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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.
Jul
20
comment Elementary equivalence of free groups
If I remember it correctly, every formula is equivalent to a boolean combination of $\forall\exists$-formulas. This is the hard part of Sela's work. So the theory is not model-complete, but you can deduce $F_X \preceq F_Y$ for finite $X \subseteq Y$ from this by combinatorial arguments.
Jul
20
comment Elementary equivalence of free groups
@AlexKruckman Yes! (2) is true and hence so is (1).
Jul
17
comment Elementary equivalence of free groups
It includes free groups on infinitely many generators. Though for them it is much easier. For infinite $X \subseteq Y$ it is easy to show that $F_X \preceq F_Y$ by Tarski-Vaught test.
Jul
16
answered Elementary equivalence of free groups
Jun
30
comment Is for every ultrahomogenous structure M the theory Th(M) model complete?
Do you want every isomorphism between any substructures to be extended to $M$ or just isomorphisms between substructures of cardinality $< |M|$?
Jun
24
revised Cardinalities of Collections of Models
fixed grammar
Jun
24
awarded  Fanatic
Jun
17
reviewed Reject suggested edit on Method to solve epidemic differential equation
Jun
17
comment On the number of countable models of complete theories of models of ZFC
My feeling is that any completion of ZFC (assuming it is consistent) would be complicated enough to have $2^{\aleph_0}$ types and hence $2^{\aleph_0}$ countable models.
Jun
13
comment Inuition regarding Lowenheim-Skolem applied to models of set theory
It might be weird, but it is not unbelievable. Consider for example $(\mathbb N, <)$ and a substructure $(\mathbb N^+, <)$. It is not an elementary substructure, yet they are isomorphic (and hence there is an elementary embedding).
Jun
13
reviewed Reject suggested edit on 'Obvious' theorems that are actually false
Jun
12
reviewed Approve suggested edit on How to prove $k!+(2k)!+\cdots+(nk)!$ has a prime divisor greater than $k!$
Jun
12
answered elementary class and abstract elementary class
Jun
4
awarded  Yearling
May
23
comment Models of infinite groups and 'Group-like' objects
$T$ is never countably categorical, for otherwise there would be finitely many $0$-definable sets in each arity. However if $a_1, a_2$ are different, then $\{(x, y) : f_{a_1}(x) = y\} \neq \{(x, y) : f_{a_2}(x) = y\}$ (and this is part of $T$).
May
7
comment Model existence for infinitary logics
Are you looking for something along the lines of The Model Existence Theorem for $L_{\omega_1, \omega}$? See e.g. Marker's notes homepages.math.uic.edu/~marker/math512-F13/inf.pdf . It is not strictly about infinitary proof systems, but it can probably be converted to a suitable form.
Mar
31
comment Definition of dicrete ordering
A ordering is usually called discrete if every element $x$ except the top one has an immediate successor $s(x)$, (i.e. $x < s(x)$ and there is nothing in between) and every element $y$ except the bottom one has an immediate predecessor $p(y)$ (i.e. $p(y) < y$ and there is nothing in between)
Mar
28
comment The unique model of cardinal $\kappa$ of a $\kappa$-categorical countable theory is saturated.
The proof you give is more or less the proof I know. I believe it is the standard proof. I can also imagine contexts where Theorem 1 and $\omega$-stable implies $\kappa$-stable can be referred as "pretty easy".
Mar
21
comment isomorphism between divisible, totally ordered, abelian groups
Also from model theoretical point of view, any theory with an order is unstable and hence has $2^\lambda$ nonisomorphic models in an uncountable cardinality $\lambda$.