All Questions
Tagged with model-theory forcing
43
questions
2
votes
0
answers
73
views
Question related to number of distinct forcing extensions of a countable model
A bit of context: the below question is motivated by roughly the following scenario: we have some countable model $\mathcal{M}$, and want to “count” the number of functions/sets $f$ such that $\...
17
votes
3
answers
1k
views
Minimum transitive models and V=L
Is there a c.e. theory $T⊢\text{ZFC}$ in the language of set theory such that the minimum transitive model of $T$ exists but does not satisfy $V=L$?
You may assume that ZFC has transitive models. ...
5
votes
1
answer
245
views
Highly improper forcings
The following question comes from a typo in an old notebook of mine (I changed what I was calling my forcing notion partway through writing the definition of properness):
Say that a forcing $\mathbb{P}...
7
votes
1
answer
369
views
How hard is it to get "absolutely" no amorphous sets?
A beautiful and surprising (to me at least) result around the axiom of choice is that, while full $\mathsf{AC}$ is preserved by forcing, a model of $\mathsf{ZF}$ + "There are no amorphous sets&...
10
votes
3
answers
925
views
Why can we assume a ctm of ZFC exists in forcing
Following Kunen's book, it makes clear that countable transitive models (ctm) exist only for a finite list of axioms of ZFC. So, why can we assume a ctm of the whole ZFC axioms exists and use it as ...
6
votes
1
answer
209
views
How often are forcing extensions of countable computably saturated models of $\mathsf{ZFC}$ computably saturated?
Recall that given a finite language $\mathcal{L}$, we say that an $\mathcal{L}$-structure is computably saturated (or recursively saturated) if for any computable set $\Sigma(\bar{x},y)$ of $\mathcal{...
2
votes
0
answers
183
views
Generalized models of set theory
The forcing method can be viewed as building a Boolean-valued model of set theory. Some generalizations include Heyting algebra/sheaf/lattice-valued model. However, it seems these generalizations are ...
12
votes
1
answer
602
views
Can $\mathcal{L}_{\omega_1,\omega}$ detect $\mathcal{L}_{\omega_1,\omega}$-equivalence?
Roughly speaking, say that a logic $\mathcal{L}$ is self-equivalence-defining (SED) iff for each finite signature $\Sigma$ there is a larger signature $\Sigma'\supseteq\Sigma\sqcup\{A,B\}$ with $A,B$ ...
5
votes
1
answer
260
views
Can the forcing-absolute fragment of SOL have a strong Lowenheim-Skolem property?
Previously asked and bountied at MSE:
Let $\mathsf{SOL_{abs}}$ be the "forcing-absolute" fragment of second-order logic - that is, the set of second-order formulas $\varphi$ such that for ...
7
votes
0
answers
283
views
Generic behavior of "polynomialish" models of $\mathsf{Q}$
(This question was originally asked and bountied at MSE - with different notation, some more explicit arguments, and topology in place of forcing.)
Suppose $\mathcal{R}=(R_i)_{i\in\mathbb{N}}$ is a ...
12
votes
2
answers
910
views
Is this definability principle consistent?
(Below I'm thinking only about computably axiomatizable set theories extending $\mathsf{ZFC}$ which are arithmetically, or at least $\Sigma^0_1$-, sound.)
Say that a theory $T$ is omniscient iff $T$ ...
7
votes
2
answers
673
views
Can second-order logic identify "amorphous satisfiability"?
Recall that a set is amorphous iff it is infinite but has no partition into two infinite subsets. I'm interested in the possible structure (in the sense of model theory) which an amorphous set can ...
8
votes
1
answer
446
views
Intuition behind Boolean-valued models of set theory
$\DeclareMathOperator\Card{Card}$The book Forcing Eine Einführung in die Mathematik der Unabhängigkeitsbeweise by Hoffmann provides an intuition behind boolean valued models of set theory which I will ...
3
votes
1
answer
418
views
Forcing, a technical detail
In the snippet below from Shelah's book P&I Forcing,
in the definition 5.2(2)
I do not follow why in this sentence [naturally extended to include $N\prec (H(\mu^\dagger),\epsilon),\mu\in N$] $N$ ...
14
votes
1
answer
506
views
Is there an infinitary sentence which is absolutely not second-order expressible?
This is a "forcing-absolute" followup to this question, whose answer was largely unsatisfying. The question is:
Suppose $V=L$. Is there an $\mathcal{L}_{\infty,\omega}$-sentence $\varphi$ ...