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Results tagged with continuum-hypothesis
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user 1946
Questions about the continuum hypothesis, or where the continuum hypothesis or its negation plays a role. This tag is also suitable, by extension, to refer to the generalized continuum hypothesis and related issues.
174
votes
Accepted
Solutions to the Continuum Hypothesis
Since you have already linked to some of the contemporary
primary sources, where of course the full accounts of those
views can be found, let me interpret your question as a
request for summary accoun …
- 236k
43
votes
What is the most "concrete-feeling" equivalent formulation of the Continuum Hypothesis that ...
Here are a few of my favorite characterizations of the continuum hypothesis:
Sierpiński (1951) proved that CH is equivalent to the assertion that there is a partition of the plane into two sets $\mat …
31
votes
Why does CH imply that there is a unique ultrapower of $\mathbb{N}$?
The point is that the ultrapower of any structure $\mathcal{M}$ by
a nonprincipal ultrafilter $\mu$ on $\mathbb{N}$ is countably
saturated, that is, it realizes any finitely satisfiable $n$-type with …
- 236k
23
votes
Accepted
Is it consistent with ZFC that the real line is approachable by sets with no accumulation po...
It is a very nice question, but unfortunately, this is impossible.
Each member $s\in S$ must be countable, since uncountable sets have accumulation points. And since the hierarchy is accumulating as y …
- 236k
21
votes
Accepted
When $2^\alpha = 2^\beta$ implies $\alpha=\beta$ ($\alpha,\beta$ cardinals)
François gives the correct affirmative answer. For the negative side, the usual method of proving that the negation of the Continuum Hypothesis is consistent with ZFC is to use the method of forcing t …
- 236k
18
votes
Is it possible to define cardinals that are distinct from either the $\aleph$ numbers or $\b...
One can consistently generate new cardinals simply by combining the two methods you have mentioned.
For example, I claim that it is relatively consistent with ZF that the cardinal $\aleph_1+\beth_1$ …
- 236k
17
votes
CH in non-set theoretic foundations
One consideration will be that the CH statement can become ambiguous in weaker foundations, since statements that are equivalent in ZFC are not always equivalent in weaker theories.
For example, witho …
- 236k
16
votes
How many well-orders of reals are there?
Every well order of the real numbers has some order type between $\mathfrak{c}$ and $\mathfrak{c}^+$, and for any given order type arising, every permutation of $\mathbb{R}$ induces another well order …
- 236k
14
votes
Accepted
Continuum Hypothesis and the fact that every co-finite topological space, with uncountable u...
Nice question!
I claim that this property does not necessarily imply CH. As Todd
guessed in his comment, the answer is related to certain cardinal
characteristics of the continuum.
Specifically, let …
- 236k
12
votes
Accepted
Ground Axiom and behaviors of continuum function
In the paper The ground axiom is consistent with $V\ne{\rm HOD}$ (J. D. Hamkins, J. Reitz, W.H. Woodin, PAMS 136(8):2008), we prove that the ground axiom is consistent with $V\neq\text{HOD}$, and rema …
- 236k
12
votes
Uniqueness results that follow from CH
Under CH, we have saturated models of size continuum of any consistent first-order theory in a countable language, and for a complete theory these are unique by the back-and-forth method.
(In my paper …
12
votes
A New Continuum Hypothesis (Revised Version)
Let me provide a large cardinal lower bound for the statement in question 1.
Since $\kappa\lt\kappa+1\lt 2^\kappa$ for $\kappa>1$, the statement implies that there is
a universal failure of the GCH a …
- 236k
10
votes
Complete resolutions of GCH
$\newcommand\Ord{\text{Ord}}$Easton's theorem allows us to control the continuum function on the infinite regular cardinals, and in particular, on the infinite successor cardinals, in a very flexible …
- 236k
9
votes
Accepted
The First Failure of GCH in Large Cardinals Smaller than Measurables
The statement is not true.
Theorem. If $\kappa$ is strongly unfoldable and the GCH holds below $\kappa$, then it holds at $\kappa$ also.
This is just because the strongly unfoldable cardinals are $\ …
- 236k
8
votes
Making all cardinals countable and its HOD
$\newcommand\gHOD{\text{gHOD}}
\newcommand\HOD{\text{HOD}}
\newcommand\ZFC{\text{ZFC}}
\newcommand\GCH{\text{GCH}}$
I believe that it was Peter Koepke who first proposed that we
should investigate t …
- 236k