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Apr
16
awarded  Custodian
Apr
16
reviewed Close using two way simplex method
Apr
16
reviewed Close Cryptography Combinatorics question
Apr
16
reviewed Approve Are there arbitratily long runs of consecutive integers n that are NOT of the form $n = p^k$ or $n = 2p^k$ for some $k>0$ and $p$ an odd prime?
Apr
16
reviewed Approve Prove: if $a$ and $b$ are algebraic, then $a + b$, $a - b$ and ab are also algebraic
Apr
11
comment Saturated model for Th(Z,+,-,0,1)?
I am not sure, but I think $\hat {\mathbb Z}$ the profinite completion of $\mathbb Z$ would be an $\omega$-saturated model.
Dec
9
awarded  Caucus
Nov
27
answered Number of automorphisms of saturated models
Nov
10
comment Uncountably Categorical Theories and Embeddings
It is basically true. For the details see the section 6.3 of Tent and Ziegler "A course in model theory".
Oct
23
comment Marker Exercise 2.5.10: universal part of a theory and supermodel
Since $\bar a$ does not occur in $T$, the relation $T \models \lnot \phi(\bar a)$ implies $T \models \forall \bar x \lnot \phi(\bar x)$.
Oct
19
comment A fragment of Exercise 1.3.4 in _Shorter Model Theory_ by Hodges
It is not true, precisely because of the reason you stated. E.g. take $\mathcal B = \mathcal A$ and let $\phi$ be the identity on $S$. Then extend $\phi$ arbitrarily to $A$. Then $\phi$ satisfies the condition of your statement, but not the conclusion (in general).
Sep
30
awarded  Explainer
Sep
25
comment Question from Hodges' textbook Shorter Model Theory
@PrimoPetri, yes $\exists x \chi$ is false in the empty structure. Did I say something that contradicts it?
Sep
25
answered Question from Hodges' textbook Shorter Model Theory
Sep
15
comment Indiscernible subsequences
If the theory has order, then the pairs $\{a_i, a_j\}$ are divided into to parts depending on whether $a_i < a_j$ or not. An indiscernible sequence must be homogeneous, showing that $\kappa \to (\kappa)^2_2$. Thus $\kappa$ must be weakly compact. I think using a similar argument we can show that $\kappa$ is Ramsey. I don't know if being Ramsey is sufficient.
Sep
10
awarded  Nice Answer
Aug
24
comment Elementary Model Theory
You probably mean $\hat {\mathfrak A} \models \phi(\bar a_1, ..., \bar a_n)$ entails $\mathfrak A \models \phi(\bar a_1, ..., \bar a_n)$. If this is not clear, then try to prove it by induction on $\phi$. The proof might depend on exact definitions van Dalen uses. So you are better of including them in the question.
Aug
7
comment Let $h: A \to B$ be a weak homomorphism. Is h$[A]$ a substructure of $B$?
Correct, because it will contain the interpretations of all constant symbols and be closed under the interpretations of functional symbols.
Aug
6
comment Let $h: A \to B$ be a weak homomorphism. Is h$[A]$ a substructure of $B$?
No, in general $\mathfrak B$ can have additional constant or functional symbols and $h[\mathfrak A]$ may not be closed under them.
Jul
23
comment Decision and the Uncountable Spectrum
There are uncountably many complete theories. You can't have an algorithm that takes as input an arbitrary complete theory.