$\beta \mathbb {N}$ is compact I want to show the compactness of the set $\beta \mathbb N := \{ \mathcal U  \mid \mathcal U \text{ is an ultrafilter on } \mathbb N \}$, with the topology induced by the basis $U_M = \{\mathcal U \in \beta \mathbb N \mid M \in \mathcal U \}$.
Since $\beta \mathbb N \subset \prod\limits_{r \in \mathcal{P}(\mathbb N)}\{0,1\}$  which is compact by Tychonoff, we can just show that $\beta \mathbb N$ is closed in this product. But I have no idea, how can I do it ? 
Please give me just hint and not all the solution if possible, thanks. 

Suppose that $x\in{^{\wp(\Bbb N)}\{0,1\}}\setminus\beta\Bbb N$, and let $\mathscr{X}=\{A\subseteq\Bbb N:x(A)=1\}$. Then $\mathscr{X}$ is not an ultrafilter on $\Bbb N$, so it must satisfy one of the following conditions:


*

*$\varnothing\in\mathscr{X}$;  

*there are $A,B\subseteq\Bbb N$ such that $B\supseteq A\in\mathscr{X}$, but $B\notin\mathscr{X}$;  

*there are $A,B\in\mathscr{X}$ such that $A\cap B\notin\mathscr{X}$;  

*there is an $A\subseteq\Bbb N$ such that $A\notin\mathscr{X}$ and $\Bbb N\setminus A\notin\mathscr{X}$.


Show that each of these conditions defines an open set in the product ${^{\wp(\Bbb N)}\{0,1\}}$. (Indeed, for a specific choice of $A$ and $B$ each of the middle two conditions defines a basic open set in the product. Similarly, for a specific choice of $A$ the last defines a basic open set in the product.) It follows at once that $\beta\Bbb N$ must be closed in ${^{\wp(\Bbb N)}\{0,1\}}$.
