A topological space is compact if and only if every ultrafilter is convergent.
While I was reading the proof of the one Side of theorem above, there is something I could not understand. Following is the proof of of the one side of the theorem.
Let $X$ be compact and assume that $\mathcal{F}$ is the ultrafilter on $X$ without a limit point. Then for each $x\in X$, there exists an open neighborhood $U_{x}$ of it such that each $U_{x}$ does not contain any member of $\mathcal{F}$. Since $\mathcal{U}=\{U_{x} : x\in X\}$ is an open cover of $X$, there exists a finite subfamily $\{U_{x_{i}}: i=1,2,…,n\}$ of $\mathcal{U}$ such that $X=\bigcup_{i=1}^{n} U_{x_{i}}$. Let $A\in\mathcal{F}$ be fixed. Then $A=(A\cap U_{x_{1}})\cap (A\cap U_{x_{2}})…(A\cap U_{x_{n}})\in\mathcal{F}$ and thus there exists an $i\in\{1,2,…,n\}$ such that the subset $A\cap U_{x_{i}}$ is in $\mathcal{F}$ which is a contradiction.
The thing that I could not understand, why there exists $i\in\{1,2,…,n\}$ such that $A\cap U_{x_{i}}$ must be in $\mathcal{F}$? If you clarify this, it would highly be appreciated. Thank you.
Since $\mathcal{F}$ is an ultrafilter, for every $B \subset X$, exactly one of $B$ and $X\setminus B$ is in $\mathcal{F}$.
If none of the $A \cap U_{x_i}$ is in $\mathcal{F}$ then for each $i$, $X \setminus (A \cap U_{x_i}) \in \mathcal{F}$. $\mathcal{F}$ is closed under intersections so that $$ X \setminus A = \bigcap_{i = 1}^n X\setminus (A \cap U_{x_i}) \in \mathcal{F}.$$
Since only one of $A$ and $X \setminus A$ can be in $\mathcal{F}$, this implies that $A \not \in \mathcal{F}$ which is a contradiction.