# difference between convergence along free ultrafilter and common convergence

Suppose $$\mathcal{F}$$ is any free ultrafilter on $$\beta\mathbb{N}\setminus\mathbb{N}$$,$$(x_n)$$ is a sequence of complex numbers.

My question is:What is the deference between the limit along $$\mathcal{F}$$ ,$$lim_{\mathcal{F}}x_n$$ and $$lim_{n\to \infty}x_n$$?If $$lim_{n\to\infty}x_n\not \to 0$$,can we conclude that $$lim_{\mathcal{F}}x_n\not\to 0$$?

• The sequence $(0,1,0,1,0,1,0,1,0,1,\dots)$ doesn't converge. It converges with respect to every ultrafilter $\mathcal F$ on $\mathbb N$. You should figure out what it converges to, in terms of $\mathcal F$. And then you should probably try to prove that every bounded sequence of complex numbers converges with respect to every ultrafilter on $\mathbb N$. – Andreas Blass Dec 7 '18 at 2:23

No, I'll use Andreas' elementary example from the comments: the sequence $$x_n = 0$$ for $$n$$ even, $$x_n= 1$$ for $$n$$ odd, does not converge in $$\mathbb{R}$$ in the usual topology.
However, if $$\mathcal{F}$$ is a free ultrafilter on $$\mathbb{N}$$, then either $$E= \{2n: n \in \mathbb{N}\}$$ or $$O = \{2n+1: n \in \mathbb{N}\}$$ is in $$\mathcal{F}$$ (as $$O \cup E = \mathbb{N}$$ and we have an ultrafilter). If the former, then for any neighbourhood $$U$$ of $$0$$ we have that $$\{n : x_n \in U\} \supseteq E$$ so $$\{n : x_n \in U\} \in \mathcal{F}$$, and so $$\lim_\mathcal{F} x_n = 0$$ and if the latter we similarly see that $$\lim_\mathcal{F} x_n = 1$$. In any case, the non-convergent sequence $$(x_n)_n$$ does converge along any free ultrafilter. So the proposed implication does not hold.