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Timeline for Complexity of ultrafilter limits

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Jul 23, 2019 at 18:19 comment added Paolo Leonetti Sorry for the delay. You have $$ \omega\text{-}\lim a_{u_n}=0\,\,\,\,\text{ if and only if }\,\,\,\,\omega\text{-}\lim \frac{1}{u_n}\sum_{i\le u_n}a_i=0, $$ this is clear. Here $\omega$ is the image of $\mathscr{F}$ through $v$, which should be defined as $\{S\subseteq \mathbf{N}: v^{-1}(S) \in \mathscr{F}\}$. However, I still don't get why $$ \Phi(L\cap P)=\{a \in \{0,1\}^{\mathbf{N}}: \omega\text{-}\lim a_n=0\}. $$ Am I missing something easy?
Jan 18, 2019 at 9:20 vote accept Paolo Leonetti
Jul 24, 2019 at 7:34
Jan 16, 2019 at 13:46 comment added YCor @PaoloLeonetti In my previous comment I have two sequences differing by $o(1)$. So if one $\mathscr{F}$-converges to 0, so does the other.
Jan 16, 2019 at 13:28 comment added Paolo Leonetti Yes, this is clear. However, assume that $v(\mathscr{F})\text{-}\lim a_{u_n}=0$, i.e., $\{n \in \mathbf{N}: a_{u_n}=0\}=v(F)$, for some $F \in \mathscr{F}$ ($\star$). Then it is claimed that $\mathscr{F}\text{-}\lim \sum_{i\le n}\frac{a_i}{n}=0$, that is, $$\left\{n\in \mathbf{N}: \sum_{i\le n}\frac{a_i}{n} \le \varepsilon \right\} \in \mathscr{F}$$ for all $\varepsilon>0$. How does it follow from ($\star$)?
Jan 16, 2019 at 1:02 comment added YCor Because for $a\in P$, $\frac{1}{u_n}\sum_{i=1}^{u_n}a_i$ is equal to $a_{u_n}+o(1)$.
S Jan 15, 2019 at 11:09 history suggested Paolo Leonetti CC BY-SA 4.0
Typo in the definition of $L$
Jan 15, 2019 at 10:25 review Suggested edits
S Jan 15, 2019 at 11:09
Jan 15, 2019 at 9:57 comment added Paolo Leonetti Thanks for your answer. One more question: may you explain why $\Phi(L\cap P)$ is the set of sequence in $\{0,1\}^{\mathbf{N}}$ which are $\omega$-convergent to $0$?
Jan 14, 2019 at 21:03 history answered YCor CC BY-SA 4.0