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.
The problem is a bit earlier actually:
This isn't true, because doesn't just ignore here (since ). I think the route is to say "Let . Then must treat and identically, meaning that either they are equal, or the frame is biextensionally equivalent to one where they are equal."
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.
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.
and observe that
This cannot be true. I can prove in general whenever by observing that the agent cardinalities on each side differ.
I haven't yet figured out why it's true under - I'll keep trying, but let me know if there's a quick argument for why this holds. (Default next step for me would be to see if I can restrict attention to the world then do something similar to my other comment.)
Because and are not a partition of the world here.
EDIT: but what we actually need in the proof is where the do result in a partition, so I think this will work out the same as the other comment. I'm still not convinced about biextensional equivalence between the frames without the rest of the product.
And it seems we do actually need in the proof to justify:
Thus it suffices to show that .
Without it, we have to show instead.
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.
Let . Thus, we also have that
I'm not seeing why this follows. I'll look for a counterexample, but in the meantime maybe there's a simple explanation for why we can combine the product of two assumes as an assume of the union? (I think the only relevant assumption in this context is that the s partition the world; but I might be missing some other important assumption.)
EDIT: I can see how maybe this will follow from the definition of observability of a partition from subsets (which we are also assuming) and the fact that is closed under union... will try to figure that out. -- Yep I think this works out. Sorry for the confusion.
Next, assume that 's agent can observe according to the additive definition. We will show that 's agent can observe .
I might be misunderstanding this, but the proof suggests you're actually assuming the assuming definition here, not the additive definition. In which case we may be missing the proof of implication of any of the other definitions from the additive definition.
Definition: We say that 's agent can observe a finite partition of if for all functions , there exists an element such that for all , .
Claim: This definition is equivalent to the definition from subsets.
This doesn't hold in the degenerate case , since then we have an empty function but no elements of . (But the definition from subsets holds trivially.)
This is the first post in the sequence that I fully read since the Introduction. So I'm not going to be able to say anything really useful about the proofs. Still, I was curious about the philosophical aspects of these definitions, so I read this post anyway.
That being said, I still think that I understood some part of the definitions, after checking terms from previous posts. My handwavy understanding of your definitions is
Is there something really wrong here?
Also, I'm curious if you have an interpretation of the differences between internalizing-externalizing definitions and the others, just like your section on updatelessness compared additive and multiplicative definitions. (Really cool section philosophically, by the way!)
Seems right, except I don't use the word "product" for the multiplicative definition.
I don't have much to say about the internalizing-externalizing definition philosophically. One thing to say is that I think the condition that observes is a weaker notion of observability, that might actually agree with philosophical intuition more, and the internalizing-externalizing definition might be easier to interpret if you are thinking in terms of this condition.
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?)
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
With the other problem resolved, I can confirm that adding an escape clause to the multiplicative definitions works out.
This is the eleventh post in the Cartesian frames sequence. Here, we compare eight equivalent definitions of observables, which emphasize different philosophical interpretations.
Throughout this post, we let be a Cartesian frame over a nonempty set , we let be a finite partition of , and we let send each element of to its part in .
The condition that is finite is an important one. Many of the definitions below can be extended to infinite partitions, and the theory of observability for infinite partitions is probably nice, but we are not discussing it here. The condition that is nonempty is just ruling out some degenerate cases
The definitions in this post will talk about when a finite partition of is observable in . This will make some of the definitions more elegant, and it is easy to translate back and forth between the new definitions of the observability of a finite partition and the old definitions of the observability of a subset.
Definition: We say 's agent can observe a finite partition of if for all parts , . We let denote the set of all finite partitions of that are observable in .
Claim: For any nonempty strict subset , 's agent can observe if and only if agent can observe .
Proof: If 's agent can observe , then clearly 's agent can observe . If agent can observe , then since observability is closed under complements, 's agent can observe , and so can observe .
1.1. Example
In "Introduction to Cartesian Frames," we gave the example of an agent that can choose between unconditionally carrying an umbrella, unconditionally carrying no umbrella, carrying an umbrella iff it's raining, and carrying an umbrella iff it's sunny:
Here, , so the partition is observable in , where and .
