bio | website | people.su.se/~cesp |
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location | Stockholm, Sweden | |
age | ||
visits | member for | 4 years, 2 months |
seen | 8 hours ago | |
stats | profile views | 1,277 |
I'm a graduate student deeply interested in foundations (logic, set theory, model theory, metamathematics) and category theory.
Apr 12 |
awarded | Enlightened |
Apr 12 |
awarded | Nice Answer |
Feb 14 |
awarded | Yearling |
Jan 12 |
comment |
A question regarding the Hahn-Banach theorem
It's clear that such a measure must take values within $0$ and $1$, but in order to even state countable additivity you would need $B$ to have countable joins and meets, which is not necessarily going to happen. Regarding $WKL_0$, you need to specify which base theory and which statement of Hahn-Banach theorem you are considering. $ZF$ is too strong since it proves Weak König's lemma, and the general formulation of Hahn-Banach theorem as in the above equivalence involves arbitrary vector spaces not necessarily definable in second order arithmetic. |
Jan 9 |
answered | A question regarding the Hahn-Banach theorem |
Dec 7 |
comment |
What are the current views on consistency of Reinhardt cardinals without AC?
...But it is conceivable that a large cardinal hierarchy could be mirrored in an intuitionistic setting where more notions of large cardinals can be defined without inconsistency (they will have to be non-classical though). |
Dec 7 |
comment |
What are the current views on consistency of Reinhardt cardinals without AC?
This is just speculative, but even if some large cardinals are found to be inconsistent with ZF, there could be weaker set theories where appropriate versions are consistent. I'm thinking more precisely of Aczel's constructive set theory CZF, based on intuitionistic logic, over which appropriate notions of regular, inaccessible and even Mahlo cardinals are defined (see "Inaccessibility in constructive set theories and type theories" by Rathjen-Griffor-Palmgren). In this case, the addition of AC turns the theory in just ZFC plus the usual notions...(cont.) |
Dec 6 |
comment |
$\zeta(0)$ and the cotangent function
Given the fact that what you end up explaining why the series, considered as a Laurent series, predicts also the values of $\zeta$ at the negative even numbers, I think this proof is as simple as it reasonably can be expected to be. This because you are forced to consider the analytic continuation of $\zeta$ to the left half plane, not just to the critical strip, so it's reasonable that the functional equation is invoked. |
Nov 5 |
comment |
Mysterious quotes (at least for me)
There is a very particular way in which those quotes make sense to me, but I'm afraid is not a mathematical way so I can't really answer :) |
Oct 16 |
revised |
Is it possible to formulate the axiom of choice as the existence of a survival strategy?
added 1981 characters in body |
Oct 15 |
answered | Is it possible to formulate the axiom of choice as the existence of a survival strategy? |
Sep 25 |
comment |
An interpretation of not-Con(PA)
Careful! All that the the 2nd incompleteness theorem tells you is that $Con(PA)$ is unprovable, not that it is independent, and this provided you already accept the consistency of PA. If you apply that theorem to your theory $PA^+$ instead, you can only conclude that $Con(PA^+)$ is unprovable in $PA^+$, because in fact $\neg Con(PA^+)$ is actually provable there. |
Sep 24 |
awarded | Autobiographer |
Sep 22 |
awarded | Curious |
Sep 21 |
comment |
On a weak tree property for inaccessible cardinals
Excellent, thanks! That was exactly what I was looking for. |
Sep 21 |
accepted | On a weak tree property for inaccessible cardinals |
Sep 21 |
asked | On a weak tree property for inaccessible cardinals |
Aug 29 |
answered | construction of nonmeasurable sets |
Aug 17 |
comment |
How short can we state the Axiom of Choice?
It seems to me that François' solution is near optimal, considering that all usual equivalents of AC (which are gather in Rubin & Rubin's book rather than in Jech's), use notions which, when unwound in the language, are considerably larger. |
Aug 13 |
comment |
Relationship between fragments of the axiom of choice and the dependent choice principles
Oh, good, then I won't have to go to the construction myself :) I just had read that it was Solovay who proved result 3 in this model. So now the real challenge is result 4, since the book only mentions permutation models for that. |