Show that $Cl_{\beta\mathbb{N}}\{0,2,4,...\}$ is open in $\beta\mathbb{N}$ Let $\beta\mathbb{N}$ denote the Stone-Cech compactification of the space $\mathbb{N}$ of the natural numbers with the discrete topology and let $E$ denote the set of even natural numbers. Show that the closure $Cl_{\beta\mathbb{N}}(E)$ is open in $\beta\mathbb{N}$.
I understand the universal property of the Stone-Cech compactification. I understand that the Stone-Cech compactification is the maximal compactification and I understand the sense in which it is maximal. I also understand the standard construction of the Stone-Cech compactification as the closure of the image of an imbedding of your topological space into $[0,1]^C$ where $C$ is the set of continuous function from your space to $[0,1]$. However, I don't know how to determine when a subset of the Stone-Cech compactification is open nor do I know what any of the elements of $\beta X\setminus X$ look like for a space $X$.
 A: Hint: Consider the "parity" function $f:\mathbb{N}\to\left\{0,1\right\}$,
$$f(x)=\begin{cases}0&\text{ if }x\text{ is even}\\
1&\text{ if }x\text{ is odd}\end{cases}$$
and remember that $\left\{0,1\right\}$ is compact.
A: Attempted solution to my problem:
I claim that $Cl_{\beta\mathbb{N}}(E)$ and $Cl_{\beta\mathbb{N}}(\mathbb{N}\setminus E)$ are compliments of one another in $\beta\mathbb{N}$. First, note that because $E\subset\mathbb{N}$, $Cl_{\beta\mathbb{N}}E=Cl_{[0,1]^C}(E)\cap\beta\mathbb{N}=Cl_{[0,1]^C}(E)\cap Cl_{[0,1]^C}(\mathbb{N})=Cl_{[0,1]^C}(E)$ which I will simply denote $\overline{E}$, and similarly $Cl_{\beta\mathbb{N}}(\mathbb{N}\setminus E)=Cl_{[0,1]^C}(\mathbb{N}\setminus E)=\overline{\mathbb{N}\setminus E}$.
Assume that $(x_f)_{f\in C}\in\overline{E}\cap\overline{\mathbb{N}\setminus E}$ and consider the parity function $g\in C$ which sends even natural numbers to $0$ and odd natural numbers to $1$. If $x_g\neq 0$, then the open set $\prod_{f\in C}U_f$ of $[0,1]^C$ defined by $U_f=[0,1]$ if $f\neq g$ and $U_g=(0,1]$ is a neighborhood of $(x_f)_{f\in C}$ which is disjoint from $E$ contradicting the assumption that $(x_f)_{f\in C}\in\overline{E}$, so we must have $x_g=0$. But now, the open set $\prod_{f\in C}V_f$ defined by $V_f=[0,1]$ if $f\neq g$ and $V_g=[0,1)$ is a neighborhood of $(x_f)_{f\in C}$ which is disjoint from $\mathbb{N}\setminus E$ contradicting the assumption that $(x_f)_{f\in C}\in\overline{\mathbb{N}\setminus E}$. Thus, by way of contradiction, it must be the case that $\overline{E}\cap\overline{\mathbb{N}\setminus E}=\emptyset$.
Finally, we see that $\overline{E}\cup\overline{\mathbb{N}\setminus E}=\overline{E\cup(\mathbb{N}\setminus E)}=\overline{\mathbb{N}}=\beta\mathbb{N}$. Thus, we conclude that $\overline{E}$ and $\overline{\mathbb{N}\setminus E}$ are compliments of one another. Since they are both closed, they are also both open.
