Why is the closure of every countable discrete subspace of $\beta\mathbb{N}$ homeomorphic to $\beta\mathbb{N}$?

Question

I remember seeing a proof of this but I can't seem to reproduce it.

(By $\beta\mathbb{N}$ I mean the Stone-Cech compactification of $\mathbb{N}$.)

Answer 1

If $A = \{a_n: n \in \omega\}$ is countable and discrete, then $h(a_n) = n$ defines a homeomorphism between $A$ and $\mathbb{N}$.

Now let $f: A \rightarrow [0,1]$ be continuous.

Then $\hat{f}:= f \circ h^{-1}: \mathbb{N} \rightarrow [0,1]$ is continuous. So by the extension property of $\beta \mathbb{N}$ ,we have some $\beta \hat{f}: \beta \mathbb{N} \rightarrow [0,1]$ that extends $\hat{f}$. I claim that $\beta \hat{f}|\overline{A}$ extends $f$, which is pretty clear as $f(a_n) = \hat{f}(n)$. So as the extension property characterises $\beta X$ among compactifications, and as $\overline{A}$ is clearly a compactification of $A$ (compact Hausdorff and $A$ is dense in it), we are done:

$$\overline{A} \simeq \beta A \simeq \beta \mathbb{N}$$

Finally note that every infinite Hausdorff space has a countable discrete subspace. So every infinite closed subset $C$ of $\beta \mathbb{N}$ contains a homeomorphic copy of $\beta \mathbb{N}$. (Find an infinite discrete $A \subset C$ which is possible, and $\overline{A} \subset C$.)