Ramana Kumar

Comments

Time in Cartesian Frames

I have something suggestive of a negative result in this direction:

Let  be the prime-detector situation from Section 2.1 of the coarse worlds post, and let  be the (non-surjective) function that "heats" the outcome (changes any "C" to an "H"). The frame  is clearly in some sense equivalent to the one from the example (which deletes the temperature from the outcome) -- I am using my version just to stay within the same category when comparing frames. As a reminder, primality is not observable in  but is observable in .
Claim: No frame of the form  is biextensionally equivalent to 
Proof Idea

The kind of additional observability we get from coarsening the world seems in this case to be very different from the kind that comes from externalising part of the agent's decision.

Eight Definitions of Observability

With the other problem resolved, I can confirm that adding an  escape clause to the multiplicative definitions works out.

Eight Definitions of Observability

Using the idea we talked about offline, I was able to fix the proof - thanks Rohin!
Summary of the fix:
When  and  are defined, additionally assume they are biextensional (take their biextensional collapse), which is fine since we are trying to prove a biextensional equivalence. (By the way this is why we can't take , since we might have  after biextensional collapse.) Then to prove , observe that for all  which means , hence  since a biextensional frame has no duplicate columns.

Eight Definitions of Observability

I presume the fix here will be to add an explicit  escape clause to the multiplicative definitions. I haven't been able to confirm this works out yet (trying to work around this), but it at least removes the  counterexample.

Eight Definitions of Observability

How is this supposed to work (focusing on the  claim specifically)?

and so

Thus, .

Earlier,  was defined as follows:

given by  and 

but there is no reason to suppose  above.

Time in Cartesian Frames

It suffices to establish that 

I think the  and  here are supposed to be  and 

Eight Definitions of Observability

Indeed I think the  case may be the basis of a counterexample to the claim in 4.2. I can prove for any (finite)  with  that there is a finite partition  of  such that 's agent observes  according to the assuming definition but does not observe  according to the constructive multiplicative definition, if I take 

Eight Definitions of Observability

Let 

nit:  should be  here

and let  be an element of .

and the second  should be . I think for these  and  to exist you might need to deal with the  case separately (as in Section 5). (Also couldn't you just use the same  twice?)

Eight Definitions of Observability

UPDATE: I was able to prove  in general whenever  and  are disjoint and both in , with help from Rohin Shah, following the "restrict attention to world " approach I hinted at earlier.
 

Eight Definitions of Observability

this is clearly isomorphic to , where , where . Thus, 's agent can observe  according to the nonconstructive additive definition of observables.

I think this is only true if  partitions , or, equivalently, if  is surjective. This isn't shown in the proof. Is it supposed to be obvious?

EDIT: may be able to fix this by assigning any  that is not in  to the frame  so it is harmless in the product of s -- I will try this.

 

Load More