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 C=(A,E,⋅) be a Cartesian frame over a nonempty set W, we let V={S1,…,Sn} be a finite partition of W, and we let v:W→V send each element of W to its part in V.
The condition that V 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 W is nonempty is just ruling out some degenerate cases
1. Definition from Subsets
The definitions in this post will talk about when a finite partition V of W is observable in C. 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 C's agent can observe a finite partition V of W if for all parts Si∈V, Si∈Obs(C). We let Obs′(C) denote the set of all finite partitions of W that are observable in C.
Claim: For any nonempty strict subset S⊂W, C's agent can observe S if and only if C'sagent can observe {S,(W∖S)}.
Proof: If C's agent can observe {S,(W∖S)}, then clearly C's agent can observe S. If C'sagent can observe S, then since observability is closed under complements, C's agent can observe W∖S, and so can observe {S,(W∖S)}. □
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:
C0= rsunu↔ru↔s⎛⎜ ⎜ ⎜⎝urusnrnsurnsnrus⎞⎟ ⎟ ⎟⎠
Here, Obs(C0)={{},{ur,nr},{us,ns},W}, so the partition V={R,S} is observable in C0, where R={ur,nr} and S={us,ns}.
As we go through the definitions in this post, we will repeatedly return to C0 and show how to understand C0'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.
2. Conditional Policies Definition
Definition: We say that C's agent can observe a finite partition V of W if for all functions f:V→A, there exists an element af∈A such that for all e∈E, f(v(af⋅e))⋅e=af⋅e.
Claim: This definition is equivalent to the definition from subsets.
Proof: We work by induction on the number of parts in V. Since W is nonempty, V has at least one part. If V={W} has one part, we clearly have that C's agent can observe V under the definition from subsets. For the conditional policies definition, we also have that C's agent can observe V, since we can take af=f(W), and thus, for all e∈E,
f(v(af⋅e))⋅e=f(W)⋅e=af⋅e.If V={S1,…,Sn} has n parts, consider the partition V′={S1∪S2,S3,…,Sn} which unions together the first two parts S1 and S2 of V. Let v′:W→V′ send each element of W to its part in V′.
First, assume that C's agent can observe V according to the definition from subsets. Then, since observability of subsets is closed under unions, C's agent can also observe V′ under the definition from subsets, and thus also under the conditional policies definition.
Given a function f:V→A, let f′:V′→A be given by f′(S1∪S2)=f(S2), and f′(Si)=f(Si) on all other inputs. Since C's agent can observe V′ under the conditional policies definition, we can let af′ be such that for all e∈E, f′(v′(af′⋅e))⋅e=af′⋅e.
Choose an af∈A such that af∈if(S1,f(S1),a′f), which we can do because S1 is observable in C. Observe that for all e∈E, we have that if af⋅e∈S1, then
f(v(af⋅e))⋅e=f(S1)⋅e=af⋅e,if af⋅e∈S2, we have af⋅e=af′⋅e, and thus
f(v(af⋅e))⋅e=f(S2)⋅e=f′(S1∪S2)⋅e=f′(v′(af′⋅e))⋅e=af′⋅e=af⋅e,and finally if af⋅e∈Si for some i≠1,2, we still have have af⋅e=af′⋅e, and thus
f(v(af⋅e))⋅e=f(Si)⋅e=f′(Si)⋅e=f′(v′(af′⋅e))⋅e=af′⋅e=af⋅e.Thus, C's agent can observe V according to the conditional policies definition.
Conversely, if C's agent can observe V according to the conditional policies definition, then to show that C's agent can observe V according to the definition from subsets, it suffices to show that the agent can observe Si for all Si∈V. Thus, we need to show that for any a0,a1∈A, there exists an a2∈A with a2∈if(Si,a0,a1).
Indeed, if we let f:V→A send Si to a0, and send all other inputs to a1, then we can take an af such that for all e∈E, f(v(af⋅e))⋅e=af⋅e. But then, if af⋅e∈Si, then
af⋅e=f(v(af⋅e))⋅e=f(S1)⋅e=a0⋅e,and otherwise,
af⋅e=f(v(af⋅e))⋅e=a1⋅e.Thus, C's agent can observe V according to the definition from subsets. □
2.1. Example
Let C0=(A,E,⋅) be defined as in the §1.1 example, with R={ur,nr}, S={us,ns}, and V={R,S}.
