\mathbb{N}\}$$ as another topological compactification.

Denote by $$\alpha \mathbb{N}$$ and $$\beta \mathbb{N}$$ the one-point and the Stone-Cech compactifications of $$\mathbb{N}$$, respectively. Let $$\mathbb{N}^*=\beta \mathbb{N}\setminus \mathbb{N}$$, the "points at infinity" if you want. If $$X\subset \mathbb{N}$$, denote by $$\overline{X}$$ its closure in $$\beta \mathbb{N}$$.

Assume that $$\gamma \mathbb{N}$$ is a compactification of $$\mathbb{N}$$ and $$f:\beta \mathbb{N}\to \gamma \mathbb{N}$$ continuous with $$f|_\mathbb{N}: \mathbb{N}\to \mathbb{N}$$ the identity. This is the extension of the inclusion of $$\mathbb{N}\hookrightarrow \gamma \mathbb{N}$$ to $$\beta \mathbb{N}$$ given by the universal property of $$\beta\mathbb{N}$$. For convenience, let's refer to $$\mathbb{N}$$ as a subset of both $$\beta\mathbb{N}$$ and of $$\gamma\mathbb{N}$$ by identifying it with its inclusions in both.

Assume that $$x,y\in \mathbb{N}^*$$ are such that $$x\neq y$$ and $$f(x)=f(y)$$. This way $$\gamma \mathbb{N}$$ is not $$\beta \mathbb{N}$$. Assume also that $$u,v\in \mathbb{N}^*$$ are such that $$f(u)\neq f(v)$$, such that $$\gamma \mathbb{N}$$ is not $$\alpha \mathbb{N}$$ either.

Since $$\gamma\mathbb{N}$$ is Hausdorff, pick $$U', V'\subset \gamma \mathbb{N}$$ open such that $$f(u)\in U'$$, $$f(v)\in V'$$ and $$U'\cap V'=\emptyset$$. Let $$U=f^{-1}(U')$$ and $$V=f^{-1}(V')$$. These are open in $$\beta \mathbb{N}$$. Take $$A=U\cap \mathbb{N}$$, $$B=V\cap \mathbb{N}$$. Then $$u\in\overline{A}\subset U$$ and $$v\in\overline{B}\subset V$$. In particular $$A,B$$ are infinite (countable) and disjoint.

Take $$X,Y\subset \mathbb{N}$$ such that $$X\cap Y=\emptyset$$, $$x\in \overline{X}$$, and $$y\in\overline{Y}$$. We can use two disjoint open subsets of $$\beta\mathbb{N}$$ that contain $$x$$ and $$y$$ and take their intersection with $$\mathbb{N}$$.

Since $$A,B,X,Y$$ are countable, with $$A,B$$ disjoint, and $$X,Y$$ disjoint, we can take a bijection $$g:\mathbb{N}\to \mathbb{N}$$ such that $$g(X)\subset A$$ and $$g(Y)\subset B$$. Let $$\beta g:\beta \mathbb{N}\to \beta \mathbb{N}$$ be its continuous extension to $$\beta \mathbb{N}$$. We have $$\beta g(x)\in \overline{\beta g(X)}\subset \overline{A}$$ and $$\beta g(y)\in\overline{\beta g(Y)}\subset \overline {B}$$.

If $$\gamma \mathbb{N}$$ is a topological compactification, then there is $$h:\gamma \mathbb{N}\to\gamma \mathbb{N}$$ such that $$h|_{\mathbb{N}}=g$$.

Now, for all $$t\in \mathbb{N}$$ we have $$f(\beta g(t))=f(h(t))=h(t)=h(f(t))$$. Since $$\mathbb{N}$$ is dense in $$\beta \mathbb{N}$$, then $$f\circ \beta g=h\circ f:\beta \mathbb{N}\to \gamma \mathbb{N}$$

On the other hand $$f(\beta g(x))\in f(\beta g(\overline{X}))\subset f(\overline{A})\subset f(U)$$ and $$f(\beta g(y))\in f(\beta g(\overline{Y}))\subset f(\overline {B})\subset f(V)$$. Since $$U\cap V=\emptyset$$, we would have $$f(\beta g(x))\neq f(\beta g(y))$$ However, $$f(\beta g(x))=h(f(x))=h(f(y))=f(\beta g(y))$$ contradicts that.

• You begin with “[…] this way $\gamma\Bbb N$ is not $\beta\Bbb N$”. But it might be the case that $f$ injects into $\gamma\Bbb N$ but $\gamma\Bbb N$ is “larger” than $\beta\Bbb N$. Right? Sep 6, 2022 at 14:34
• @FShrike No, $f$ doesn't inject into $\gamma\mathbb{N}$, the opposite. $\beta\mathbb{N}$ is the Stone-Cech compactification. All others we get by gluing some of the infinite points, the one-point compactification $\alpha\mathbb{N}$ by gluing all infinite points. The assumption is that $\gamma\mathbb{N}$ is in between. $x,y$ get glued, and $u,v$ don't get glued.
– plop
Sep 6, 2022 at 14:37
• And, unless I’m blind, you don’t use $h$. Sep 6, 2022 at 14:40
• @FShrike That's one of the basic properties of the Stone-Cech compactification of being maximal. Imagine $f:X\to Y$ a compactification, with $X$ Tychonoff, and let $\beta f:\beta X\to Y$ be the map given by the universal property of $\beta X$. Take a point $p\in Y$ and a net $x_\alpha\in X$ converging to $p$. Passing to a subnet we can assume that $x_\alpha\to q\in\beta X$. Then by continuity of $\beta f$ we must have $\beta f(q)=p$.
– plop
Sep 6, 2022 at 15:14
• @FShrike Yes, and not only that Hausdorff too is a tacit assumption here. Otherwise we can just double the infinite point of $\alpha\mathbb{N}$ and put them both in all the same open sets.
– plop
Sep 6, 2022 at 15:32