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Elliot Glazer
  • Member for 5 years, 1 month
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12 votes
Accepted

Is this set theory equivalent to ZFC?

12 votes
Accepted

Is every set being cardinal definable consistent with ZF + negation of Choice?

11 votes
Accepted

How much choice is necessary to prove this statement?

11 votes

Is Global Choice conservative over Zermelo with Choice?

11 votes
Accepted

Is $\in$-induction provable in first order Zermelo set theory?

11 votes
Accepted

Does ZF minus infinity imply collection?

10 votes

What are some reasonable-sounding statements that are independent of ZFC?

9 votes
Accepted

Can one exhibit an explicit Kuratowski infinite set without invoking Replacement?

9 votes
Accepted

Example of a $\Pi^2_2$ sentence?

9 votes
Accepted

Can a Vopenka cardinal be supercompact?

8 votes
Accepted

Can we make ZF − infinity + “all ordinals are finite” as strong as ZFC?

7 votes
Accepted

Products of Cohen forcings

7 votes
Accepted

Is "There exists an unbounded non-measurable set but no bounded non-measurable set" consistent with $\mathsf{ZF}$?

6 votes
Accepted

Stronger negation of AC given by rejecting "infinite hat" puzzles

5 votes
Accepted

Does $\text{AC}_{\text{WO}}$ prove $\Theta \neq \aleph_{\omega+1}?$

4 votes

Consistency of embedding cardinals in linear orderings

4 votes
Accepted

Is there a maximal translation-invariant extension of Lebesgue measure?

4 votes

If $X$ a subset of ordinals, is $M[X]$ a generic extension?

4 votes
Accepted

Strong form of $\mathtt{PSP}$ for $K_\sigma$ sets

2 votes

Extending the product measure on $2^\omega$

2 votes

Is "There exists an unbounded non-measurable set but no bounded non-measurable set" consistent with $\mathsf{ZF}$?

2 votes

Can a Vitali set be Lebesgue measurable? (ZF)

1 vote

Can a Vitali set be Lebesgue measurable? (ZF)