A={u,n,u↔r,u↔s} is a four-element set, and V={R,S} is a two-element set, so there are sixteen functions f:V→A. For each function, there is a possible agent af∈A that satisfies f(v(af⋅e))⋅e=af⋅e for all e∈E. We can illustrate the sixteen functions and the corresponding af∈A in a sixteen-row table:
Since there is an af∈A for each function, C0's agent can observe V according to the conditional policies definition.
3. Additive Definitions
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 C's agent can observe a finite partition V={S1,…,Sn} of W if there exist C1,⋯Cn, Cartesian frames over W, with Ci◃⊥Si such that C≃C1&…&Cn.
This can also be strengthened to a constructive version of the additive definition, which we will call the assuming definition.
Definition: We say C's agent can observe a finite partition V={S1,…,Sn} of W if C≃AssumeS1(C)&…&AssumeSn(C).
Claim: These definitions are equivalent to each other and the definitions above.
Proof: We assume that n≥2, and that A is nonempty. The case where n=1 and the case where A={} are trivial.
If C's agent can observe V according to the assuming definition of observables, then it can also clearly observe V according to the additive definition, since AssumeS1(C)◃⊥S1.
Next, assume that C's agent can observe V according to the additive definition. We will show that C's agent can observe S1. Consider the pair of Cartesian frames C1 and C2&…&Cn. Observe that C1◃⊥S1 and that C2&…&Cn◃⊥W∖S1, and that C≃C1&(C2&…&Cn). Thus, S1 is observable in C. Symmetrically, Si is observable in C for all i=1,…n, and thus V is observable in C according to the definition from subsets.
Finally, assume that C's agent can observe V according to the conditional policies definition (and also the definition from subsets). We will show that C≃C1&…&Cn, where Ci=AssumeSi(C).
We have C1&…&Cn=(An,E1⊔⋯⊔En,⋆), where Ci=(A,Ei,⋅i), and ⋆ is given by (a1,…,an)⋆e=ai⋅e, where e∈Ei.
First observe that for every e∈E, there is a unique i∈{1,…,n} such that e∈Ei. This is because there exists an a0∈A, and from the definition from subsets, C's agent can observe each Si, and so given an e∈E, if a0⋅e∈Si, it must be the case that for all a∈A, a⋅e∈Si. Thus, we have that that E=E1⊔⋯⊔En.
We construct (g0,h0):(An,E,⋆)→C and (g1,h1):C→(An,E,⋆) which compose to something homotopic to the identity in each order. Let g1:A→An be the diagonal, given by g1(a)=(a,…,a). Let h0 and h1 be the identity on E. Let g0 be given by g0(a1,…,an)=af, where f:V→A is given by f(Si)=ai, and af satisfies f(v(af⋅e))⋅e=af⋅e for all e∈E, which is possible by the conditional policies definition.
To see that (g1,h1) is a morphism, observe that for all a∈A and e∈E,
g1(a)⋆e=(a,…,a)⋆e=a⋅e=a⋅h1(e).To see that (g0,h0) is a morphism, observe that for all (a1,…,an)∈A, and e∈E, if we let f:V→A be given by f(Si)=ai, we have
g0(a1,…,an)⋅e=f(v(g0(a1,…,an)⋅e))⋅e=f(Si)⋅e=ai⋅e=(a1,…,an)⋆e=(a1,…,an)⋆h0(e),where i is such that e∈Ei. The fact that (g0,h0) and (g1,h1) compose to something homotopic to the identity in both orders follows from the fact that h0∘h1 and h1∘h0 are the identity on E. Thus, C≃AssumeS1(C)&…&AssumeSn(C), and so V is observable in C according to the assuming definition. □
3.1. Example
Let C0 be defined as in the previous examples, with R={ur,nr} and S={us,ns}. By the assuming definition, there exist two frames
C1=AssumeR(C0)= run(urnr)
and
C2=AssumeS(C0)= sun(usns)
such that C0≃C1&C2.
