# every bounded sequence has limit along an ultrafilter

###### Theorem 1.

Let $\mathcal{F}$ be an ultrafilter on $\mathbb{N}$ and $(x_{n})$ be a real bounded sequence. Then $\operatorname{\mathcal{F}\text{-}\lim}x_{n}$ exists.

###### Proof.

Let $(x_{n})$ be a bounded sequence. Choose $a_{0}$ and $b_{0}$ such that $a_{0}\leq x_{n}\leq b_{0}$. Put $c_{0}:=\frac{a_{0}+b_{0}}{2}$. Then precisely one of the sets $\{n\in\mathbb{N};x_{n}\in\langle a_{0},c_{0}\rangle\}$, $\{n\in\mathbb{N};x_{n}\in\langle c_{0},b_{0}\rangle\}$ belongs to the filter $\mathcal{F}$. (Their union is $\mathbb{N}$ and the filter $\mathcal{F}$ is an ultrafilter.) We choose $\langle a_{1},b_{1}\rangle$ as that subinterval from $\langle a_{0},c_{0}\rangle$ and $\langle c_{0},b_{0}\rangle$ for which $C:=\{n\in\mathbb{N};x_{n}\in\langle a_{1},b_{1}\rangle\}$ belongs to $\mathcal{F}$.

Now we again bisect the interval $\langle a_{1},b_{1}\rangle$ by putting $c_{1}=\frac{a_{1}+b_{1}}{2}$. Denote $A:=\{n\in\mathbb{N};x_{n}\in\langle a_{1},c_{1}\rangle\}$, $B:=\{n\in\mathbb{N};x_{n}\in\langle c_{1},b_{1}\rangle\}$. It holds $B\cup A\cup(\mathbb{N}\setminus C)=\mathbb{N}$. By the alternative characterization of ultrafilters we get that one of these sets is in $\mathcal{F}$. The set $\mathbb{N}\setminus C$ doesn’t belong to $\mathcal{F}$, therefore it must be one of the sets $A$ and $B$. We choose the corresponding interval for $\langle a_{2},b_{2}\rangle$.

By induction we obtain the monotonous sequences $(a_{n})$, $(b_{n})$ with the same limit $\lim\limits_{n\to\infty}a_{n}=\lim\limits_{n\to\infty}b_{n}:=L$ such that for any $n\in\mathbb{N}$ it holds $\{n\in\mathbb{N};x_{n}\in\langle a_{1},b_{1}\rangle\}\in\mathcal{F}$.

We claim that $\operatorname{\mathcal{F}\text{-}\lim}x_{n}=L$. Indeed, for any $\varepsilon>0$ there is $n\in\mathbb{N}$ such that $\langle a_{n},b_{n}\rangle\subseteq(L-\varepsilon,L+\varepsilon)$, thus $\{n\in\mathbb{N};x_{n}\in\langle a_{n},b_{n}\rangle\}\subseteq A(\varepsilon)$. The set $\{n\in\mathbb{N};x_{n}\in\langle a_{1},b_{1}\rangle\}$ belongs to $\mathcal{F}$, hence $A(\varepsilon)\in\mathcal{F}$ as well. ∎

Note that, if we modify the definition of $\mathcal{F}$-limit in a such way that we admit the values $\pm\infty$, then every sequence has $\mathcal{F}$-limit along an ultrafilter $\mathcal{F}$. (The limit is $+\infty$ if for each neighborhood $V$ of infinity, the set $\{n\in\mathbb{N};x_{n}\in V\}$ belongs to $\mathcal{F}$. Similarly for $-\infty$.)

## References

• 1 M. A. Alekseev, L. Yu. Glebsky, and E. I. Gordon, On approximations of groups, group actions and Hopf algebras, Journal of Mathematical Sciences 107 (2001), no. 5, 4305–4332.
• 2 B. Balcar and P. Štěpánek, Teorie množin, Academia, Praha, 1986 (Czech).
• 3 K. Hrbacek and T. Jech, Introduction to set theory, Marcel Dekker, New York, 1999.
Title every bounded sequence has limit along an ultrafilter EveryBoundedSequenceHasLimitAlongAnUltrafilter 2013-03-22 15:32:26 2013-03-22 15:32:26 kompik (10588) kompik (10588) 4 kompik (10588) Theorem msc 40A05 msc 03E99 Ultrafilter