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I have read everything filter-related at the basic/medium level that can be found but still can't get a solid understanding of them. It confuses me a lot the idea of "big sets" or "something like a 0-1 measure"

If an ultrafilter on a set $X$ contains a finite set then it is principal but what if we extend the filter $\{A\subset X : Y\subseteq X\}$ generated by some finite $Y$ (with at least two elements)? Then the ultrafilter contains $Y$ but is not principal.

Where is my reasoning wrong? In the "extending the filter to an ultrafilter" step? How is a filter extended to an ultrafilter?

Also, given some ultrafilter $U$ of the natural numbers and a bounded sequence $(x_n)_{n=1}^\infty$ of real numbers then we say that $(x_n)_{n=1}^\infty$ converges to some $L$ if, for every $\varepsilon>0$ we have $$\{\,n : |x_n-L|<\varepsilon\,\}\in U$$ I understand this definition when $U$ is the cofinite filter but why is it necessary to extend it to an ultrafilter? If the ultrafilter is principal I think I understand it (not very sure though)

I don't really know how to ask specific questions about all the chaos. For example when using neighborhood filters and convergence you want smaller and smaller neighborhoods but i think of filters as collections of big sets.

Thanks

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    $\begingroup$ I suggest you expand your question by adding the definitions as you understand them. For instance, then you recognize that the second paragraph contains no reasoning at all, jus an incorrect claim. $\endgroup$ – Moishe Kohan Jul 4 at 17:48
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    $\begingroup$ Why do you think the ultrafilter in the second paragraph is not principal? $\endgroup$ – Eric Wofsey Jul 4 at 18:31
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    $\begingroup$ My main recommendation to you would be to not just read about filters, but find a text with exercises and try to do the exercises one by one. That will help you develop an understanding starting from the beginning. $\endgroup$ – Eric Wofsey Jul 4 at 18:33
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If we have the extension of the supersets of $Y$ to an ultrafilter, then for sure this ultrafilter $\mathcal{U}$ will be fixed. This is because $Y$ is a union of its singletons, and if an ultrafilter contains $Y$ (a finite union of its singletons) it contains one of those singletons, and so is point-fixed, i.e. of the form $\{A \subseteq X: p \in A\}$ for some $p \in Y$.

If the ultrafilter to define an ultralimit were fixed (on $n_0 \in \Bbb N$) say, then the ultralimit would be just the term $x_{n_0}$ of the sequence. Not very limit-like. So ultralimits are only taken, really, for free ultrafilters, i.e. ones that extend the cofinite filter. The cofinite filter limits correspond to usual limits, suing tails of sequences. We go to ultrafilters because then we can show that a sequence in a compact $X$ always has an ultralimit, but not always a convergent subsequence; the theory is much nicer.

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