As we go through the definitions in this post, we will repeatedly return to and show how to understand 's observables in terms of our new definitions.
Before presenting fundamentally new definitions, we will modify our two old definitions to be about finite partitions instead of subsets.
Definition: We say that 's agent can observe a finite partition of if for all functions , there exists an element such that for all , .
Claim: This definition is equivalent to the definition from subsets.
Proof: We work by induction on the number of parts in . Since is nonempty, has at least one part. If has one part, we clearly have that 's agent can observe under the definition from subsets. For the conditional policies definition, we also have that 's agent can observe , since we can take , and thus, for all ,
If has parts, consider the partition which unions together the first two parts and of . Let send each element of to its part in .
First, assume that 's agent can observe according to the definition from subsets. Then, since observability of subsets is closed under unions, 's agent can also observe under the definition from subsets, and thus also under the conditional policies definition.
Given a function , let be given by , and on all other inputs. Since 's agent can observe under the conditional policies definition, we can let be such that for all , .
Choose an such that , which we can do because is observable in . Observe that for all , we have that if , then
if , we have , and thus
and finally if for some , we still have have , and thus
Thus, 's agent can observe according to the conditional policies definition.
Conversely, if 's agent can observe according to the conditional policies definition, then to show that 's agent can observe according to the definition from subsets, it suffices to show that the agent can observe for all . Thus, we need to show that for any , there exists an with .
Indeed, if we let send to , and send all other inputs to , then we can take an such that for all , . But then, if , then
and otherwise,
Thus, 's agent can observe according to the definition from subsets.
2.1. Example
Let be defined as in the §1.1 example, with , , and .
is a four-element set, and is a two-element set, so there are sixteen functions . For each function, there is a possible agent that satisfies for all . We can illustrate the sixteen functions and the corresponding in a sixteen-row table:
Since there is an for each function, 's agent can observe according to the conditional policies definition.
Next, we give an additive definition of observables. This is a version of our categorical definition of observables from "Controllables and Observables, Revisited," modified to be about finite partitions.
Definition: We say 's agent can observe a finite partition of if there exist , Cartesian frames over , with such that .
This can also be strengthened to a constructive version of the additive definition, which we will call the assuming definition.
Definition: We say 's agent can observe a finite partition of if .
Claim: These definitions are equivalent to each other and the definitions above.
Proof: We assume that , and that is nonempty. The case where and the case where are trivial.
If 's agent can observe according to the assuming definition of observables, then it can also clearly observe according to the additive definition, since .
Next, assume that 's agent can observe according to the additive definition. We will show that 's agent can observe . Consider the pair of Cartesian frames and . Observe that and that , and that . Thus, is observable in . Symmetrically, is observable in for all , and thus is observable in according to the definition from subsets.
Finally, assume that 's agent can observe according to the conditional policies definition (and also the definition from subsets). We will show that , where .
We have , where , and is given by , where .
First observe that for every , there is a unique such that . This is because there exists an , and from the definition from subsets, 's agent can observe each , and so given an , if , it must be the case that for all . Thus, we have that that .
We construct and which compose to something homotopic to the identity in each order. Let be the diagonal, given by . Let and be the identity on . Let be given by , where is given by , and satisfies for all , which is possible by the conditional policies definition.
To see that is a morphism, observe that for all and ,
To see that is a morphism, observe that for all , and , if we let be given by , we have
where is such that . The fact that and compose to something homotopic to the identity in both orders follows from the fact that and are the identity on . Thus, , and so is observable in according to the assuming definition.
3.1. Example
Let be defined as in the previous examples, with and . By the assuming definition, there exist two frames
and
such that .
This example both illustrates the idea behind the additive definitions, and shows the construction used in the assuming definition. This is also the same example we provided to illustrate products of Cartesian frames in "Additive Operations on Cartesian Frames."
