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 ...

42
votes

### What is the most "concrete-feeling" equivalent formulation of the Continuum Hypothesis that you can think of?

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 $\...

Community wiki

42
votes

Accepted

### What is the most "concrete-feeling" equivalent formulation of the Continuum Hypothesis that you can think of?

Since we can force $CH$ and $\neg CH$ between transitive models of ZFC, we know that $CH$ cannot be equivalent to a statement that is absolute between transitive models, including arithmetic ...

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33
votes

Accepted

### If someone can prove Goldbach conjecture assuming the continuum hypothesis, do we consider the conjecture proved?

Because the Goldbach conjecture is an arithmetic statement, it is absolute between any two models which agree on the natural numbers.
Now, given any model of $\sf ZFC$, $M$, there is a forcing ...

29
votes

Accepted

### On the probability of the truth of the continuum hypothesis

For any $E$ modelling ZFC and for each $n\in\omega$, we must have $(n,n)\not\in E$. Therefore $M_{ZFC}$ is contained in the cylinder set defined by $(0,0)\not\in E,\dots,(n,n)\not\in E$ which has ...

25
votes

### Unnecessary uses of the Continuum Hypothesis

Theorem: The space $\mathbb N^*$ of non-principal ultrafilters on $\mathbb N$ is not homogeneous.
Using CH, it is fairly straightforward to prove there is a special kind of ultrafilter called a $P$-...

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25
votes

### What is the most "concrete-feeling" equivalent formulation of the Continuum Hypothesis that you can think of?

Morayne showed the following in this paper:
Theorem. The continuum hypothesis is equivalent to the existence of a surjection $F : \mathbb{R} \to \mathbb{R}^2$ such that for every $x \in \mathbb{R}$, ...

Community wiki

25
votes

Theorem (Erdős [1964]). The following assertion is equivalent to $\mathsf{CH}$: There exists an uncountable family $ℱ$ of entire analytic functions such that for each $z ∈ ℂ$, the set of values $\{ f(...

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24
votes

Accepted

### Does $V = \textit{Ultimate }L$ imply GCH?

In his slide Absolutely ordinal definable sets John Steel writes:
At the same time, one hopes that V = ultimate L will yield a detailed ﬁne structure theory for V, removing the incompleteness that ...

24
votes

A cloud around $x$ in $\Bbb R^2$ is a set $C$ such that for all lines $\ell$ through $x$, $C\cap\ell$ is finite. A cloud is a cloud around some point.
Theorem: (Komjáth, Schmerl) the continuum ...

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23
votes

Accepted

### Must uncountable standard models of ZFC satisfy CH?

Remarks (2) and (3) are added in this edit.
What Cohen's quoted proof outline is leaving implicit is the following statement in which $\mathrm{Con}(T)$ means "$T$ is consistent".
$(*)$ ...

21
votes

Accepted

### Can GCH fail everywhere every way?

No. An early nontrivial constraint on the $\beth$ function comes from Kőnig's Theorem, that for all infinite $\kappa$, $\mathrm{cf}(2^\kappa)>\kappa$. This implies that we cannot have $\beth_\...

19
votes

Accepted

### Is $[0,1]$ a disjoint union of $\aleph_1$ compact subsets with empty interior?

It is independent of ZFC. As you mention, CH implies the answer is yes. A different axiom, MA+$\neg$CH, implies the answer is no.
A little more precisely, there is a cardinal number denoted $\mathrm{...

18
votes

### Is it possible to define cardinals that are distinct from either the $\aleph$ numbers or $\beth$ numbers?

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 $\aleph_1+\beth_1$ ...

15
votes

### Does anyone understand this comment about the continuum hypothesis?

You might read "if we do not allow generic functions to exist then the continuum hypothesis is obviously true" as a reference to the fact that Borel sets satisfy CH, or possibly to the fact that CH ...

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, ...

14
votes

Accepted

### Continuum Hypothesis and the fact that every co-finite topological space, with uncountable underlying set , is contractible

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 ...

14
votes

This is a Hilbert-hotel variant I came up with that I'm putting in a philosophy of math book I'm currently writing.
In the infinite binary tree shown, imagine that it is a hotel, and infinitely many ...

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13
votes

Accepted

### Can the cardinal $2^{\aleph_0}$ be order-embedded in ${\cal P}(\omega)/(\text{fin})$?

Yes! The fact that this is consistent is originally due to Laver. Later, Baumgartner, Frankiewicz, and Zbierski strengthened Laver's result to the following theorem:
Theorem: Is it consistent that $\...

13
votes

The continuum hypothesis is equivalent to the assertion that for any set $X\subseteq\mathbb R$ either White or Black has a winning strategy in the following infinitely long game $G(X)$.
At the $n^\...

Community wiki

13
votes

My favorite equivalence is that CH fails if and only if whenever we color $\mathbb{R}$ with countably many colors, we can find monochromatic (and distinct) $x$, $y$, $u$, and $v$ such that $x+y = u + ...

Community wiki

11
votes

### Can GCH fail everywhere every way?

When working in ZF, one can have more freedom. See
An Easton-like Theorem for Zermelo-Fraenkel Set Theory with the Axiom of Dependent Choice and An Easton-like theorem for Zermelo-Fraenkel Set Theory ...

11
votes

Accepted

### Bernstein's proof of the continuum hypothesis

This is really a long comment. This paper has been reviewed twice by zbMATH: one by H. B. Curry, which is not informative; another by W. Ackermann, which is in German. The following is the (...

Community wiki

10
votes

### Last Status of Feferman's Conjecture on Indefinite Value of Continuum

Feferman has posted an updated version of his paper:
http://math.stanford.edu/~feferman/papers/CH_is_Indefinite.pdf
Towards the end, he mentions that Michael Rathjen has proved Feferman's conjecture,...

10
votes

### Solutions to the Continuum Hypothesis

In Aug. 2020, I gave a talk at Wuhan with the title "How many real numbers are there?", taking into account my result with D. Asperó on MM++ => (*). There is a recording: https://m....

10
votes

### Is it still an open problem whether $\mathbb{R}^\omega$ is normal in the box topology?

I'm transcribing here some things from the comments section, so that this question can be marked as answered:
The problem is still open.
This survey by Roitman and Williams, from 2015, said that it ...

Community wiki

10
votes

### Continuum hypothesis and cardinality of infinite tree paths

If the nodes add a fixed amount of new nodes at each level, then the number of infinite paths seems to be countable.
It does not seem so for me. Even if the number of nodes $k_n$ on level $n$ ...

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^{\...

10
votes

The global dimension of the ring $\prod_{n = 1}^\infty \mathbb{F}_2$ is $k+1$ if and only if $2^{\aleph_0} = \aleph_k$. Note that the given ring, in another guise, is the power set of the set $\...

Community wiki

9
votes

### Does $V = \textit{Ultimate }L$ imply GCH?

During this year's conferene on inner model theory in Münster, Gabriel Goldberg proved that the so-called Ultrapower Axiom implies that $\mathrm{GCH}$ holds above a supercompact cardinal (and since ...

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