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| bio | website | iecn.u-nancy.fr/~gaillapy |
|---|---|---|
| location | Nancy, France | |
| age | 62 | |
| visits | member for | 2 years, 9 months |
| seen | 23 hours ago | |
| stats | profile views | 1,816 |
Email: pierre.yves.gaillard at gmail.com
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May 6 |
awarded | Caucus |
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May 1 |
comment |
Proving a function is continuous for a fixed variable Dear @Tsotsi: This is explained on the last line of the answer. (Sorry for answering your comment so late; I've been busy.) |
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Apr 30 |
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Proving a function is continuous for a fixed variable Dear @Tsotsi: Thanks for your interest. Could you tell me the first thing that is unclear? |
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Apr 15 |
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Error in argument regarding the Cayley Hamilton theorem Dear @MarcvanLeeuwen: Thank you very much! Sorry for not being able to make less typos! |
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Apr 6 |
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Square root of a matrix Dear Marc: I've just edited the answer. Thank you again for your interest! |
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Apr 6 |
revised |
Square root of a matrix rewrote the answer |
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Apr 6 |
revised |
Holomorphic function of a matrix rewrote the answer |
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Apr 4 |
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Square root of a matrix Dear Marc: Thanks for your comment. I'll answer it as soon as possible, but I'll be very busy today and tomorrow. I'll try to get back to you Saturday (Paris time) at the latest. |
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Mar 18 |
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Cantor's diagonal argument without equality Are you sure you're not assuming the consistency of ZF? |
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Mar 18 |
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Cantor's diagonal argument without equality I fully agree. My question is: Can one prove that $\mathcal T$ is consistent using (in our meta-language) accepted logical notions? ($\mathcal T$ being again the theory with specification and no other axiom.) |
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Mar 18 |
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Cantor's diagonal argument without equality ... consistent. |
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Mar 18 |
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Cantor's diagonal argument without equality Do you agree that one cannot exclude a priori the possibility that $\mathcal T$ is consistent, while ZF is not? |
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Mar 18 |
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Cantor's diagonal argument without equality I'm sure you know much more than I about models. It seems to me that, according to Cohen, the notion of model involves that of equality. Anyway, in theory $\mathcal T$ (with specification only), we have infinitely many distinct terms, but no notion of object. And, again, the statement $\mathcal P(A)\not\subset A$ (suitably translated) holds in $\mathcal T$. |
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Mar 18 |
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Cantor's diagonal argument without equality Recall that $\mathcal T$ assumes only a single Axiom schema: that of specification. Are you saying that the other axioms of ZF are provable in $\mathcal T$? |
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Mar 18 |
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Cantor's diagonal argument without equality The notion of monoid is not crucial here. The point is: Do you agree that the consistency of $\mathcal T$ can be translated into a combinatorial statement, or, if you prefer, into a standard mathematical statement? |
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Mar 18 |
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Cantor's diagonal argument without equality How do you state and prove $0=0$ (or $0\not=1$) in $\mathcal T$? |
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Mar 18 |
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Cantor's diagonal argument without equality Dear Trevor: I also agree with your first sentence. I'm not wondering whether ZFC proves $\mathcal T$ to be consistent. Do you agree that the consistency of $\mathcal T$ can be translated into a combinatorial statement (in term of free monoids)? If you do, then you can ask whether this statement can be proved (like any statement about monoids). |
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Mar 18 |
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Cantor's diagonal argument without equality Dear Trevor: I'm happy to see that we agree on the first point. I dropped the completeness aspect in the new version. It was a bad idea of mine to talk about completeness. I also agree with the phrase: "without equality we cannot define objects at all". That's precisely my point! Do you agree that, without equality, we can still prove $\mathcal P(A)\not\subset A$? |
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Mar 18 |
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Cantor's diagonal argument without equality Dear Trevor: Thank you for your answer. Unfortunately, I don't understand it. Here are two questions: (1) I think an inconsistent theory is complete. Do you agree with this? (2) How do you define the cardinal number 0 without the Axiom of extensionality? |
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Mar 18 |
revised |
Cantor's diagonal argument without equality edit clearly indicated |
