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 |
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.
4
votes
A question regarding strong cardinals and measure sequence
Let me explain in a few more details the argument. Let $j_0: V \to V$ and for each $n$, let $j_{n+1}: V \to M_n \simeq M_{n+1}=Ult(M_n, j_n(E))$. Then:
1) $j_{n+1}=j \circ j_n,$
2) $V_{\kappa+2}^\al …
5
votes
Accepted
Slim Kurepa tree at a singular strong limit cardinal of uncountable cofinality
The following is proved by Erdos-Hajnal-Milner in ``On sets of almost disjoint subsets of a set. Acta Math. Acad. Sci. Hungar 19 1968 209–218'', from which the required result follows
Theorem. ssume …
10
votes
Accepted
Does "Every infinite set is splittable" imply $\mathsf{AC}$?
The answer is no and it follows from the following:
It is consistent that $AC$ fails but for all infinite cardinals $\kappa, 2 \cdot \kappa=\kappa.$
The above result is proved by Sageev:
Sagee …
5
votes
Accepted
Is it consistent that the gaps between cardinals $\kappa$ and $2^\kappa$ "get larger and lar...
Yes, in the Foreman-Woodin model for the global failure of $GCH$ your statement is true.
See The generalized continuum hypothesis can fail everywhere.
3
votes
Rigid structure which is generically homogeneous
Yes, it is at least consistent. See:
Fuchs, Gunter, Club degrees of rigidity and almost Kurepa trees. Arch. Math. Logic 52 (2013), no. 1-2, 47–66.
In this paper a (very) rigid Souslin tree $T$ is co …
8
votes
Accepted
Axiom of choice and the equality between second-order constructible universe and HOD
The equality $L_{SO}=HOD$ can not be proved just in $ZF$. This is proved in the paper ``The consistency of the theory $ZF+L^1\neq HOD$'' by Szczepaniak.
Here $L^1$ refers to what you named $L_{SO}$. …
7
votes
What is the status of the assertion "There are arbitrarily large cardinals with the tree pro...
Let me address your question in the comments.
If one starts with a weakly compact cardinal $\kappa,$ then there exists a forcing extension in which all cardinals and cofinalities are preserved (in pa …
14
votes
Accepted
Is it consistent that $2^{(\cdot)}$ is "surjective" on the class of uncountable ordinals?
By Konig's lemma $cf(2^\kappa) > \kappa,$ so for example $\aleph_\omega$ can never be of the form $2^\kappa$ for any $\kappa.$
9
votes
Accepted
Does every Aronszajn tree has a Suslin or a Special subtree?
It is consistent that the answer is no. The following is proved in Beaodouin's thesis ``On uncountable trees and linear orders'', as Theorem 1.10:
Theorem. Assume $\kappa^{<\kappa}=\kappa$ and $\D …
8
votes
Accepted
Applications of "model-theoretic" forcing
Model theoretic forcing is extended to wider languages with some applications, for example:
1- Model theoretic forcing in analysis which gives an exposition of forcing for metric structures.
2- Shel …
12
votes
Accepted
Can there be a tree of height $\omega_2$ having all levels countable, with no cofinal branch?
The following theorem of Kurepa answers the question.
Theorem (Kurepa) Suppose that $\kappa$ is regular, $\lambda< \kappa$, and $T$
is a $\kappa$-tree each of whose levels has cardinality less …
2
votes
The transcendence degree of $\mathbb R$ after adding a Cohen
I think the following argument might be easier than Hamkins answer. It also does not use AC. So suppose $g$ is a Cohen real. We want to show that there are continuum many mutually generic Cohen reals …
18
votes
What is the large cardinal strength of the assertion that every $\kappa$-complete filter on ...
Let us call a cardinal $\kappa, \kappa-$compact, if every $\kappa-$complete filter on $\kappa$ extends to a $\kappa-$complete ultrafilter.
The following is proved in Gitik's paper On measurable cardi …
8
votes
Boolean Valued Models of PA
The paper "Partially definable forcing and bounded arithmetic" by
A. Atserias and M. Müller presents a very general framework of forcing for models of (week) arithmetic.
Its presentation is more clos …
9
votes
What can the extremely large cardinals tell us about small sets?
The paper Generic $I_0$ at $\aleph_\omega$ by Vincenzo Dimonte might be of interest to you. It introduces the notion of being generic $I_0$ at $\aleph_\omega$ (Def. 3.1 of the paper), and proves sever …