bio | website | boolesrings.org/asafk |
---|---|---|
location | Israel | |
age | 29 | |
visits | member for | 4 years, 9 months |
seen | 14 mins ago | |
stats | profile views | 9,056 |
Born and raised in Israel. Ph.D. student in the Hebrew university in Jerusalem; studying set theory.
Apr 14 |
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Does the Brouwer fixed point theorem admit a constructive proof?
@Emil: It wasn't a constructive comment. :-) |
Apr 13 |
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Does the Brouwer fixed point theorem admit a constructive proof?
Now I'm not sure if you got that I was just joking or not. :| |
Apr 13 |
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Does the Brouwer fixed point theorem admit a constructive proof?
This is not a constructive proof, since it relies on the law of excluded middle, that either LLPO is constructive or it is not. What happens if this information cannot be computed? :-) |
Apr 10 |
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Constructing unnatural transformations
Is it customary, after constructing an unnatural transformation to shout in a trembling voice "It's alive! ALIVE!!!" or do you guys don't do that? |
Apr 9 |
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What's the minimum amount of knowledge to start doing research?
Au contraire. If you know a lot about a subject, you also know what are the open problems, and you have some sort of intuition as to how to attack them. |
Apr 9 |
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How do I apply the Boolean Prime Ideal Theorem?
The point I didn't make in my previous comment is that these proofs are often very different from one another. They might appeal to various different equivalents of $\sf BPI$ (e.g. the compactness theorem of first-order logic; or the Tychonoff theorem for Hausdorff spaces). |
Apr 9 |
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How do I apply the Boolean Prime Ideal Theorem?
The easiest way to learn how to do these proofs, is to read the papers. To find the papers, simply open Google Scholar or MathSciNet and try and search for keywords like "Hahn-Banach Prime Ideal" etc. |
Apr 8 |
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Logical strength of “choice functions exist for well-ordered families”?
$\sf DC_\kappa\implies AC_\kappa, W_\kappa$ and that is the only implication, without adding more assumptions. |
Apr 8 |
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Logical strength of “choice functions exist for well-ordered families”?
Well, only the minor addition that it does not imply that every set is comparable with $\aleph_1$ in cardinality (which is weaker than $\sf DC_{\omega_1}$ of course). I actually linked to your question regarding these proofs from the summer, below in the comments of the question. |
Apr 8 |
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Logical strength of “choice functions exist for well-ordered families”?
I actually outlined a proof of what Vika wrote in her answer, in a question she had asked some time ago. mathoverflow.net/questions/178473/… |
Apr 8 |
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Logical strength of “choice functions exist for well-ordered families”?
It was 37 degrees centigrade today. Tomorrow is 21 degrees. The weather is crazy. I also have to grade homework assignments. All these things drained my will to stay awake... :-) |
Apr 8 |
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Logical strength of “choice functions exist for well-ordered families”?
I literally snoozed and missed answering the question. :-) |
Apr 7 |
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Is there some absoluteness between the Boolean valued universe $V^{B}$ and $V$?
Some first remark, $\check x$ is a name for an object in $V[G]$ which is the canonical name for the object $x$ from $V$, and moreover $\check x^G=x$. So you can just write $V\models\phi(x)\iff V[G]\models\phi(x)$. If you want to put emphasis on generic objects, then $\phi(\dot x^G)$ is preferable. |
Apr 7 |
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The role of the rigid relation principle ($RR$) in the Kunen inconsistency
Let me nitpick again, and point out that the definition $j(\dot x^G)=j(\dot x)^G$ requires more justification: (1) $G$ is usually not in $W$, so you argue about a definition in $V[G]$. Then by the fact that the class itself is symmetric we have that it is a class of $W$ and it works out. (2) Again with hereditarily symmetric names. It's not sufficient just that $\operatorname{sym}(\dot x)=\operatorname{sym}(j(\dot x))$, you also need to ensure that all the names which appear in $j(\dot x)$ are already hereditarily symmetric. But this is a first-order property (with parameters) so it works too. |
Apr 7 |
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The role of the rigid relation principle ($RR$) in the Kunen inconsistency
Two remarks: (1) $fix(\dot x)$ is usually reserved for the pointwise stabilizer, and $\operatorname{sym}(\dot x)$ is more appropriate. But this doesn't matter. (2) You need to require in the last paragraph that $\dot x$ is hereditarily symmetric name, not just that its stabilizer is in the filter, otherwise it cannot possibly be a class of $W$ (since it will include objects not in $W$). |
Apr 7 |
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The role of the rigid relation principle ($RR$) in the Kunen inconsistency
It should be remarked that this absolutely does not generalize to $\sf ZF$. The reason is that here $j$ was a class in $V$; so it was easily extended to a class of the symmetric extension. On the other hand, in $\sf ZF$ we don't even know if the ground model is necessarily a class of the symmetric extension. So the fact that a class of names is stable under automorphisms does not necessarily means that it will be a class of the symmetric model. (In fact, this is probably the biggest open problems in symmetric extensions.) |
Apr 7 |
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The role of the rigid relation principle ($RR$) in the Kunen inconsistency
edited body |
Apr 7 |
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Dedekind-finite arithmetic vs natural numbers arithmetic
@Joel: Because it's not quite an answer. In particular, I have no idea if it is consistent that there is such Diophantine equation. |
Apr 7 |
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Dedekind-finite arithmetic vs natural numbers arithmetic
@bof: Yes, the exponentiation is not cardinal exponentiation, but of course it coincides with it when the power is finite. What is it exactly? Some binary operation I suppose. I haven't investigated Sageev's model yet. My guess is that you somehow quantify this as some sort of limit exponentiation (e.g. one might notice that the set of all finite injective sequences of a Dedekind-finite set, is also Dedekind-finite, so there is some weak exponentiation to be done here. But it seems to be closer to factorial than to exponentiation. So in short, I have no idea.) |
Apr 7 |
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Dedekind-finite arithmetic vs natural numbers arithmetic
Well, Dedekind-finite cardinals need not be linearly ordered, so they can be very different. Sageev proved that assuming an inaccessible cardinal exists, there is an extension of the universe where all Dedekind-finite cardinals are comparable (in a nontrivial way, of course) and the axiom of choice for families of finite sets hold. Then, by a theorem of Ellentuck, it follows that the Dedekind-finite cardinals are a model of true arithmetic, if I remember correctly. |