Another way of thinking about the additive definition of observables: Recall "Committing, Assuming, Externalizing, and Internalizing" §3.2 (Committing and Assuming Can Be Defined Using Lollipop and Tensor), where we saw that . This means that (up to isomorphism) we can restate as , i.e.,
.
This (equivalent) framing makes it easier to keep track of what "assuming" is doing categorically, so that we can see what interfaces between frames we are relying on when we say that something is "observable" using an additive definition.
Our multiplicative definitions will depend on a notion of agents being powerless outside of a subset.
4.1. Powerless Outside of a Subset
Definition: Given a subset of , we say that 's agent is powerless outside if for all , and all , if , then .
To say that 's agent is powerless outside is to say that the if the world is at all dependent on 's agent, then the world must be in .
Here are some lemmas about being powerless outside of a subset, which we will use later.
Lemma: If 's agent is powerless outside and , then 's agent is powerless outside .
Proof: Trivial.
Lemma: If and 's agents are both powerless outside , then 's agent is powerless outside .
Proof: Let , and let . Consider some and . We will use the fact that if then , and the fact that if then . Observe that if , then
Now, we are ready for our first truly new definition of the observability of a finite partition.
4.2. Multiplicative Definitions of Observables
Definition: We say that 's agent can observe a finite partition of if , where each 's agent is powerless outside .
Again, we also have a constructive version of this definition:
Definition: We say that 's agent can observe a finite partition of if , where , where .
Claim: These definitions are equivalent to each other and equivalent to the definitions above.
Proof: First, observe that if 's agent can observe according to the constructive version of the multiplicative definition, it can also observe according to the nonconstructive version of the multiplicative definition, since the agent of is clearly powerless outside .
Next, we show that if 's agent can observe according to the nonconstructive multiplicative definition, it can also observe according to the definition from subsets. Let , where each 's agent is powerless outside . It suffices to show that 's agent can observe , since the definition from subsets is equivalent to the additive definition, and thus closed under biextensional equivalence. Thus, it suffices to show that 's agent can observe for all . We will show that 's agent can observe , and the rest will follows by symmetry.
Let , and let . We start by showing that 's agent is powerless outside . We have that the agents of are all powerless outside , since being powerless outside something is closed under supersets. Thus we have that 's agent is powerless outside , since being powerless outside is closed under tensor.
Thus, we have , with 's agent powerless outside and 's agent powerless outside . Given an arbitrary , we will show that , and thus show that 's agent can observe .
It suffices to show that for all , if , then , and if , then . Indeed, if , then, since 's agent is powerless outside , we have
Similarly, if , then, since 's agent is powerless outside , we have
Thus, 's agent can observe , so 's agent can observe according to the definition from subsets.
Finally, we assume that 's agent can observe according to the assuming definition, and show that 's agent can observe according to the constructive version of the multiplicative definition.
We work by induction on , the number of parts. The case where is trivial. Let . Thus, we also have that , and so by induction, we have that , where and are as in the constructive multiplicative definition. Thus, it suffices to show that
First, observe that we have , where (C), , and . Let . Let .
Observe that , , and , and observe that . Thus it suffices to show that .
Let , let , and let where , , and are all given by if , if , and otherwise.
Let . Let ), where
Observe that for any , there is a , given by and . This is clearly a morphism, since
Similarly, for any , there is a morphism given by and . Finally, for any , there is a morphism , given by , which is also clearly a morphism.
We show that these are in fact all of the morphisms in . Indeed, let be a morphism in , let be an element of , and let be an element of . Let
If , then , so given any ,
, and so
Similarly, for any , and so .
If , then , and . Let . Given any ,
, so
Given any ,
, and so
Thus, .
Finally, if , we similarly have , where .
We construct a pair of morphisms
and
by letting and be the identity on , letting be given by as above. Since we have shown that is surjective, we let be any right inverse to . It is easy to show that both of these are morphisms by the construction of , and they compose to something homotopic to the identity in both orders since and are the identity of .
Thus , so 's agent can observe according to the constructive multiplicative definition, completing the proof.
You may have noticed that the last part of the proof would have been much simpler if distributed over , but does not in general distribute over . ( distributes over and distributes over .)
