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Model theory is the branch of mathematical logic which deals with the connection between a formal language and its interpretations, or models.

1 vote
1 answer
363 views

Convergence of the harmonic series in larger fields

The Harmonic series is well known and its divergence was proven back in the middle ages. I've taken an introductory course in model theory so I know a bit about RCF and some properties of it. We did …
Asaf Karagila's user avatar
  • 39.9k
5 votes

Is there an inner model between two distinct inner models of ZFC?

The answer to the second question is negative. Take, for example, Sacks forcing over $L$. The result is $L[r]$ where $r$ is a real number and the following property is true: $$\forall x(x\in L\lor r\i …
Asaf Karagila's user avatar
  • 39.9k
4 votes

Joint Forcing Extension Property

Note that it is consistent that $\sf ZFC$ doesn't have weak JFEP either. Suppose that there are exactly two different height of transitive models, that is some $\alpha<\beta$ such that $L_\alpha$ and …
Asaf Karagila's user avatar
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6 votes
Accepted

Is there a lower bound on the size of a supertransitive model of ZFC?

If $M$ is supertransitive and satisfies $\sf ZFC$, then $\omega\in M$, and more importantly, $V_\omega\in M$. Now by recursion, if $\alpha$ is an ordinal in $M$, then $V_\alpha\in M$ as well. Therefor …
Asaf Karagila's user avatar
  • 39.9k
11 votes

Taking a proper class as a model for Set Theory

Yes, that is true. But note that in its nature statements like $\operatorname{Con}(T)$ are meta-theoretic statements. So when we say that $V$ is a model of $\sf ZF$, we mean that in the meta-theory it …
Asaf Karagila's user avatar
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10 votes
1 answer
431 views

What is first-order logic with Dedekind-finite sets of variables?

The usual set up of first-order logic is with an infinite reservoir of variables which we can use in formulas. This is one of the annoying reasons why we need to put $\aleph_0$ into the cardinal equat …
Asaf Karagila's user avatar
  • 39.9k
8 votes
Accepted

Is there a 'Constructible Universe' that is a submodel of a non-transitive model of $ZF$?

What $\sf ZF$ actually proves is that there is a class called $L$, and for every axiom of $\sf ZFC$, the relativization of that axiom holds in $L$. Therefore, the fact we are talking about non-transi …
Asaf Karagila's user avatar
  • 39.9k
2 votes
Accepted

Are externally pointwise definable models of ZFC subject to the same limitations of the inte...

Obviously not. Any model living inside a pointwise definable model is going to be "externally pointwise definable" for obvious reasons. Now take any generic, symmetric, or otherwise extension of the m …
Asaf Karagila's user avatar
  • 39.9k
7 votes
Accepted

Elementary countable submodels in Gödel's universe

Very clearly not. Take some countable elementary submodel $M_0$ of $L_{\omega_2}$, and take $M_1$ to be another one, but with $M_1$ a end extension of $M_0$. We can find such models by first finding t …
Asaf Karagila's user avatar
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3 votes

What axioms (other than choice) have a taming effect on the ordering of cardinalities?

I don't have anything in mind which "tames" the cardinals structure. I also think that you don't fully understand the consequences of $\sf AD$ on the cardinal structure if you say that it tames the st …
Asaf Karagila's user avatar
  • 39.9k
3 votes
Accepted

If a theory speaks of sets that cannot be forced to be parameter free definable, then does t...

If $T$ is a theory which proves "there is no extension of the model to a model of $\sf ZFC$ without adding ordinals", then there is no extension of models of $T$ by a class forcing to a pointwise defi …
Asaf Karagila's user avatar
  • 39.9k
13 votes
Accepted

Can local $0^\#$ exists in L?

Take a countable elementary submodel of $L_\kappa[0^\#]$, code that into a real, note that "There is a real coding a well-founded model of $V=L[0^\#]$" is a $\Sigma^1_2$ statement, remember what Shoen …
Asaf Karagila's user avatar
  • 39.9k
26 votes

What are some nice uses of ultraproducts/ultrapowers?

One of my favourite applications is proving that $\sqrt2$ is irrational using ultraproducts. This requires knowing a nontrivial fact, that there are infinitely many prime numbers $p$ such that $x^2\no …