12,012 reputation
339102
bio website boolesrings.org/asafk
location Israel
age 29
visits member for 4 years, 9 months
seen 1 min ago

Born and raised in Israel. Ph.D. student in the Hebrew university in Jerusalem; studying set theory.


9h
comment 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
comment 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 physics-related journal.
10h
answered Applications of set theory in physics
1d
comment 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
comment 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
comment 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
comment 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
comment 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
comment 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
comment 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
comment How many subsets of $[0,1)$ are there modulo null sets?
I don't think so.
Mar
20
comment 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
comment 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?