TITLE: If $X$ has the discrete topology then the Stone-Čech compactification is zero dimensional.
QUESTION [5 upvotes]: If $X$ has the discrete topology then the Stone-Čech compactification is zero-dimensional.
I would like to show the above to be true. If I can show that every closure (in $\beta X$) of an open set (in $\beta X$) is open (in $\beta X$), then I would be done.
Here zero-dimensional means that the continuous simple functions on $\beta X$ are dense in  $\beta X$.

REPLY [4 votes]: The fact is a simple observation. I use the construction of $\beta X$ by zero sets of $X$.
Let $X$ be an infinite discrete space. Consider $\mathcal{P}(X)$ -- the algebra of all subsets of $X$. It is easy to see that $Stone(\mathcal{P}(X))$ is $\beta X$ (so $\beta X$ must be zero-dimensional as the Stone space of every Boolean algebra is). Indeed, every subset of a discrete space is a zero set, hence $\mathcal{P}(X)=\mathcal{Z}(X)$ (zero sets of $X$), so ultrafilters in $\mathcal{Z}(X)$ are the same as ultrafilters of algebra $\mathcal{P}(X)$, and a basic set $U_A=\{p: A\in p\}$ of the Stone space is a basic set $U_{X\setminus A}=\{p: X\setminus A\not\in p\}$ of the topology of $\beta X$. As $A$ is an arbitrary subset of $X$, $X\setminus A$ is also an arbitrary subset of $X$, thus those two bases are equal.