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| bio | website | |
|---|---|---|
| location | ||
| age | 21 | |
| visits | member for | 1 year, 7 months |
| seen | 21 hours ago | |
| stats | profile views | 102 |
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May 15 |
comment |
Number of automorphisms of saturated models @tomasz Makes sense, I'll try thinking about this approach. Two questions, perhaps related: why is saturation needed? How could you get such a long indiscernible sequence? (I'm not too familiar with indiscernibles, but I remember something called the Standard Lema?) |
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May 13 |
asked | Number of automorphisms of saturated models |
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May 9 |
awarded | Caucus |
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May 6 |
accepted | Classification of models |
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May 5 |
comment |
Classification of models Oh, and I get why arbitrary types turn out to not be so complicated. But my previous comment stands. Even after showing $T$ is model complete, what does that have to do with saturation? |
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May 5 |
comment |
Classification of models @AlexKruckman I don't see why they are saturated. In $\mathfrak{A}_0$, there is no element that realizes (the type generated by) the formula $P_1(x)$, since $P_1$ is empty. |
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Apr 22 |
comment |
Classification of models I'm a little confused. Why are you saying "All four" of these models are saturated? |
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Apr 18 |
revised |
Classification of models added an extra condition on the theory, to make the models always infinite |
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Apr 18 |
revised |
Classification of models added homework tad |
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Apr 18 |
asked | Classification of models |
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Apr 8 |
accepted | Two homogenous structures realizing the same types are isomorphic |
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Mar 31 |
revised |
Two homogenous structures realizing the same types are isomorphic Added a possible approach to a solution, based on a comment. |
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Mar 31 |
asked | Two homogenous structures realizing the same types are isomorphic |
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Mar 30 |
comment |
Nullstellensatz equivalence question I.e. (1)$\implies (2)$ is vacuously true, as (2) is always true. |
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Mar 30 |
comment |
Nullstellensatz equivalence question @xyzzyz The statement we officially have as the Nullstellensatz is: Let $F$ be an alg. closed field. If $I$ is an ideal of $F[\vec x]$, and $g\in F[\vec x]$ is s.t. g ∈ I(V(I)), then there is an $n\in \mathbb{Z}^+$ s.t. $g^n\in I$. Regardless of this, though... as you point out, my proof shows that statement (2) in my original question is true, independently of the truth of the Nullstellensatz, right? |
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Mar 30 |
asked | Nullstellensatz equivalence question |
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Mar 30 |
comment |
Can a strictly increasing bounded function have a sequence that diverges? @robjohn Yes, but maybe he meant a sequence which doesn't converge, but not necessarily goes to $\infty$. That's why I said we should have cleared up what he meant by "diverges" in the title of the question. Not that it really matters too much now, anyway. |
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Mar 29 |
comment |
Can a strictly increasing bounded function have a sequence that diverges? Perhaps the OP meant to say Is there a sequence $(x_n)$ of real numbers such that $f(x_n)\to\infty$? Then there are some subtleties to the meaning of diverges, and whether he means that or that really $f(x_n)\to\infty$. |
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Mar 22 |
accepted | Proof of Robinson's test |
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Mar 5 |
asked | Proof of Robinson's test |
