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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 10 months |
| seen | 2 days ago | |
| stats | profile views | 14 |
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Apr 5 |
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Existence of a singleton set (AC for a singleton set) No, you read the question wrong. I don't want the existence of $\{x\}$, but of a $\{v\}$ for $v\in x$. |
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Feb 3 |
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Definabilty of two functions on natural numbers Oh, alright. I have to read into that.. |
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Feb 2 |
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polar decomposition for non square matrices I think it should be $Px\mapsto Ax$ on $ran(P)$. Is that correct? Thanks for the answer. |
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Feb 2 |
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polar decomposition for non square matrices Fixed, thank you. |
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Dec 22 |
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Understanding the model-theoretic proof of Hilbert's Nullstellensatz Ah, I get it now. The definition of elementary substructure allows using free variables in a formula which is supposed to be equivalent in both structures as long as they are interpreted as some fixed elements of the substructure. Somehow I missed that, I was thinking only of sentences. |
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Dec 22 |
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Understanding the model-theoretic proof of Hilbert's Nullstellensatz But what about the coefficients? I mean, how can you e.g. state that something is a zero of $\pi X +1\in \mathbb{C}[X]$ |
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Nov 4 |
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Existence of elementary substructures of a uncountable structure over a countable language $\alpha_i \subseteq M_i\subseteq \alpha_{i+1}$, so $\alpha = \bigcup_i M_i$ is a substructure of $\kappa$. ($M_i$ can be chosen to be of card. $max(card(\alpha_i),card(\tau))<\kappa$ by Löw-Skl, so $\sup M_i<\kappa$) For $i<j$, since both $M_i$ and $M_j$ embed elementarily in $\kappa$, $M_i$ embeds elementarily in $M_j$. By a theorem on chains, all $M_i$ emb.el. in $\alpha$. Since for each $a_1,..,a_n\in \alpha$, an index $i$ with $a_1,..,a_n\in M_i$ exists and $M_i$ emb.el. in $\kappa$, a formula $\varphi(a_1,..,a_n)$ is true in $\alpha$ iff in $\kappa$. So $\alpha$ el.emb. in $\kappa$. |
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Nov 4 |
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Existence of elementary substructures of a uncountable structure over a countable language Thanks for the quick answer. |
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Oct 13 |
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For every axiomatic system in first order logic there exists an equivalent independent system Thanks for the proof. I am currently just learning the ropes in logic and this answer was helpful. Will mark as accepted even though the uncountable case isn't covered. |
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Oct 12 |
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For every axiomatic system in first order logic there exists an equivalent independent system Yeah, or you can remove axioms which follow from the (remaining) others one by one until arriving at an independent system. But the case of infinitely many axioms seems to be more difficult. |
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Jul 28 |
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Intersection of 2 $p$-simplices is a finite union of some $p$-simplices Unsuccessfully tried to prove the existence. Could you give me a hint on how to approach this? |
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Jul 25 |
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Intersection of 2 $p$-simplices is a finite union of some $p$-simplices Thank you for your answer. It was all very understanable. I will have a look at a proof for the theorem you cited, it doesn't seem that difficult. |
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Jul 25 |
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Intersection of 2 $p$-simplices is a finite union of some $p$-simplices I have next to no knowledge of convex geometry, so I was somewhat hoping for an elegant elementary proof. But I like the systematic approach you suggested as well. Now I edited the question to an interesting more general one. (I don't know if it is much more difficult to prove.) |