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 AssumeS(C)≅1S⊗C. This means that (up to isomorphism) we can restate C0≃C1&C2 as C0≃(1R⊗C0) &(1S⊗C0), i.e.,
⎛⎜ ⎜ ⎜⎝urusnrnsurnsnrus⎞⎟ ⎟ ⎟⎠ ≃(ur nr)⊗⎛⎜ ⎜ ⎜⎝urusnrnsurnsnrus⎞⎟ ⎟ ⎟⎠ &(us ns)⊗⎛⎜ ⎜ ⎜⎝urusnrnsurnsnrus⎞⎟ ⎟ ⎟⎠.
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.
4. Multiplicative Definitions
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 S of W, we say that C's agent is powerless outside S if for all e∈E, and all a0,a1∈A, if a0⋅e∉S, then a0⋅e=a1⋅e.
To say that C's agent is powerless outside S is to say that the if the world is at all dependent on C's agent, then the world must be in S.
Here are some lemmas about being powerless outside of a subset, which we will use later.
Lemma: If C's agent is powerless outside S and T⊇S, then C's agent is powerless outside T.
Proof: Trivial. □
Lemma: If C and D's agents are both powerless outside S, then C⊗D's agent is powerless outside S.
Proof: Let D=(B,F,⋆), and let C⊗D=(A×B,hom(C,D∗),⋄). Consider some (a0,b0),(a1,b1)∈A×B and (g,h)∈hom(C,D∗). We will use the fact that if a0⋅h(b0)∉S then a0⋅h(b0)=a1⋅h(b0), and the fact that if b0⋆g(a1)∉S then b0⋆g(a1)=b1⋆g(a1). Observe that if (a0,b0)⋄(g,h)∉S, then
(a0,b0)⋄(g,h)=a0⋅h(b0)=a1⋅h(b0)=b0⋆g(a1)=b1⋆g(a1)=(a1,b1)⋆(g,h).□
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 C's agent can observe a finite partition V={S1,…,Sn} of W if C≃C1⊗⋯⊗Cn, where each Ci's agent is powerless outside Si.
Again, we also have a constructive version of this definition:
Definition: We say that C's agent can observe a finite partition V={S1,…,Sn} of W if C≃C1⊗⋯⊗Cn, where Ci=AssumeSi(C)&1Ti, where Ti=(W∖Si)∩Image(C).
Claim: These definitions are equivalent to each other and equivalent to the definitions above.
Proof: First, observe that if C's agent can observe V according to the constructive version of the multiplicative definition, it can also observe V according to the nonconstructive version of the multiplicative definition, since the agent of AssumeSi(C)&1Ti is clearly powerless outside Si.
Next, we show that if C's agent can observe V according to the nonconstructive multiplicative definition, it can also observe V according to the definition from subsets. Let C≃D=C1⊗⋯⊗Cn, where each Ci's agent is powerless outside Si. It suffices to show that D's agent can observe V, since the definition from subsets is equivalent to the additive definition, and thus closed under biextensional equivalence. Thus, it suffices to show that D's agent can observe Si for all i=1,…,n. We will show that D's agent can observe S1, and the rest will follows by symmetry.
Let C1=(A1,E1,⋅1), and let D1=(B1,F1,⋆1)=C2⊗⋯⊗Cn. We start by showing that D1's agent is powerless outside W∖S1. We have that the agents of C2,…,Cn are all powerless outside W∖S1, since being powerless outside something is closed under supersets. Thus we have that D1's agent is powerless outside W∖S1, since being powerless outside W∖S1 is closed under tensor.
Thus, we have D=(A1×B1,hom(C,D∗),⋄)=C1⊗D1, with C1's agent powerless outside S1 and D1's agent powerless outside W∖S1. Given an arbitrary (a1,b1),(a2,b2)∈A1×B1 , we will show that (a1,b2)∈if(S1,(a1,b1),(a2,b2)), and thus show that D's agent can observe S1.
