All Questions
Tagged with forcing soft-question
7 questions
14
votes
2
answers
678
views
Are there interesting examples of theorems proved using ‘height’ extensions?
It's well known that forcing is more than a tool for proving independence: We can prove theorems and formulate axioms in theories like $\mathsf{ZFC}$ by moving to forcing extensions (e.g. $\mathfrak{p}...
- 236k
10
votes
3
answers
1k
views
Philosophy of forcing and ctm
I asked a similar question on SE before and received an answer. Not completely convinced, I decided to ask it here with some modifications. Note: I understand how forcing works and how it proves ...
- 613
15
votes
3
answers
1k
views
What did Paul Cohen mean by saying that generic sets of natural numbers have "no asymptotic density?"
In Paul Cohen's original 1963 paper on forcing, The independence of the Continuum Hypothesis, published in PNAS, he gives his general proof sketch of how he intends to create a model of ZFC that doesn'...
- 35.5k
19
votes
0
answers
905
views
What examples of existence forcing proofs are there?
Forcing proofs tend to be fairly constructive, in the sense that if I claim that there is a forcing that does something, I usually prove this by constructing that forcing.
There are only a handful of ...
- 39.7k
5
votes
2
answers
363
views
Formal Definition of Finite Conditions
Forcing with finite conditions is a common concept used by set theorists. I was thinking about its meaning, but I couldn't find any exact definition of it. At the first glance it seemed to me that ...
- 48.8k
8
votes
2
answers
789
views
Why is this theorem (about $L(P(\omega_1))^V$ and $L(P(\omega_1))^{V[G]}$) nice?
I was recently told that the following (due to M. Viale) is a nice theorem:
Suppose there are arbitrarily large supercompacts, and $\mathrm{MM}$ holds in $V$. Let $G$ be generic for a proper ...
- 44.4k
2
votes
2
answers
465
views
When forcing with a poset, why do we order the poset in the order that we do?
In forcing, we take a collection of forcing conditions and impose a partial order on them. The convention is that if $p$ is stronger than $q$, then we say $p < q$. This is perfectly fine, but it ...
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