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This question already has an answer here:

Let $X$ be a set. If $X$ is finite then all ultrafilters on $X$ are principal, i.e. have the form $\{A \subseteq X : x \in A\}$ for some $x\in X$.

But now suppose $X$ is infinite, say $X=\mathbb N$. Is there any concrete example of a non-principal ultrafilter on $X$? And does one need the axiom of choice to prove their existence?

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marked as duplicate by Kyle, Community May 4 '16 at 17:10

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    $\begingroup$ I assume that you read all the posts in math.stackexchange.com/questions/tagged/axiom-of-choice+filters before posting this, yes? $\endgroup$ – Asaf Karagila May 4 '16 at 14:58
  • $\begingroup$ Using AC (but we don't need the full AC) one can prove that any filter can be extended to an ultrafilter. In particular, the filter of cofinite subsets of the natural numbers can be extended to an ultrafilter, which cannot be principal. $\endgroup$ – André Nicolas May 4 '16 at 14:59
  • $\begingroup$ Asaf: Roughly. But I don't fully understand what they're talking about $\endgroup$ – Gabriel Nivasch May 4 '16 at 15:03
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There is no explicit example of a non-principal ultrafilter over the natural numbers without appealing to choice.

We know this because there are models of $\sf ZF$ where every ultrafilter over the natural numbers is principal. In fact there are models where every ultrafilter over any set is principal.

The proofs are quite technical and require understanding of forcing and symmetric extensions, or relative constructibility.

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