Questions about the generalized continuum hypothesis.

**10**

votes

**1**answer

256 views

### Precipitous ideals and GCH

It is well known that ZFC + "There is a measurable cardinal" is equiconsistent with ZFC + "There is a precipitous ideal on $\omega_1$." Is ZFC + "There is a measurable $\kappa$ such that $2^\kappa ...

**4**

votes

**2**answers

233 views

### Is injectivity of $2^{(\ldots)}$ weaker than $\mathsf{GCH}$? [duplicate]

The following statement cannot be proven in $\mathsf{ZFC}$:
(S) : If $A, B$ are sets with $|A| < |B|$, then $2^{|A|} = |{\cal P}(A)| < |{\cal P}(B)| = 2^{|B|}$.
Obviously, ...

**10**

votes

**1**answer

203 views

### Is there a (first-order) sentence which admits $(\aleph_2,\aleph_0)$ iff a Kurepa tree exists?

In Chang and Keisler's Model Theory I came across the following theorem (Theorem 7.2.13):
Theorem There exists a (first-order) sentence $\sigma$ such that for all infinite cardinals $\alpha$, ...

**3**

votes

**1**answer

181 views

### Failure of GCH at indescribable cardinals

Can $\Pi^m_n$ indescribable cardinal be the first one where $\text{GCH}$ fails?
Hauser showed in
Hauser,K.: Indescribable cardinals and elementary embeddings.
J. Symb. Logic 56, 439457 (1991)
that ...

**5**

votes

**4**answers

360 views

### do behavior of gimel or GCH determine all infinte products of cardinals?

Let $Card$ be the class of infinite cardinals and $p\colon Card^2\to Card$ be given by $(\kappa,\lambda)\mapsto\kappa^\lambda$.
Assuming GCH it is known that $p(\kappa,\lambda)$ is either $\kappa$ (if ...

**3**

votes

**1**answer

287 views

### The canonical forcing of the GCH and direct limits.

The motivation for this question is that I am working through an exercise to force the GCH (generalized continuum hypothesis) over a model of ZFC and obtain a model of ZFC where GCH holds.
The ...

**9**

votes

**4**answers

1k views

### Failure of the GCH

What is the (currently known) consistency strength of global failure of the GCH?
I do not have access to the exact statement of the original Foreman-Woodin result. My searches seem to indicate that ...

**22**

votes

**4**answers

2k views

### When $2^\alpha = 2^\beta$ implies $\alpha=\beta$ ($\alpha,\beta$ cardinals)

Sorry if this is a silly question. I was wondering, under what axioms of set theory is it true that if $\alpha$,$\beta$ are cardinals, and $2^\alpha=2^\beta$, then $\alpha=\beta$? Do people use these ...