In this case, however, does distribute over . I do not plan on going over it now, but there is actually an interesting relationship between observables and cases where distributes over .
4.3. Example
Let be defined as in the previous examples, with and . Let , so that and . By the multiplicative definitions of observables, there then exist two frames
and
such that .
Here, is an agent that treats the "makes decisions when it's sunny" part of itself as though it were an external process. Similarly, externalizes its ability to make decisions when it's rainy.
This example illustrates both multiplicative definitions, and also shows the construction used in the constructive multiplicative definition.
Appealing again to the fact that , we also have the option of restating here as . In words, this says that is (biextensionally equivalent to) a team consisting of:
4.4. Updatelessness
The relationship between observables' additive and multiplicative definitions is interesting. You can think of the additive definition as updateful, while the multiplicative definition is updateless.
The in the additive definition are basically given a promise that the world will end up in . The in the multiplicative definition, however, are instead given a promise that their choices have no effect on worlds outside of .
I think the updateless factorization is better, and thus prefer the multiplicative definition in spite of the fact that it is more complicated.
When an updateless agent observes something, it becomes the version of itself that only affects the worlds in which it makes that observation. When an updateful agent observes something, we assume that all the worlds in which it does not make that observation do not exist. The fact that the additive and multiplicative definitions above are equivalent illustrates the equivalence of the updateful and updateless views in the simple cases where there is true observation. However, they diverge as soon as you want to try to approximate observation. The updateless view approximates better, as it makes sense to think of a subagent that has only a very small effect on worlds in which it does not make the observation that it makes.
Also, note that the in the additive definition are not subagents of , but they are additive sub-environments. The in the multiplicative definition are multiplicative subagents of .
Next, we have the nonconstructive internalizing-externalizing definition of observables.
Definition: We say that 's agent can observe a finite partition of if either or is biextensionally equivalent to something in the image of .
Again, we have a constructive version of this definition.
Definition: We say that 's agent can observe a finite partition of if either or .
Claim: These definitions are equivalent to each other and to the definitions above.
Proof: The case where is trivial, so we assume that is nonempty. Clearly if 's agent can observe under the constructive internalizing-externalizing definition, then is also observable in under the non-constructive version.
Next, assume that is in the image of (up to biextensional equivalence). Recall that the image of up to biextensional equivalence is exactly those Cartesian frames such then is nonempty and for all and we have . Thus, , where is of this form. Let send each element to the unique such that for all , and let be the image of . Then, , where , and .
Let , and let . Then, we clearly have that , where . But this is clearly isomorphic to , where , where . Thus, 's agent can observe according to the nonconstructive additive definition of observables.
Finally, we assume that 's agent can observe according to the nonconstructive additive definition of observables, and we show that 's agent can observe according to the constructive internalizing-externalizing definition. Let , where . Let , and without loss of generality, let , where and .
First, we show that . Let . Observe that (since is nonempty), , where , where .
We construct
and
as follows. Let . Let , where , and is chosen arbitrarily for . Let , and , where . Clearly, and are inverses.
To see that is a morphism, observe that for all and , we have
where .
To see that is a morphism, observe that for all , and for all , we have
It is clear that and are both homotopic to the identity, since and are both the identity.
Now, we have that , and so we also have dually that . Thus, .
The thing that is going on here is that when internalizes , the agent of then has the full ability to choose how goes (among ways of going that were possible in ). might have other choices than just choosing how goes. If it does, then it can freely entangle those other choices with the choice of however it wants.
When then externalizes , it loses all control over . However, it preserves the ability to entangle all of its other choices with the way that goes. This ability for the agent to entangle its choices with is exactly what it means to say " is observable."
5.1. Example
Let be defined as in the previous examples, with . By the internalizing-externalizing definitions, there exists a frame
,
which is biextensionally equivalent to
.
We then have that
,
which is isomorphic to .
This example illustrates both internalizing-externalizing definitions, and also shows the construction used in the constructive definition.
In our next post, we'll conclude the sequence by showing how to formalize agents that learn and act over time using Cartesian frames.