Any bijection from $\Bbb N$ to itself transforms an ultrafilter on $\Bbb N$ to another (isomorphic) ultrafilter. Any two principal ultrafilters are isomorphic in that sense.
For free ultrafilters on $\Bbb N$, there are $2^{2^{\aleph_0}}$ of them. Since there are $2^{\aleph_0}$ bijections of $\Bbb N$ to itself, there are also $2^{2^{\aleph_0}}$ isomorphism classes of free ultrafilters on $\Bbb N$. So lots of free ultrafilters must be nonisomorphic to each other.
Question: Can you give an explicit example or construction of two free ultrafilters on $\Bbb N$ that are not isomorphic? Assume ZFC.
(Added at the suggestion of @bof in the comments below, in case the question proves too difficult to answer directly):
- Give an explicit example of two filters such that no free ultrafilter extending one of them can be isomorphic to a free ultrafilter extending the other.
- Can you state a property, preserved by isomorphism, possessed by some but not all free ultrafilters?
(1) is as good as the original question as far as I am concerned.
from Hot Weekly Questions - Mathematics Stack Exchange
PatrickR
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