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 |
Forcing is a method first used to prove the continuum hypothesis is independent of the classical axioms of set theory
5
votes
2
answers
411
views
Question about prompt names of ordinals
So:
The following concept is due to Shelah and I have some issues with a claim using this notion:
Suppose that $\nu$ is a limit ordinal and that $P_\nu$ is an iteration of forcing notions. …
2
votes
1
answer
320
views
Gluing functions together in the generic extension
However a (probably very easy) question came up:
Assume that we force with a Boolean algebra of the form $P(A) / I$, where $A$ is an arbitrary set and $I$ is an ideal over $A$; i.e. our forcing conditions …
8
votes
1
answer
466
views
Properness of quotient forcing
It is well known that if $P$ is a proper notion of forcing and $\Vdash_P \dot{Q} \text{ is proper}$ then the iteration $P \ast \dot{Q}$ is proper. … My intuition would tell me no, we could probably add in the first step a new stationary set, which gets killed after the second forcing. …
4
votes
2
answers
613
views
Examples of stationary set preserving forcings that are not semiproper?
A notion of forcing $P$ is called stationary set preserving iff each stationary subset of $\omega_1$ remains stationary in $V^P$. … It is standard to show that semiproper (and of course proper) notions of forcing are stationary set preserving. …
15
votes
0
answers
605
views
Cohen/Random reals over intermediate models in countable support iterations
I am also highly interested in the dual situation, where we replace Cohen forcing with Random forcing in the above. Can we guarantee that there is a real $r$ which is Random over all $V^{P_n}$'s? …
3
votes
Complexity of the statement 'P is proper'
I think that I might have found a solution to this rather dispensable question. I will sketch it:
Consider the following characterization of properness:
$P$ is proper iff for all $\lambda > 2^{|P|}$ …
3
votes
1
answer
583
views
Equivalent definitions of $(M,P)$-genericity
I started to read the chapter 31 in Jechs book about proper forcing. … Jech gives the following definition:
Let $(P,<)$ be a fixed notion of forcing, $\lambda > 2^{|P|}$ and let $M \prec (H_{\lambda}, \in, <, ..)$, where $H_{\lambda}$ is the set of all sets whose transitive …
8
votes
Size of stationary sets
There exists a stationary subset of $P_{\omega_1} (\omega_2)$ of size $\aleph_2$. This is a result of Baumgartner and you can find a proof for this here: Why is this set stationary?
I don't dare to a …
3
votes
3
answers
448
views
Complexity of the statement 'P is proper'
Assume that $(P,\le)$ is a notion of forcing. …
5
votes
Examples of ZFC theorems proved via forcing
Later a more elementary proof was found, using no 'metamathematical' concepts as forcing or ultrapowers of the universe, similar to the history of the proofs of Silvers theorem. …
6
votes
1
answer
320
views
Elementary chains in forcing extensions of $M_1$
Now suppose that $\mathbb{P}$ is a forcing notion of size $< \delta$, which preserves $\omega_1$ and that $G$ is a generic filter for the forcing. … Note that the statement is true if $\mathbb{P}$ is just the trivial forcing, however I would be interested in the case when we collapse cardinals down to $\omega_1$. …
5
votes
1
answer
498
views
The closure of a generic ultrapower
Let $I$ be a normal ideal on $P_{\kappa} (\lambda)$. Let $V$ denote our ground model. Now we force with the $I$-positive sets, and if $G$ is the resulting generic filter, it can be shown that $G$ is a …
8
votes
What ccc forcings add a Suslin tree?
If we start with $L$ as our ground model then whenever $T$ is a Suslin tree, the forcing $\mathbb{P}_T$ which shoots a branch through $T$ will always introduce new Suslin trees. … Thus in $L$, $\mathbb{P}_T$ is an example of a forcing which has the ccc, adds Suslin trees and does not add a Cohen real. …
10
votes
Kunen's use of Countable Transitive Models
It is true that you can use Löwenheim Skolem to get a countable model $M$ of ZFC assuming $Con(ZFC)$. But to use Mostowski you need additionally the well foundedness of that model, which doesn't have …
6
votes
1
answer
268
views
$\omega_2$-sequence of Suslin trees
Is it possible to have an $\omega_2$-length sequence of ($\omega_1$-)Suslin trees such that if one builds the product of finitely many trees in that sequence, one ends up with a Suslin tree again?
Th …