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.
 A: 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.
