Transfinite epistemic logic puzzle challenge: Cheryl's birthday on steroids

[AllTalk]

http://jdh.hamkins.org/transfinite-epistemic-logic-puzzle-challenge/

[fsk]

tl;dr version

He constructs a weird order on the rational numbers, and they play the game on that ordering. He also has steps for limit ordinals.

For example:

Cheryl: I have given each of you an ordinal. I didn't give both of you the same ordinal. Do you know if you have the smallest one?

Bob: I can't prove I have the smallest. (I.e., I don't have 0.)

Albert: With that information, I can't prove I have the smallest. (I.e., I don't have 1.)

Bob: I don't have 2.

Cheryl: You can go on forever without figuring out who has smaller.

Bob: I don't have the smallest. (I.e., I don't have omega.)

Albert: I don't have the smallest. (I.e., I don't have omega+1).

Cheryl: You can do this 100 times and it wouldn't help.

Bob: I don't have the smallest. (I.e., I don't have 100 * omega.)

Albert: I know I have the smallest now! (I.e., I have 100 * omega+1)

The weird ordering he places on the rational numbers is equivalent to playing this way with ordinals.

[AllTalk → fsk]

I think it is just the usual order on the rational numbers, but since he uses only some of the rationals, it does amount to using ordinals.

But, I think your solution is off by at least a factor of omega.

[fsk → AllTalk]

I see, he has limit ordinals in his embedded order with the sequence 1 - (1/2)^n, with 1 being the equivalent of omega.

You probably even could get an embedded omega squared with

1 - (1/2)^n - (1/3)^m with m>n

So 1/2 = omega

3/4 = 2 * omega

7/8 = 3 * omega

and 1 is omega * omega

He might have omega squared in his version, but I didn't read it that carefully.

Now play the version where the secret ordinal is an uncountable ordinal or a strongly inaccessible cardinal.