# free ultrafilter

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 $$x$$"?

• My guess (not confident enough to make it an answer) would be: $x_* \omega = \{ S \mid x^{-1}(S) \in \omega \}$ is an ultrafilter on $\mathbb{C}$, and we're requiring that ultrafilter not to have limit $x$. So, that would mean that the neighborhood filter $\mathcal{N}_x \not\subseteq x_* \omega$, i.e. there is some $\epsilon > 0$ such that $\{ n \in \mathbb{N} \mid |x_n - x| < \epsilon \} \notin \omega$. – Daniel Schepler Nov 16 '18 at 22:26
• Since $\omega$ is often synonymous to $\Bbb N$, I would to nominate this to the prize of "worst notation ever". :-) – Asaf Karagila Nov 16 '18 at 23:16
• Yes, write $\mathcal{F}$ for the filter, and use $\lim_{\mathcal{F}} x_n$ instead, as is more usual. – Henno Brandsma Nov 17 '18 at 5:45

## 1 Answer

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.