Existence of ultrafilters in a compact subset.

Question

(If $X\neq \varnothing$, $Y$ is a compact topological space and $f:X\to Y$ is a function, then for the ultrafilter $U$ on $X$, $\lim_{U}f$ exist.)

Let $\mathbb{U}$ be the set of ultrafilters over $\mathbb{N}$. Let $U\in \mathbb{U}$ and $x\in\ell^\infty(\mathbb{N})$. I know that $x\in\ell^\infty(\mathbb{N})$ (i.e. $x:\mathbb{N}\to\mathbb{C}$ is limited) implies that $\overline{x (\mathbb{N})}$ is compact. Why does $\lim_{U}x$ exist?

In my case, I have $x:\mathbb{N}\to\mathbb{C}$, but $\mathbb{C}$ is not compact.

Answer 1

$x$ is a function from $\mathbb{N}$ to $\overline{x(\mathbb{N})}$, which is compact. Hence, $\lim_U x$ exists.

Answer 2

For any $A\in U$ let $K_A=\overline{\{x_n:n\in A\}}$. Each $K_A$ is compact and non-empty, hence $K:=\cap_{A\in U}K_A$ is non-empty and compact. If $K$ is a singleton $\{p\}$ then $\lim_{U}x=p$ and we are done.

Suppose toward a contradiction that there were two different points $p$ and $q$ in $K$. Take disjoint open neighborhoods $V$ of $p$ and $W$ of $q$. Let $A_p=\{n:x_n\in V\}$ and $A_q=\{n:x_n\in W\}$. Then $A_p$ and $A_q$ are disjoint, hence at most one of them belongs to $U$ (since $U$ is a filter and $A_p\cap A_q=\varnothing\notin U$). Say $A_q\notin U$. Then, since $U$ is an ultrafilter, we have that $\Bbb N\setminus A_q\in U$. Hence $K\subseteq\overline{\{x_n:n\in(\Bbb N\setminus A_q)\}}\subseteq \Bbb C\setminus W$, a contradiction since $q\in W$.

The result follows from what you wrote on the first line of your post, that $\lim_{U}f$ exists, I just wrote some details to make it easier to follow what exactly that statement says in this context. As you have noted, $\overline{x(\mathbb{N})}$ is compact. The main property of an ultrafilter $U$ used in the proof is that for every $A\subseteq U$ exactly one of $A$ and $U\setminus A$ belongs to $U$. If $U$ were only a filter, not necessarily an ultrafilter, then there might be some $A$ such that neither $A$ nor $U\setminus A$ belongs to $U$. This happens for example if $U=\{\Bbb N\}$ and $A$ is the set of all even natural numbers.