It suffices to show that for all (g,h):C→D∗, if (a1,b2)⋄(g,h)∈S1, then (a1,b2)⋄(g,h)=(a1,b1)⋄(g,h), and if (a1,b2)⋄(g,h)∉S1, then (a1,b2)⋄(g,h)=(a2,b2)⋄(g,h). Indeed, if (a1,b2)⋄(g,h)∈S1, then, since D1's agent is powerless outside W∖S1, we have
(a1,b2)⋄(g,h)=b2⋆1g(a1)=b1⋆1g(a1)=(a1,b1)⋄(g,h).Similarly, if (a1,b2)⋄(g,h)∉S1, then, since C1's agent is powerless outside S, we have
(a1,b2)⋄(g,h)=a1⋅1h(b2)=a2⋅1h(b2)=(a2,b2)⋄(g,h).Thus, D's agent can observe S1, so C's agent can observe V according to the definition from subsets.
Finally, we assume that C's agent can observe V according to the assuming definition, and show that C's agent can observe V according to the constructive version of the multiplicative definition.
We work by induction on n, the number of parts. The case where n=1 is trivial. Let C≃AssumeS1(C)&…&AssumeSn(C). Thus, we also have that C≃AssumeS1∪S2(C)&AssumeS3(C)&…&AssumeSn(C), and so by induction, we have that C≃(AssumeS1∪S2(C)&1T1∩T2)⊗C3⊗⋯⊗Cn, where Ci and Ti are as in the constructive multiplicative definition. Thus, it suffices to show that
AssumeS1∪S2(C)&1T1∩T2≃C1⊗C2=(AssumeS1(C)&1T1)⊗(AssumeS2(C)&1T2).First, observe that we have C≃D1&D2&D3, where D1=AssumeS1(C), D2=AssumeS2(C), and D3=AssumeS3(C)&…&AssumeSn(C). Let Di=(Bi,Fi,⋆i). Let Ri=Image(Di).
Observe that T1=R2∪R3, T2=R1∪R3, and T1∪T2=R3, and observe that AssumeS1∪S2(C)≃D1&D2. Thus it suffices to show that (D1&1R2∪R3)⊗(D2&1R1∪R3)≃D1&D2&1R3.
Let D1&1R2∪R3=(B1,F1⊔R2⊔R3,∙1), let D2&1R1∪R3=(B2,F2⊔R1⊔R3,∙2), and let D1&D2&1R3=(B1×B2,F1⊔F2⊔R3,∙3) where ∙1, ∙2, and ∙3 are all given by b∙if=b⋆1f if f∈F1, b∙if=b⋆2f if f∈F2, and b∙if=f otherwise.
Let H=hom(B1&1R2∪R3,(D2&1R1∪R3)∗). Let (D1&1R2∪R3)⊗(D2&1R1∪R3)=(B1×B2,H,∙4)), where
(b1,b2)∙4(g,h)=b1∙1h(b2)=b2∙2g(b1).Observe that for any f1∈F1, there is a (gf1,hf1)∈H, given by gf1(b1)=b1⋅1f1 and hf1(b2)=f1. This is clearly a morphism, since
b1∙1hf1(b2)=b1∙1f1=b1⋅1f1=gf1(b1)=b2∙2gf1(b1).Similarly, for any f2∈F2, there is a morphism (gf2,hf2)∈H given by gf2(b1)=f2 and hf2(b2)=b2⋅2f2. Finally, for any r∈R3, there is a morphism (gr,hr)∈H, given by gr(b1)=hr(b2)=r, which is also clearly a morphism.
We show that these are in fact all of the morphisms in H. Indeed, let (g,h) be a morphism in H, let b1 be an element of B1, and let b2 be an element of b2. Let
r=b2∙2g(b1)=b1∙1h(b2).If r∈R3, then g(b1)=h(b2)=r, so given any b′1∈B1,
b2∙2g(b′1)=b′1∙1h(b2)=r∈R3, and so
g(b′1)=b2∙2g(b′1)=r.Similarly, for any b′2∈B2, h(b′2)=r and so (g,h)=(gr,hr).
