every bounded sequence has limit along an ultrafilter
Let be a bounded sequence. Choose and such that . Put . Then precisely one of the sets , belongs to the filter . (Their union is and the filter is an ultrafilter.) We choose as that subinterval from and for which belongs to .
Now we again bisect the interval by putting . Denote , . It holds . By the alternative characterization of ultrafilters we get that one of these sets is in . The set doesn’t belong to , therefore it must be one of the sets and . We choose the corresponding interval for .
By induction we obtain the monotonous sequences , with the same limit such that for any it holds .
We claim that . Indeed, for any there is such that , thus . The set belongs to , hence as well. ∎
Note that, if we modify the definition of -limit in a such way that we admit the values , then every sequence has -limit along an ultrafilter . (The limit is if for each neighborhood of infinity, the set belongs to . Similarly for .)
- 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|
|Date of creation||2013-03-22 15:32:26|
|Last modified on||2013-03-22 15:32:26|
|Last modified by||kompik (10588)|