Suppose $(x_n)_n$ is a bounded sequence of complex numbers, there must exist a accumulation point, say $x_0$, thus we can find a free ultrafilter $\mathcal{F}$ on $\Bbb N$ such that $\lim_{\mathcal{F}}x_n=x_0$.

  1. Can we find a free ultrafiler $\omega$ on $\Bbb N$ such that $\lim_{\omega}x_n\not \to x_0$.

  2. For any point $c\in \Bbb C$, can we construct a free ultrafilter $\omega$ on $\Bbb N$ such that $\lim_{\omega}x_n\not \to c$?

  • $\begingroup$ I suppose that this is what you meant, but one should be a bit careful with formulation such as construct a free ultrafilter. Probably more appropriate formulation is prove that free ultrafilter with the required properties exists. (Since existence of a free ulftrafilter cannot be shown in ZF, you cannot expect an explicit description of such things. So any proof will contain some step which is non-contructive. For example, it might rely on Zorn's lemma.) $\endgroup$ – Martin Sleziak Dec 9 '18 at 9:59

No: if $(x_n)$ actually converges (as an ordinary sequence) to $x_0$, then it also converges to $x_0$ with respect to every free ultrafilter.

More generally, the set of limits of a sequence with respect to free ultrafilters is exactly the set of accumulation points of the sequence. You seem to already be aware of one direction of this implication; for the other direction, if $(x_n)$ converges to $c$ with respect to a free ultrafilter $\omega$, then for every neighborhood $U$ of $c$ the set $\{n:x_n\in U\}$ is in $\omega$ and so in particular is infinite.

  • $\begingroup$ You mean the cofinte filter $\mathcal{F_0}$ on $\Bbb N$ is contained in any free ultrafilter $\mathcal{F}$ on $\Bbb N$,so if $x_n\to x_0$,then for any neighborhood $U_{x_0}$ of $x_0$,we have $\{n\in \Bbb N:x_n\in U_{x_0}\}\in \mathcal{F_0}\subset\mathcal{F}$.Is my thought correct? $\endgroup$ – mathrookie Dec 9 '18 at 8:50
  • $\begingroup$ No. We know that $\{n\in\mathbb{N}:x_n\in U_{x_0}\}\in\mathcal{F}$, not that it is in $\mathcal{F}_0$. But every element of $\mathcal{F}$ is infinite, so that's all we need. $\endgroup$ – Eric Wofsey Dec 9 '18 at 9:02
  • $\begingroup$ What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct? $\endgroup$ – mathrookie Dec 9 '18 at 9:35
  • $\begingroup$ According to the above answer,If point $c$ is not the acculation point of the sequence,can we deduce that for any free ultrafilter $\mathcal{F}$ on $\Bbb N$,$lim_{\mathcal{F}}x_n \not \to c$? $\endgroup$ – mathrookie Dec 9 '18 at 9:44

Just to sketch a quick answer:

1) If your sequence happened to actually converge to $a$ (which is my notation for $x_0$, I dislike indices for objects which are not components of a certain family indexed by a certain set of indices), then the limit along any free ultrafilter will still be $a$.

2) You will only be able to find such ultrafilters for points $c \notin \overline{\{x_n\}_{n \in \mathbb{N}}}$. As pointed out in one of the answers above, the set of limits along ultrafilters coincides with the set of points adherent to the given sequence $x$.

  • $\begingroup$ If $c$ is not a accumulation point ,can we find the free ultrafilter $\mathcal{F}$ on $\Bbb N$ ?how to construct it? $\endgroup$ – mathrookie Dec 9 '18 at 9:00
  • $\begingroup$ what kind of ultrafilter? do you mean such that convergence to $c$ along the ultrafilter will not occur? If so, matters are quite simple for the limit along any ultrafilter $\mathscr{F}$ will necessarily exist (the sequence induces an ultrafilter in the compact space which is the closure of its bounded set of terms) and it necessarily belongs to the said closure! So it's not a matter of constructing a special ultrafilter to avoid convergence to $c$, that will automatically happen with any ultrafilter whatsoever, as long as $c$ is not adherent to $\{x_{n}\}_{n \in \mathbb{N}}$. $\endgroup$ – ΑΘΩ Dec 9 '18 at 10:20

For 1, $x_0$ being an accumulation point of the sequence means that all sets $\hat{U}:=\{n \in \omega: x_n \in U\}$ are infinite, where $U$ ranges over the neighbourhoods of $x_0$, and obey the FIP. So they extend to some ultrafilter, and $(x_n)$ converges along this ultrafilter. So yes, if you really meant converge (as you should have).

An ultrafilter limit is unique. So if it converges to $c$ it won't converge to any other $c'$.

  • $\begingroup$ I know that an ultrafilter limit is unique.But if $(x_n)$ has a accumulation point $c$,there exists free ultrafilter $\mathcal{F}$ on $\Bbb N$ such that $lim_{\mathcal{F}}x_n=c$.Is the free ultrafilter unique?Does there exists another free ultrfilter $\mathcal{F^{\prime}}$ such that $lim_{\mathcal{F^{\prime}}}x_n=lim_{\mathcal{F}}x_n=c$? $\endgroup$ – mathrookie Dec 9 '18 at 23:39
  • $\begingroup$ @mathrookie there could be another one, it depends on the sequence sometimes. The sequence $0,1,0,1,0,1,\ldots$ has many ultrafilters along which it converges to $0$, likewise for $1$, and none to other point, e.g. $\endgroup$ – Henno Brandsma Dec 10 '18 at 4:38
  • $\begingroup$ .I have another question:If $x_n\not \to c$,does there exist free untafiler $\omega$ on $\Bbb N$ such that $lim_{n\to \omega}x_n\not \to c$? $\endgroup$ – mathrookie Dec 17 '18 at 11:00

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.