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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.
47
votes
Accepted
Hilbert's alleged proof of the Continuum Hypothesis in "On the Infinite"
The article "Hilbert and Set Theory" by Dreben and Kanamori devotes Section 7 to this argument and an analysis of its flaws. Dreben and Kanamori use the translation provided by van Heijenoort, so that …
15
votes
Accepted
Almost everywhere “filling” of the continuum by the first uncountable cardinal without CH
The existence of such an $f$ is consistent with $\mathsf{ZFC+\neg CH}$. Suppose $\mathfrak{c}=\aleph_2$ and every set of size $\le\aleph_1$ is null (this is consistent with $\mathsf{ZFC}$; it follows, …
10
votes
Consistency of $c=2^{\aleph_0}=2^{\aleph_1}=\ldots=2^{\aleph_n}\ldots$, for every $n<\omega$
The answer is yes. Starting for simplicity with a model of $\mathsf{ZFC+GCH}$, if we force to add $\aleph_{\omega+1}$-many Cohen reals we will get (by the usual "nice names" analysis) $2^{\aleph_0}=2^ …
9
votes
Mathematical Evidence Backing $|\mathbb{R}|=\aleph_2$
A partial answer to (2): my recollection is that the strongest viewpoint Godel ever put forth on the value of the continuum was that it should be $\aleph_2$, this being captured in his manuscript "Som …
8
votes
Very Large Cardinal Axioms and Continuum Hypothesis
Here's a brief sketch of why, assuming $ZFC+I_0$ is consistent, so is $ZFC+CH+I_0$. (This is just Levy-Solovay.)
Suppose $\lambda$ is $I_0$ - that is, there is a nontrivial elementary embedding $j$ o …
4
votes
Accepted
What's the consistency status/strength of this limitation principle?
This principle is inconsistent: consider the formula $\theta(x)$ = "$x^+$ is the smallest infinite cardinal at which $\mathsf{CH}$ fails." The formula $\theta$ cannot hold on more than one infinite ca …
3
votes
Bernstein's proof of the continuum hypothesis
Ackermann's above-cited review focuses on the second of Bernstein's proposed rules. However, even the first ("axiom of identity") is broken. Gentling tweaking the language, this axiom reads:
$(\star) …
2
votes
Accepted
Continuum function maximum
The intricacies of arithmetic at singular cardinals notwithstanding, I think you're looking for something which doesn't exist.
Given any cardinals $\kappa,\lambda$, singular or regular, there is a (se …