If r∈R1, then g(b1)=r, and h(b2)∈F1. Let f1=h(b2). Given any b′1∈B1,
b2∙2g(b′1)=b′1∙1h(b2)=b′1∙1f1∈R1, so
g(b′1)=b2∙2g(b′1)=b′1∙1f1=b′1⋅1f1.Given any b′2∈B2,
b1∙1h(b′2)=b′2∙2g(b1)=b′2∙2r=r∈R1, and so
h(b′2)=b1∙1h(b′2)=r.Thus, (g,h)=(gf1,hf1).
Finally, if r∈R2, we similarly have (g,h)=(gf2,hf2), where f2=g(b1)∈F2.
We construct a pair of morphisms
(g0,h0):(B1×B2,H,∙4)→(B1×B2,F1⊔F2⊔R3,∙3)and
(g1,h1):(B1×B2,F1⊔F2⊔R3,∙3)→(B1×B2,H,∙4),by letting g0 and g1 be the identity on B1×B2, letting h0:F1⊔F2⊔R3→H be given by h0(f)=(gf,hf) as above. Since we have shown that h0 is surjective, we let h1 be any right inverse to h0. It is easy to show that both of these are morphisms by the construction of (gf,hf), and they compose to something homotopic to the identity in both orders since g0∘g1 and g1∘g0 are the identity of B1×B2.
Thus (D1&1R2∪R3)⊗(D2&1R1∪R3)≃D1&D2&1R3, so C's agent can observe V 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 C0 be defined as in the previous examples, with R={ur,nr} and S={us,ns}. Let TX=(W∖X)∩Image(C0), so that 1TR=1S and 1TS=1R. By the multiplicative definitions of observables, there then exist two frames
C1=AssumeR(C) & 1S= rusnsr→ur→n(urusnsnrusns)
and
C2=AssumeS(C) & 1R= surnrs→us→n(usurnrnsurnr)
such that C0≃C1⊗C2.
Here, C1 is an agent that treats the "makes decisions when it's sunny" part of itself as though it were an external process. Similarly, C2 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 AssumeS(C)≅1S⊗C, we also have the option of restating C0≃C1⊗C2 here as C0 ≃ ((1R⊗C0) & 1S) ⊗ ((1S⊗C0) & 1R). In words, this says that Agent(C0) 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 Ci in the additive definition are basically given a promise that the world will end up in Si. The Ci in the multiplicative definition, however, are instead given a promise that their choices have no effect on worlds outside of Si.
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 Ci in the additive definition are not subagents of C, but they are additive sub-environments. The Ci in the multiplicative definition are multiplicative subagents of C.
5. Internalizing-Externalizing Definitions
Next, we have the nonconstructive internalizing-externalizing definition of observables.
Definition: We say that C's agent can observe a finite partition V of W if either A={} or C is biextensionally equivalent to something in the image of ExternalV∘InternalV.
Again, we have a constructive version of this definition.
Definition: We say that C's agent can observe a finite partition V of W if either A={} or C≃ExternalV(InternalV(C)).
Claim: These definitions are equivalent to each other and to the definitions above.
Proof: The case where A={} is trivial, so we assume that A is nonempty. Clearly if C's agent can observe V under the constructive internalizing-externalizing definition, then V is also observable in C under the non-constructive version.
Next, assume that C is in the image of ExternalV∘InternalV (up to biextensional equivalence). Recall that the image of InternalV up to biextensional equivalence is exactly those Cartesian frames (B,F,⋆) such then F is nonempty and for all f0,f1∈F and b∈B, we have v(b⋆f0)=v(b⋆f1). Thus, C≃ExternalV(B,F,⋆), where (B,F,⋆) is of this form. Let vB:B→V send each element b∈B to the unique vb∈V such that v(b⋆f)=vb for all f∈F, and let VB be the image of vB. Then, ExternalV(B,F,⋆)=(B/X,X×F,⋄), where X={{b∈B | vB(b)=v′} | v′∈VB}, and q⋄(x,f)=q(x)⋆f.
Let VB={v1,…,vm}, and let Bi={b∈B | vB(b)=vi}. Then, we clearly have that ExternalV(B,F,⋆)≅(B1×⋯×Bm,VB×F,∙), where (b1,…,bm)∙(vi,f)=bi⋆f. But this is clearly isomorphic to D1&…&Dm, where Di=(Bi,F,⋆i), where b⋆if=b⋆f. Thus, C's agent can observe V according to the nonconstructive additive definition of observables.
