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10 votes
Accepted

Is the class of models of ZFC minus Replacement for which Replacement fails axiomatisable?

No, the compactness theorem plus a result specific to $\mathsf{ZFC}$ prevents this. Suppose $T$ were such a set of sentences. Then the union of $T$ and the replacement scheme is unsatisfiable, so by ...
Noah Schweber's user avatar
7 votes
Accepted

Theory $\mathcal{T}$ with predicate $P$ that is satisfied by countably many elements in every model of $\mathcal{T}$?

Assuming you meant countable as “countably infinite”, then no, with the same proof as the proof that the entire model cannot be restricted to being countable: Add to the language uncountably many ...
David Gao's user avatar
  • 7,430
5 votes

Compactness Theorem for First Order Logic

In fact, there is a general tactic we can use here to "reduce" classes to sets. Rather than consider the class(es) of all models, look at the sets of models of minimal rank in the cumulative ...
Noah Schweber's user avatar
5 votes

Theory $\mathcal{T}$ with predicate $P$ that is satisfied by countably many elements in every model of $\mathcal{T}$?

Not unless your theory implies $P$ is finite. Otherwise, you can use the same argument as the upward Lowenheim-Skolem theorem, only stipulating that the new constants are in $P$, and get a model where ...
spaceisdarkgreen's user avatar
4 votes
Accepted

Is the first-order theory of the class of well-ordered sets the same as the first-order theory of the class of ordinals?

Yes, since every ordinal is a well-ordered set, and every well-ordered set is isomorphic to an ordinal. So the classes are the same (up to isomorphism).
Alex Kruckman's user avatar
3 votes

An explicit axiomatization of the theory of the class of all $Z_n$

In very recent work, Derakhshan and Macintyre have shown that $\mathrm{Th}(K)$ is decidable. This answers a question of Ax from 1968. Since every decidable theory $T$ is recursively axiomatizable (...
Alex Kruckman's user avatar
3 votes
Accepted

Do rational inequalities preserve componentwise convex combinations?

When you say "I have two relations $R(a_1,\dots,a_n)$ and $R(b_1,\dots,b_n)$", presumably you mean that $R(x_1,\dots,x_n)$ is a formula in your restricted class (a conjunction of atomic ...
Alex Kruckman's user avatar
3 votes
Accepted

Inaccessible cardinal in standard transitive models of ZFC

No. If there is any transitive model containing inaccessibles, there will be (by using Lowenheim-Skolem to get a countable elementary submodel and then Mostowski collapsing it) a countable transitive ...
spaceisdarkgreen's user avatar
2 votes

Non-isomorphic countable models of $\text{Th }(\mathbb{R},<, I)$

Clearly we have that $(\mathbb{Q},<,I)$ is a model of the theory since $(\mathbb{Q},<)\cong (\mathbb{R},<)$ and $\mathbb{Q}$ contains all the same integers that $\mathbb{R}$ contains. This ...
Alex Kruckman's user avatar
2 votes
Accepted

Axiomatization of the theory of finite structures in a signature consisting only of function symbols

Since every finite structure is interpretable in a finite $\{*\}$-structure (basically: a binary relation symbol, let alone a binary function symbol, is already "expressively complete"), the ...
Noah Schweber's user avatar
2 votes
Accepted

Every element realizing a nonempty type in a $\omega$-saturated model is definable

In this case, "definable" means "definable with parameters from $M$". So a finite tuple of elements $\bar{a}$ from $M$ is always definable by the formula $\bar{x} = \bar{a}$. In ...
Mark Kamsma's user avatar
  • 13.1k
1 vote

Why are types important in model theory?

A good place to get started is with compactness arguments. When doing a compactness argument to construct new models of a theory, one typically adds in a new constant symbol (or several) and specifies ...
TomKern's user avatar
  • 2,969
1 vote

Can we sheaf-theoretically force a violation of the continuum hypothesis in a (nice) topos which is *bicomplete*?

$\require{AMScd}$Well, I don't have a full answer since maybe there is something about "internal" completeness still to be said, but someone hinted the following to me - which more or less ...
FShrike's user avatar
  • 41.8k
1 vote
Accepted

A question on the proof of the cardinality property of Stone space

If $q$ is a type containing $\phi$ and $q\neq p$, then there is some $\psi\in q$ such that $\psi\notin p$, so $q\in [\phi\land \psi]$.
Alex Kruckman's user avatar
1 vote
Accepted

Understanding Open Mapping Theorem for Extensions of Types in Model Theory

I agree with you that there is an argument missing from the proof. Marker has neglected to show that $\psi\in p$, and this is where we need to use that $p'$ is an nonforking extension of $p$. $\...
Alex Kruckman's user avatar
1 vote

Are normal theories categorical in uncountable cardinalities?

There is no implication in either direction between normality and $\aleph_1$-categoricity. The theory of an equivalence relation with two classes, both of which are infinite, is normal but not $\...
Alex Kruckman's user avatar
1 vote
Accepted

Inaccessible cardinals and consistency of ZFC

For all $\phi$ in $ZFC$, we have : $ZFC+\exists \kappa'(\kappa' inaccessible)\vdash \forall\kappa(\kappa inaccessible\rightarrow \phi^{V_{\kappa}})$ This makes sense and is true, provided we ...
spaceisdarkgreen's user avatar
1 vote
Accepted

questions about proving ACF(Algebraically Closed Fields) has quantifier elimination

I assume we are working with the language of rings: $L = \{+,\times,-,0,1\}$. Let $M$ be a field, and let $M'$ be an $L$-substructure. Then $M'$ is a subring of $M$, so it is an integral domain. Let $...
Alex Kruckman's user avatar
1 vote

Introductory text on logic for those interested in the intersection of logic, algebra, and topology?

Vickers Topology via Logic treats topology thru the lens of logic. Its a fairly easy read with low prereqs, but maybe not the best for foundation building. For a general 1-year sequence (say model ...
emesupap's user avatar
  • 698

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