bio  website  boolesrings.org/asafk 

location  Israel  
age  29  
visits  member for  4 years, 9 months 
seen  1 min ago  
stats  profile views  8,937 
Born and raised in Israel. Ph.D. student in the Hebrew university in Jerusalem; studying set theory.
9h

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Applications of set theory in physics
Well, obviously they don't do mathematics in the same way mathematicians do. I am also sure that any quantum theorist, the purest of the pure, would still be able to find some excuse why this is not a real issue. But it does show something about the way physicists [ab]use very abstract mathematics. If that's not important... I don't know what can be said to be important from set theory to physicists. 
10h

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Applications of set theory in physics
@CarloVonSchnitzel: I asked Magidor about this paper once, and he said that they are trying to get it published in a physicsrelated journal. 
10h

answered  Applications of set theory in physics 
1d

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A classic cardinal characteristic of the continuum in disguise?
You really aim at this to be a temporary name, eh? :) 
Mar 24 
comment 
Uninteresting questions with interesting answers
Somewhat subjective, isn't it? I find the Whitehead problem to be very uninteresting, but the answer itself is quite interesting. 
Mar 23 
revised 
Antichain on $\mathcal{P}(\omega)/fin$ of cardinality $2^{\aleph_0}$?
added 2 characters in body 
Mar 23 
revised 
Antichain on $\mathcal{P}(\omega)/fin$ of cardinality $2^{\aleph_0}$?
added 88 characters in body 
Mar 23 
answered  Antichain on $\mathcal{P}(\omega)/fin$ of cardinality $2^{\aleph_0}$? 
Mar 22 
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How many subsets of $[0,1)$ are there modulo null sets?
@Alexander: I look up which models are these two, and those are the models were every set is Lebesgue measurable (Solovay's classic model, and Shelah's model separating the perfect set and Baire properties from Lebesgue measurability). I don't know the argument, but I'll see if I can find somewhere that it is written. I doubt the Consequences dictionary would have added that without reason. That's new to me, so thanks! 
Mar 22 
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How to use Holder's Inequality to prove two Banach spaces are equal $(\mathbb R^n,\\cdot\_3)*= (\mathbb R^n,\\cdot\1.5)$
@András: Please don't recommend this when the question is this badly written. It will get closed there just as well. 
Mar 21 
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Is there a bijection between R ans R²
This is not a suitable question for MathOverflow. It might have been suitable for math.stackexchange.com, but it is badly written, the title doesn't match the body, and both answers have been answered about a zillion times before there (yes and no, respectively). So please search that website if you're interested in the answer. In the meantime, I'm voting to close this. 
Mar 21 
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Is there a natural bijection from $\mathbb{N}$ to $\mathbb{Q}$?
Related: math.stackexchange.com/q/424654/622 and math.stackexchange.com/q/7643/622 
Mar 21 
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Inaccessible cardinal and $\Sigma_1$ reflection
I wouldn't avoid it. I'd just skip it, since $\mathcal P(\alpha)$ is its own transitive closure. 
Mar 21 
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Inaccessible cardinal and $\Sigma_1$ reflection
Sorry, I wasn't pinged in your original reply, so I only now see that you wrote it. The point is that since $\alpha$ is transitive, $\mathcal P(\alpha)=\operatorname{tc}(\mathcal P(\alpha))$. I don't understand why you're insisting that I am appealing to "additional" information, where your argument also appeals to that, because you need to pause and consider what in the name of Zeus is $\operatorname{tc}(\mathcal P(\alpha))$. Oh, right, it's just $\mathcal P(\alpha)$. :) 
Mar 20 
revised 
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
Fixed link. 
Mar 20 
comment 
How many subsets of $[0,1)$ are there modulo null sets?
Well. I don't know which models these are, but I'll investigate further when I return home. Thanks giving me something for tonight. I'll get back to you with some further conclusions tomorrow. 
Mar 20 
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How many subsets of $[0,1)$ are there modulo null sets?
I don't think so. 
Mar 20 
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How many subsets of $[0,1)$ are there modulo null sets?
Luckily $\mathfrak c^2=\mathfrak c$ without using choice. Unluckily, the assumption that there are $\frak c$ Borel sets does use the axiom of choice. 
Mar 19 
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How many subsets of $[0,1)$ are there modulo null sets?
Joel, it's a nice answer. You should leave it up. If you feel guilty about it, you can always make it into community wiki. 
Mar 19 
reviewed  Leave Closed When is 2 a generator for a prime modules? 