Finally, we assume that C's agent can observe V according to the nonconstructive additive definition of observables, and we show that C's agent can observe V according to the constructive internalizing-externalizing definition. Let C≃C1&…&Cn, where Ci◃⊥Si. Let Ci=(Ai,Ei,⋅i), and without loss of generality, let C=C1&…&Cn=(A,E,⋅), where A=A1×⋯×An and E=E1⊔⋯⊔En.
First, we show that InternalV(C)≃C1⊕⋯⊕Cn. Let C1⊕⋯⊕Cn=(A1⊔⋯⊔An,E1×⋯×En,⋆). Observe that (since A is nonempty), InternalV(C)≅(A×F,B/F,⋆′), where F={E1,…,En}, where (a,f)⋆q=a⋅q(f).
We construct
(g0,h0):(A1⊔⋯⊔An,E1×⋯×En,⋆)→(A×F,B/F,⋆′)and
(g1,h1):(A×F,B/F,⋆′)→(A1⊔⋯⊔An,E1×⋯×En,⋆)as follows. Let g1((a1,…,an),Ei)=ai. Let g0(ai)=((a1,…,ai,…,an),Ei), where ai∈Ai, and aj∈Aj is chosen arbitrarily for j≠i. Let h0(q)=(q(E1),…,q(En)), and h1(e1,…,en)=q, where q(Ei)=ei. Clearly, h0 and h1 are inverses.
To see that (g0,h0) is a morphism, observe that for all ai∈A1⊔⋯⊔An and q∈B/F, we have
g0(ai)⋆′q=((a1,…,ai,…,an),Ei)⋆′q=(a1,…,ai,…,an)⋅q(Ei)=ai⋅iq(Ei)=ai⋆(q(E1),…,q(En))=ai⋆h0(q),where ai∈Ai.
To see that (g1,h1) is a morphism, observe that for all ((a1,…,an),Ei)∈A×F, and for all (e1,…en)∈E1×⋯×En, we have
g1((a1,…,an),Ei)⋆(e1,…,en)=ai⋆(e1…,en)=ai⋅iei=(a1,…,an)⋅ei=(a1,…,an)⋅h1(e1,…,en)(Ei)=((a1,…,an),Ei)⋆h1(e1,…,en).It is clear that (g0,h0)∘(g1,h1) and (g1,h1)∘(g0,h0) are both homotopic to the identity, since h0∘h1 and h1∘h0 are both the identity.
Now, we have that InternalV(C1&…&Cn)≃C1⊕⋯⊕Cn, and so we also have dually that ExternalV(C1⊕⋯⊕Cn)≃C1&…&Cn. Thus, C≃ExternalV(InternalV(C)). □
The thing that is going on here is that when C internalizes V, the agent of C then has the full ability to choose how V goes (among ways of V going that were possible in C). InternalV(C) might have other choices than just choosing how V goes. If it does, then it can freely entangle those other choices with the choice of V however it wants.
When C then externalizes V, it loses all control over V. However, it preserves the ability to entangle all of its other choices with the way that V goes. This ability for the agent to entangle its choices with V is exactly what it means to say "V is observable."
5.1. Example
Let C0 be defined as in the previous examples, with V={{ur,nr},{us,ns}}. By the internalizing-externalizing definitions, there exists a frame
InternalV(C0)≅ (u,r)(u,s)(n,r)(n,s)(u↔r,r)(u↔r,s)(u↔s,r)(u↔s,s)⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝urusnrnsurnsnrus⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠,
which is biextensionally equivalent to
C1= urnrusns⎛⎜ ⎜ ⎜⎝urnrusns⎞⎟ ⎟ ⎟⎠.
We then have that
ExternalV(C1)≅ rs(r→ur,s→us)(r→nr,s→ns)(r→ur,s→ns)(r→nr,s→ur)⎛⎜ ⎜ ⎜⎝urusnrnsurnsnrus⎞⎟ ⎟ ⎟⎠,
which is isomorphic to C0.
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.