Equivalent statements to the GCH

311 Views Asked by At

The GCH is the statement that $\forall \kappa \geq \aleph_0 : 2^\kappa = \kappa^+$. That is, $\forall \alpha :2^{\aleph_\alpha}=\aleph_{\alpha+1}$.

I was told that the Generalized Continuum Hypothesis is equivalent to the following identities. I'm curious of the proof but have no idea how to work it out.

$\sum_{\mu <\kappa}2^{\mu}=\kappa$.

$\kappa^{\text{cf}(\kappa)}=\kappa^+$ for any infinite $\kappa$.

To be clear on definitions: $\kappa^+$ is a successor cardinal; $\text{cf}(\kappa)$ denotes the cofinality of $\kappa$, which is the least limit ordinal $\theta$ cut that there is an increasing sequence over $\theta$ that is cofinal in $\kappa$.

I'd appreciate it if someone could explain these proofs.

1

There are 1 best solutions below

15
On BEST ANSWER

The first one comes from the fact that $$\sum_{i\in I}\lambda_i=\max\{|I|,\sup\{\lambda_i\mid i\in I\}\},$$ which means that $\sum_{\mu<\kappa}2^\mu=\max\{\kappa,\sup\{2^\mu\mid\mu<\kappa\}\}$.

Now given any arbitrary $\lambda$, let $\kappa=\lambda^+$, easily $2^\lambda\geq\kappa$. But the above implies that $2^\lambda=\kappa=\lambda^+$. So $\sf GCH$ holds. (Note that for limit cardinals the identity only implies that they are strong limit cardinals, but not that $\sf GCH$ holds below, or at, the cardinal.)


The second one comes from the fact that $\kappa^{\operatorname{cf}(\kappa)}$ is $\kappa^\kappa$ for regular cardinals, which in turns is equal to $2^\kappa$.

For a singular $\kappa$, fix a cofinal sequence $\kappa_i$. If $A\subseteq\kappa$, then $A$ can be written as the union of its intersections with $\kappa_i$ for all $i$. This gives us the following: $$2^\kappa\leq\prod 2^{\kappa_i}=\kappa^{\operatorname{cf}(\kappa)},$$ which by the assumption equals to $\kappa^+$.