7
$\begingroup$

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.

$\endgroup$

1 Answer 1

5
$\begingroup$

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 $\kappa^+$.

$\endgroup$
15
  • $\begingroup$ For the first one, I still don't see how the identity implies the GCH. $\endgroup$ Commented May 10, 2017 at 17:32
  • 1
    $\begingroup$ I think that I spelled out pretty neatly in the last comment that it doesn't. And it is completely irrelevant here. $\endgroup$
    – Asaf Karagila
    Commented May 10, 2017 at 17:48
  • 1
    $\begingroup$ Yes, GCH implies it for limit cardinals. But you can also violate GCH by Easton's theorem and have, say $2^\kappa=\kappa^{++}$ for all successor cardinals, and still have the identity for limit cardinals. $\endgroup$
    – Asaf Karagila
    Commented May 10, 2017 at 17:51
  • 1
    $\begingroup$ Which part troubles you? $\endgroup$
    – Asaf Karagila
    Commented May 11, 2017 at 21:41
  • 1
    $\begingroup$ I've edited my answer. $\endgroup$
    – Asaf Karagila
    Commented May 12, 2017 at 7:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .