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We know that free ultrafilter exist on natural numbers. I would like to see an example of a free ultrafilter on natural numbers.

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I'm assuming you meant "free ultrafilter."

This is a very reasonable question! Unfortunately, in a very real sense we can't exhibit a concrete example of a free ultrafilter on $\mathbb{N}$ - it is consistent with ZF (= set theory without the axiom of choice) that there are no free ultrafilters on $\mathbb{N}$. Any free ultrafilter has to be "hard to define" in various precise ways (this gets a bit technical - the relevant field is "descriptive set theory").

The situation is similar with regard to a number of other kinds of object whose existence relies on the axiom of choice, like non-measurable sets (as in the Banach-Tarski paradox).

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  • $\begingroup$ Yes I meant "free ultrafilter", thank you. $\endgroup$ – asv Aug 17 '18 at 14:11
  • $\begingroup$ If I am not wrong, it is like to construct a choice function (if someone crates it, we would have that ZF implies AC but it is impossible) $\endgroup$ – asv Aug 17 '18 at 14:14
  • $\begingroup$ The hyperreal numbers are constructed by any free ultrafilter. It is possible to give a canonical hyperreal numbers also if it is not possibile to give explicitly a free ultrafilter? $\endgroup$ – asv Aug 17 '18 at 14:22
  • $\begingroup$ Do you have a reference for the consistency of ZF+ ¬UF(ℕ)? $\endgroup$ – Lukas Juhrich Jul 15 '19 at 21:05
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    $\begingroup$ @Luke See the references at this old MSE question. $\endgroup$ – Noah Schweber Jul 15 '19 at 21:11
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Edit: The question originally asked for an example of a free filter on the natural numbers.

The filter $\{X\subseteq \mathbb{N}\mid \mathbb{N}\setminus X\text{ is finite}\}$ is free. This is known as the cofinite filter, or the Fréchet filter.

Maybe you meant to ask for an example of a free ultrafilter? No truly explicit example can be given - there is no constructive proof of the existence of a free ultrafilter on $\mathbb{N}$.

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  • $\begingroup$ Yes I meant "free ultrafilter", thank you. $\endgroup$ – asv Aug 17 '18 at 14:10
  • $\begingroup$ The hyperreal numbers are constructed by any free ultrafilter. It is possible to give a canonical hyperreal numbers also if it is not possibile to give explicitly a free ultrafilter? $\endgroup$ – asv Aug 17 '18 at 14:20
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    $\begingroup$ I'm assuming that by "hyperreal numbers", you mean an ultraproduct of the ordered field $\mathbb{R}$ by a free ultrafilter on $\mathbb{N}$. If you assume the continuum hypothesis, then any two such ultraproducts are isomorphic (see here), so there is a canonical ordered field of hyperreals (up to isomorphism). But in the absence of CH, or if you want to do something like work in a language with all subsets of the reals named by predicates, the hyperreals won't be canonical. $\endgroup$ – Alex Kruckman Aug 17 '18 at 14:43

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