Levon Haykazyan
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 Apr16 awarded Custodian Apr16 reviewed Close using two way simplex method Apr16 reviewed Close Cryptography Combinatorics question Apr16 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? Apr16 reviewed Approve Prove: if $a$ and $b$ are algebraic, then $a + b$, $a - b$ and ab are also algebraic Apr11 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. Dec9 awarded Caucus Nov27 answered Number of automorphisms of saturated models Nov10 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". Oct23 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 $T \models \forall \bar x \lnot \phi(\bar x)$. Oct19 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). Sep30 awarded Explainer Sep25 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? Sep25 answered Question from Hodges' textbook Shorter Model Theory Sep15 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. Sep10 awarded Nice Answer Aug24 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. Aug7 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. Aug6 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. Jul23 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.