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2d
revised Given a finite collection of disjoint subsets of $I$ must every ultrafilter on $I$ contain exactly one?
removed a claim which was established to be false by an commentor below
2d
comment Given a finite collection of disjoint subsets of $I$ must every ultrafilter on $I$ contain exactly one?
Very nice. I take back everything I said. Thank you!
2d
comment Given a finite collection of disjoint subsets of $I$ must every ultrafilter on $I$ contain exactly one?
What I'm getting at is that going from $X_{1}$ and $X_{2}$ forming a partition of $I$ to $X_{1}$ and $X_{2}$ forming a partition of an element of $\mathfrak{U}$ makes the details in the maximality argument not work anymore. Nonetheless you can still prove this stronger statement by first proving what Michael Albanese shows above, and then deriving it as a Corollary.
2d
revised Given a finite collection of disjoint subsets of $I$ must every ultrafilter on $I$ contain exactly one?
added 317 characters in body
2d
comment Given a finite collection of disjoint subsets of $I$ must every ultrafilter on $I$ contain exactly one?
I agree with that argument (which proves the fact in my OP). In fact it is the one I was referring to in my previous comment (with the adjustment that either $A$ or $A^{c}$ may be added to $D$, you don't know in advance which one). But it doesn't work to prove the lemma mentioned above. I will make an edit to my original post to clarify the distinction which is subtle.
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comment Given a finite collection of disjoint subsets of $I$ must every ultrafilter on $I$ contain exactly one?
I agree that the induction part of the proof is indeed easy. But you are using the following lemma, which I think is the tricky part: If $X_{1}\cup X_{2}\in \mathfrak{U}$, with $X_{1}\cap X_{2} = \phi$, then exactly one of $X_{1}$ or $X_{2}$ is in $\mathfrak{U}$. This is slightly stronger than the fact I mentioned, and cannot be proved in quite the same way (using a brief maximality argument). The lemma clearly follows immediately from Michael Albanese's argument above, but I do not see an easy way to get it directly.
2d
comment Given a finite collection of disjoint subsets of $I$ must every ultrafilter on $I$ contain exactly one?
It is the intuition you describe that led me to guess that the result might be true. I could not come up with a proof, however. Thanks for your additional information!
2d
comment Given a finite collection of disjoint subsets of $I$ must every ultrafilter on $I$ contain exactly one?
A partition is what I intended to start with, but forgot to mention. Thanks to all for the great answers.
2d
accepted Given a finite collection of disjoint subsets of $I$ must every ultrafilter on $I$ contain exactly one?
2d
asked Given a finite collection of disjoint subsets of $I$ must every ultrafilter on $I$ contain exactly one?
May
24
accepted Reducing a double ultrapower to a single ultrapower
May
23
revised Reducing a double ultrapower to a single ultrapower
edited title
May
23
comment Reducing a double ultrapower to a single ultrapower
In math.lsa.umich.edu/~ablass/thesis.pdf, it is established that the cartesian product of two ultrafilters is again an ultrafilter. So that would pose as an strong candidate for $\mathfrak{W}$. But I still can't find the reduction property that I'm seeking.
May
23
asked Reducing a double ultrapower to a single ultrapower
May
11
awarded  Popular Question
Apr
9
comment Iterated Limits Along an Ultrafilter
Thank you for this!
Apr
9
accepted Iterated Limits Along an Ultrafilter
Apr
9
comment Iterated Limits Along an Ultrafilter
Note that my question was not whether this holds in general. I'm asking about sufficient and/or necessary conditions for these limits to agree.
Apr
9
comment Iterated Limits Along an Ultrafilter
Without being rigorous, I was able to vaguely come up with an example of sequences where this does not hold. But I did not work it out fully. I can do so over the next hour and report back!
Apr
9
asked Iterated Limits Along an Ultrafilter