Stone-Čech Compactification of the Natural numbers

Question

I am trying to prove that if $U$ is contained in the Stone-Čech Compactification of the natural number ($\beta N$) that the closure of $U$ is open.

I have a really hard time with even understanding what the Stone-Čech Compactification is, so I can't even start this problem. Any suggestions would be greatly appreciated.

Thank you!

Answer 1

Lemma 1 (see the Engelking's book Corollary 3.6.5): For every open and closed subset $A$ of a Tychonoff space $X$ the closure $\overline{A}$ of $A$ in $\beta X$ is open and closed.

Proof: Notice that $U \cap N$ is not empty, since $U$ is open and $N$ is dense in $\beta X$. Moreover, $U \cap N$ is open and closed in $N$, then we apply the Lemma 1 to conclude that $\overline{U\cap N}$ is open in $\beta X$. Finally, note that $\overline{U}=\overline{U\cap N}$, since $N$ is dense in $\beta N$. Thus we conclude that $\overline{U}$ is open in $\beta X$.

Answer 2

A basic (perhaps defining) property of $\beta$ tells us:

Let $X$ be completely regular Hausdorff and let $Y$ be compact Hausdorff. If $f : X \to Y$ is continuous, then there is a unique continuous extension $f^\beta : \beta X \to Y$.

So, let $U \subset \mathbb N$. Let $f : \mathbb N \to \{0,1\}$ be the characteristic function of $U$. Examine what it means for the extension $f^\beta$.

Answer 3

Like those who have answered previously (GEdgar and Paul), I'll assume you intended $U$ to be a subset of $\mathbb N$, even though the question assumes only $U\subseteq\beta\mathbb N$. (If one takes the question literally, the result is just false.) Once you see how $\beta\mathbb N$ is constructed using ultrafilters, it will be easy to check that, if $\mathbb N$ is partitioned into two pieces, then the closures of those pieces constitute a partition of $\beta\mathbb N$. In particular, $(\beta\mathbb N)-\overline U=\overline{N-U}$ is closed, and therefore $\overline U$ is open.