free ultrafilter

Question

If $\omega$ is a free ultrafilter on $\mathbb{N}$,$(x_n)$ is a sequence of complex numbers,what is the precise definition of "$lim_{\omega}(x_n)$ does not converge to $x$"?

Answer 1

First know what $x =\lim_\omega(x_n)$ means, then look at the negation:

$$x = \lim_\omega(x_n) \text{ iff } \forall O \text{ open with } x \in O: \{n: x_n \in O\} \in \omega$$

So the negation of that is that there exists some open neighbourhood $O_x$ of $x$ such that $\{n: x_n \in O_x\} \notin \omega$, but as $\omega$ is an ultrafilter (so has the property $A \notin \omega \leftrightarrow \omega\setminus A \in \omega$ for all subset $A$ of $\omega$) this is equivalent to the fact that $\{n: x_n \in X\setminus O_x\} \in \omega$.

So e.g. if $X$ were compact and we'd assume no $x$ is a limit, we'd have an open cover of $X$ of such $O_x$, hence a finite cover of these sets: $O_{x_1},\ldots,O_{x_N}$, and then we'd have contradiction as

$$\cap_{i=1}^N \{n: x_n \in X\setminus O_{x_i}\} \in \omega$$ while the left hand side is empty, as the $O_{x_i}$ form a cover.