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Aug 14, 2017 at 12:29 vote accept Ali Taghavi
Apr 20, 2017 at 12:49 answer added Mikhail Katz timeline score: 5
Apr 19, 2017 at 13:38 comment added Mikhail Katz math.stackexchange.com/a/280935/72694 is a related answer by @AndreasBlass.
Apr 19, 2017 at 12:02 comment added Mikhail Katz Some articles related to this can be found in this MSE post. There are $2^c$ ultrafilters but also $2^c$ non-isomorphic fields when $\neg CH$, so straightforward "counting" does not seem to allow one to decide.
Apr 12, 2017 at 18:11 comment added Paul McKenney Just to finish the proof, there are always nonequivalent nonprincipal ultrafilters (since there are $2^{2^{\aleph_0}}$-many nonprincipal ultrafilters on $\mathbb{N}$ and only $2^{\aleph_0}$-many bijections from $\mathbb{N}$ to itself), so under CH the answer is no. But I think the answer is probably no just in ZFC.
Apr 11, 2017 at 11:03 comment added Michael Greinecker A question that addresses the result mentioned by YCor can be found here.
Apr 11, 2017 at 9:34 history edited YCor
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Apr 11, 2017 at 9:34 comment added YCor I think it's known that under CH, all ultrapowers $\mathbb{R}_U$ are isomorphic. Maybe somebody more familiar known more about this.
Apr 11, 2017 at 9:15 history asked Ali Taghavi CC BY-SA 3.0