Ramana Kumar

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Finite Factored Sets: Conditional Orthogonality

That's right. A partial function can be thought of as a subset (of its domain) and a total function on that subset. And a (total) function can be thought of as a partition (of its domain): the parts are the inverse images of each point in the function's image.

The Blackwell order as a formalization of knowledge

Blackwell’s theorem says that the conditions under which  can be said to be more generally useful than  are precisely the situations where  is a post-garbling of .

Are the indices the wrong way around here?

Introduction to Cartesian Frames

A formalisation of the ideas in this sequence in higher-order logic, including machine verified proofs of all the theorems, is available here.

Time in Cartesian Frames

"subagent [] that could choose " -- do you mean  or  or neither of these? Since  is not closed under unions, I don't think the controllables version of "could choose" is closed under coarsening the partition. (I can prove that the ensurables version is closed; but it would have been nice if the controllables version worked.)

ETA: Actually controllables do work out if I ignore the degenerate case of a singleton partition of the world. This is because, when considering partitions of the world, ensurables and controllables are almost the same thing.
 

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 

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