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14h
comment
When can you canonically extend an ultrafilter after forcing?
Thanks! I think that might be helpful in understanding my situation a bit better!
19h
asked
When can you canonically extend an ultrafilter after forcing?
23h
revised
When $2^\alpha = 2^\beta$ implies $\alpha=\beta$ ($\alpha,\beta$ cardinals)
edited tags; edited title
2d
awarded
Announcer
Nov
23
revised
Theories introduced by a class of forcing notions
deleted 9 characters in body
Nov
21
comment
New hilbert system and theorem
math.stackexchange.com/questions/1539816/…
Nov
21
comment
Invertibility of a polynomial in a commutative ring
Those who vote to migrate this, don't. The question is likely to be closed on MSE and get bounced back here.
Nov
17
revised
How can the critical point of an elementary embedding be omega_1?
edited tags
Nov
17
comment
Removing large cardinals from an uncountable transitive model
Wooo! My idea was correct! Also, it's good seeing you around Trevor! :)
Nov
14
comment
Generating family for the Lebesgue $\sigma$-algebra
@Ashutosh: Awesome! Let me know if you find anything out! :-)
Nov
14
comment
Generating family for the Lebesgue $\sigma$-algebra
@Ashutosh: Not that I know of.
Nov
13
revised
Is there any elementary embedding characterization for $\Pi_{1}^{1}$ - reflecting cardinals?
edited title
Nov
11
comment
Removing large cardinals from an uncountable transitive model
Yeah, I meant of course the latter case. How about trying to show that in that situation there is no uncountable model without an inaccessible?
Nov
11
reviewed
Leave Closed
Non-standard numbers and exponential form of Zeta function
Nov
11
comment
Removing large cardinals from an uncountable transitive model
My usual solution when I'm stuck is to try and engineer a counterexample. Have you tried starting with $L$ with a single Mahlo and collapse it to be $\omega_1$?
Nov
10
awarded
Enlightened
Nov
10
awarded
Nice Answer
Nov
10
reviewed
Leave Open
Replacing Axiom of Choice with Axiom of Countable Choice
Nov
10
revised
Replacing Axiom of Choice with Axiom of Countable Choice
edited tags
Nov
10
answered
Replacing Axiom of Choice with Axiom of Countable Choice
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