Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with lo.logic
Search options answers only
not deleted
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
continuity points of elementary embeddings from $0^\sharp$
Claim: Suppose $i : L \to L$ is an elementary embedding. If $\kappa$ is an $L$-regular cardinal and $\sup i[\kappa] < i(\kappa)$, then $\kappa$ is a Silver indiscernible.
The claim answers your more …
- 8,099
8
votes
Accepted
Buying more absoluteness for countable transitive models?
First, it is consistent that there is a ctm but no $\Sigma^1_2$-correct ctm, for example if $V$
is the minimal model with a ctm. If there is a model of ZFC $M$
containing all the reals (e.g., if the …
- 8,099
5
votes
Accepted
How many reals can we construct by iteratively writing down truth tables for ZFC?
I think it is easier to think about the question generalized to an arbitrary transitive model of ZFC, resisting the natural urge to grasp towards the Absolute. So fix such a model $M$, and let $\math …
- 8,099
6
votes
Accepted
For ideals, does normal imply countably complete?
I think the answer is yes. It helped to take generic ultrapowers, somehow.
We may assume without loss of generality that $I$ is a normal ideal on $P(\lambda)$ where $\lambda$ is an infinite cardinal. …
- 8,099
9
votes
Accepted
Finding many subsets of $V_{\lambda+2}$ stable under $j:V\prec V$
There can be no such set $\mathcal A$. In fact, no set $\mathcal A$ with $j(\mathcal A) = \mathcal A$ and $j(x) = x$ for all $x\in \mathcal A$ can surject onto $\kappa$, and this does not require that …
- 8,099
5
votes
Accepted
Number of ultrafilters in an extender
TLDR: Yes, all the ultrafilters are different.
Suppose $M$ is a transitive class and $j : V\to M$ is an elementary embedding.
Some notation.
For any $x\in M$, let $H_x = \{j(f)(x) : f\text{ is a …
- 8,099
5
votes
Stationary correctness of ultrapowers by low order measures
I once tried and failed to answer this question in the canonical (forgive me) inner models, and your really nice observation about strong cardinals and Mitchell rank helped me finally make some progre …
- 8,099
6
votes
What happens with large singular cardinals on the far side of the HOD dichotomy?
This is independent relative to the failure of the HOD hypothesis in the presence of large cardinals.
We first give a positive answer under GCH. (Note that if it is consistent for the HOD Hypothesis …
- 8,099
11
votes
Accepted
If $U,D$ are $\kappa$-complete nonprincipal ultrafilters on $\kappa$ and $j_U(U) = j_D(D)$, ...
It is consistent with ZFC that the answer is no, but under the Ultrapower Axiom, the answer is yes, not only for $\kappa$-complete ultrafilters on $\kappa$, but also for arbitrary countably complete u …
- 8,099
5
votes
Accepted
Extendibility vs supercompactness
$2$-extendibility reflects $2^\kappa$-supercompactness. It suffices to show that any $2$-extendible $\kappa$ is $2^\kappa$-supercompact. Then if $\mathcal U$ is a normal fine $\kappa$-complete ultrafi …
- 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
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
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
4
votes
Accepted
A weak (?) form of Shelah cardinals
To answer the first three questions negatively, the key is to show that measurable weakly Shelah cardinals are limits of weakly Shelah cardinals.
To see this, suppose that $\kappa$ is weakly Shelah a …
- 8,099