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Results tagged with lo.logic
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user 102684
first-order and higher-order logic, model theory, set theory, proof theory, computability theory, formal languages, definability, interplay of syntax and semantics, constructive logic, intuitionism, philosophical logic, modal logic, completeness, Gödel incompleteness, decidability, undecidability, theories of truth, truth revision, consistency.
9
votes
Accepted
At what ordinal $\chi$ does $\mathrm{L}_\chi$ contain a surjection from $\omega$ to $\mathrm...
By replacement in $L_{\beta_0}$, there is no function from $\omega$ to $L_{\beta_0}$ that is definable over $L_{\beta_0}$ from parameters. Therefore no such function is in $L_{\beta_0+1}$. On the othe …
- 8,099
17
votes
Accepted
Must there be a proper class of Reinhardt cardinals if there is a Reinhardt cardinal?
No, if the existence of a Reinhardt is consistent, then it is consistent with a Reinhardt cardinal that the class of inaccessible cardinals is bounded in the ordinals. Indeed, if $j : V\to V$ is a non …
- 8,099
15
votes
Accepted
Consistency strength of strongly compact cardinal
Steel [1] showed that if $\square_\kappa$ fails for some singular strong limit $\kappa$, then $\text{AD}$ holds in ${L(\mathbb R)}$. Since Solovay showed a strongly compact cardinal implies the failur …
- 8,099
7
votes
Accepted
7
votes
Accepted
Ultrafilter projections and critical points of factor maps
You will always have $N = M$ and $k = \text{id}$. As Joel mentions, this uses Solovay's lemma that $M = H^M(\text{ran}(j)\cup \{\sup j[\lambda]\})$. We can use this to show that $k$ is surjective, by …
- 8,099
21
votes
Accepted
Are Berkeley cardinals easier to refute in ZFC than Reinhardt cardinals?
Yes, it is easier to refute Berkeleys than Reinhardts. There is a very simple refutation of Berkeleys in ZFC that is due to Woodin. It is part of the motivation for his contention that Berkeley cardin …
- 8,099
13
votes
Accepted
Natural set-theoretic principles implying the Ground Axiom
The existence of a proper class of supercompact cardinals $\kappa$ that are indestructible by $\kappa$-directed closed forcing implies the Ground Axiom. This is because it implies the Continuum Coding …
- 8,099
23
votes
Accepted
Is there a form of choice that can elude Kunen's inconsistency theorem?
Work of Usuba combined with work of Woodin shows that if there is a Reinhardt cardinal $\kappa$ that is a limit of Lowenheim-Skolem cardinals, then there is a forcing extension in which $\kappa$ remai …
- 8,099
7
votes
Accepted
Can there be no complexity bound on the definable elementary $V\rightarrow M$?
Yes, this is possible. If there is a proper class of measurable cardinals and $V = \text{HOD}$, then any class of ordinals $A$ is definably encoded by the iterated ultrapower $j_A : V\to M$ that hits …
- 8,099
10
votes
Accepted
Locating generic filters in the Lévy collapse
Note that the lemma doesn't show that $h$ is in $V[G]$, it assumes this. But yes, if $h\in V[G]$ is a subset of a set $X\in V$ such that $|X| < \kappa$, then for some $\beta < \kappa$, $h\in V[G\restr …
- 8,099
5
votes
Accepted
If GCH is breached the same way before a singular of uncountable cofinality, would that brea...
One can collapse $2^\lambda$ to have cardinality $\lambda^+$ without adding $\lambda$-sequences even if $\lambda$ is singular. Therefore one could start with $2^{\aleph_\alpha} = \aleph_{\alpha+2}$ fo …
- 8,099
2
votes
A Baire subset of reals that is not Suslin measurable
What you are calling Suslin measurable sets are also known as coanalytic sets (in the context of ZF + DC). The coanalytic sets are the sets obtained by applying your version of the Suslin operation to …
- 8,099
4
votes
Accepted
Can $\mathsf{Ord}$ be weakly compact from a second-order perspective?
If $\kappa$ is weakly compact and there is a wellorder of $V_{\kappa+1}$ definable over $V_{\kappa+1}$ without parameters, then second order logic is Loraxian for $V_\kappa$: the least branch through …
- 8,099
8
votes
Accepted
Is this definability principle consistent?
There is a consistent omniscient theory, at least assuming the consistency of a Woodin limit of Woodin cardinals.
The Maximality Principle (MP) asserts that if a sentence is forceable in $V$, it is fo …
- 8,099
6
votes
Accepted
Whatever happened to $L(j)$?
I recently noticed that Mitchell returned to $L[j]$ in "Applications of the covering lemma for sequences of measures," where he shows that the model is quite a bit larger than one might expect. The th …
- 8,099