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Aug
22
awarded  Nice Question
Aug
18
comment Notation Clarification: $M\odot N$ for von Neumann algebras $M$ and $N$
Is there any advantage to using $\odot$ instead of $\otimes$, aside from reserving $\otimes$ for another meaning?
Aug
18
accepted Notation Clarification: $M\odot N$ for von Neumann algebras $M$ and $N$
Aug
10
comment Notation Clarification: $M\odot N$ for von Neumann algebras $M$ and $N$
Update: I think that the author intended to just refer to the algebraic tensor product with this $\odot$ notation.
Aug
10
revised Notation Clarification: $M\odot N$ for von Neumann algebras $M$ and $N$
edited title to be more descriptive
Aug
10
asked Notation Clarification: $M\odot N$ for von Neumann algebras $M$ and $N$
Jul
13
accepted Passing complemented subspaces to duals
Jul
13
asked Passing complemented subspaces to duals
Jul
3
accepted Is $L^{1}(\Omega,\mu)$ only an algebra when $\Omega$ is a group?
Jul
2
asked Is $L^{1}(\Omega,\mu)$ only an algebra when $\Omega$ is a group?
Jun
20
comment When does a topological space inherit multiplication from a dense subspace?
Thank you Greg, this is exactly what I was seeking. Thank you!
Jun
19
comment When does a topological space inherit multiplication from a dense subspace?
Thank you for the quick answer.
Jun
19
asked When does a topological space inherit multiplication from a dense subspace?
May
26
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
May
26
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!
May
26
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.
May
26
revised Given a finite collection of disjoint subsets of $I$ must every ultrafilter on $I$ contain exactly one?
added 317 characters in body
May
26
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.
May
26
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.
May
26
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!