Re: Paper and slides on indefiniteness of CH

Dear Hugh,

I have to leave on a short trip now, and will respond in more detail as soon as I can.

You have again misunderstood the $\textsf{IMH}^\#$ !

Below are some brief responses.

On Mon, 29 Sep 2014, W Hugh Woodin wrote:

Dear Sy,

The disadvantage of your formulation of $\textsf{IMH}^\#$ is that it is not even in general a $\Sigma^1_3$ property of M and so it is appealing in more essential ways to the structure of the “hyperuniverse”.

No. It appeals only to the ordinals of “lengthenings”, not to the structure of the Hyperuniverse!

This is why the consistency proof of $\textsf{SIMH}^\#(\omega_1)$ uses substantially more than a Woodin cardinal with an inaccessible above, unlike the case of $\textsf{IMH}$ and $\textsf{SIMH}(\omega_1)$ .

OK, It seems we will just have to agree that we disagree here.

OK, so you disagree with treating width actualism with “lengthenings”, unlike Pen, Geoffrey and myself. I am missing a coherent explanation for your view.

I think it is worth pointing out to everyone that $\textsf{SIMH}^\#$ , and even the weaker $\textsf{SIMH}^\#(\omega_1)$ which we know to be consistent, implies that there is a real $x$ such that $x^\#$ does not exist.

No, that is not true. The $\textsf{IMH}^\#$ is compatible with all large cardinals. So is the $\textsf{SIMH}^\#(\omega_1)$ . What argument are you thinking of?

(even though $x^\#$ exists in the parent hyperuniverse which is a bit odd to say the least in light of the more essential role that the hyperuniverse is playing). The reason of course is that $\textsf{SIMH}^\#(\omega_1)$ implies that there is a real $x$ such that $L[x]$ correctly computes $\omega_1$ .

No, it does not. What argument do you have in mind?

Set Theory — The Search for New Axioms

Re: Paper and slides on indefiniteness of CH

Leave a Reply Cancel reply

Cancel reply