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Controllables and Observables, Revisited

12Ramana Kumar

6Scott Garrabrant

4Rob Bensinger

3Scott Garrabrant

2Scott Garrabrant

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Counterexample to your claim that (with both and not ):

Take and to be the following Cartesian frames over world with agent and environment both as the following matrices:

Then but .

(Sorry for my formatting - I haven't done LaTeX in these comments before.)

Thank You! You are correct. Oops!

I deleted that claim from the post. That was quite a mistake. Luckily it seems self-contained. This is the danger of "Trivial" proofs.

Initial versions of the theory only talked about ensurables, and the addition of (and shift of attention to) controllables came much later.

To get an intuition for morphisms, I tried listing out every frame that has a morphism going to a simple 2x2 frame

.

Are any of the following wrong? And, am I missing any?

Frames I think have a morphism going to :

Every frame that looks like a frame on this list (other than ), but with extra columns added — regardless of what's in those columns. (As a special case, this includes the five frames corresponding to the other five ensurables that can be deduced from the matrix: , , , , . If has more than four elements, then there will be additional frames / additional ensurables beyond these seven.)

Every frame biextensionally equivalent to one of the frames on this list.

This is the fourth post in the Cartesian frames sequence. Read the first post here.

Previously, we defined controllables as the sets of possible worlds an agent can both ensure and prevent, and we defined observables as the sets of possible worlds such that the agent can implement all conditional policies.

Now that we have built up more language, we can redefine controllables and observables more categorically.

## 1. Controllables

1.1. Ensurables and PreventablesThe categorical definition of ensurables is very simple.

Definition:Ensure(C) is the set of all S⊆W such that there exists a morphism ϕ:1S→C.As an example, let C0=(A,E,⋅). Recall that 1S=({b},S,⋆), where b⋆s=s for all s∈S. E.g., if S0={w0,w3,w4}, then

1S0=w0 w3 w4b(w0 w3 w4).

If there is a morphism (g,h) from 1S0 to C0, this means that:

That function h:E→S0 is exactly like the function you get by looking at that row, so a morphism (g,h):1S0→C0 is like a row in C0 that is entirely contained in S0. If there are multiple such rows, then there will be multiple distinct morphisms 1S0→C0 picking out different a∈A.

In "Biextensional Equivalence," we noted that 1S is like a passive observer who has a promise from the environment that the world will be in S. The existence of a morphism 1S→C means that there's an interface that allows a powerless bystander who has been promised S to play C's game. Since Agent(1S) only has one option, this interface must send that one option to some option for C's agent that is compatible with this promise.

Claim:This definition is equivalent to the one in "Introduction to Cartesian Frames": Ensure(C)={S⊆W | ∃a∈A,∀e∈E,a⋅e∈S}.Proof:Let C=(A,E,⋅) and let 1S=({b},S,⋆), where b⋆s=s for all s∈S. First, assume there exists a morphism (g,h):1S→C. Here, g:{b}→A and h:E→S. Consider the element g(b)∈A. It suffices to show that g(b)⋅e∈S for all e∈E. Indeed, g(b)⋅e=b⋆h(e)∈S.Conversely, assume that there exists an a∈A, such that a⋅e∈S for all e∈E. Then, there is a morphism (g,h):1S→C given by g(b)=a, and h(e)=a⋅e. This is a morphism because

g(b)⋅e=a⋅e=h(e)=b⋆h(e)for all b∈{b} and e∈E. □

Definition:Prevent(C)Claim:This definition is equivalent to the one in "Introduction to Cartesian Frames": Prevent(C)={S⊆W | ∃a∈A, ∀e∈E, a⋅e∉S}.Proof:This follows from the proof for Ensure(C), substituting W∖S for S. □Our categorical definition gives us a bunch of facts about how ensurability interacts with various operations on Cartesian frames. First, ensurability is monotonic in the existence of morphisms.

Claim:If there exists a morphism ϕ:C→D, then Ensure(C)⊆Ensure(D).Proof:If S∈Ensure(C), there exists a morphism ψ:1S→C, so we have ϕ∘ψ:1S→D, so S∈Ensure(D). □This fact justifies our interpretation of the existence of a morphism from C to D as saying that "D is at least as strong as C."

We also have that ensurables interact very strongly with sums and products. The ensurables of a product are the intersection of the original two agents' ensurables, and the ensurables of a sum are (usually) the union of the original two agents' ensurables.

This makes sense when we think of C⊕D as "there are two games, and the agent gets to choose which one we play," and C&D as "there are two games, and the environment gets to choose which one we play." The agent of C⊕D can make sure something happens if either C or D's agent could, whereas the agent of C&D can only ensure things that are ensurable across both games.

Claim:Ensure(C&D)=Ensure(C)∩Ensure(D).Proof:Since & is a categorical product, if there exists a morphism from 1S to C and a morphism from 1S to D, there must exist a morphism from 1S to C&D. Thus Ensure(C&D)⊇Ensure(C)∩Ensure(D). Conversely, since & is a categorical product, there exist projection morphisms from C&D to C and from C&D to D, so Ensure(C&D)⊆Ensure(C)∩Ensure(D). □Claim:If C≠null and D≠null, then Ensure(C⊕D)=Ensure(C)∪Ensure(D).Proof:Let C=(A,E,⋅), and let D=(B,F,⋆). Since C is not null, if E were empty, then A would have to be nonempty, and Ensure(C) would be the full set 2W. C⊕D would also have empty environment and nonempty agent, and so Ensure(C⊕D) would also be the full set 2W. Thus, we are done in the case where E is empty. Similarly, we are done in the case where F is empty. Assume E and F are nonempty.It is clear thatEnsure(C⊕D)⊇Ensure(C)∪Ensure(D) because C⊕D is a coproduct, so there are canonical injection morphisms from C to C⊕D and from D to C⊕D.

Conversely, if S∈Ensure(C⊕D), then there is a morphism (g,h):1S→C⊕D. Since g:Agent(1S)→A⊔B, and Agent(1S)={b′} is a singleton, the image of g must be entirely in A or in B. Without loss of generality, assume it is in A. Then, let f be any element of F. There is a morphism (g′,h′):1S→C given by g′(b′)=g(b′), and h′(e)=h(e,f). This is a morphism because

b′⋄h′(e)=b′⋄h(e,f)=g(b′)∙(e,f)=g′(b′)⋅e,where ⋄=Eval(1S) and ∙=Eval(C⊕D). Thus, S∈Ensure(C). □

The condition that C and D are not null is annoying. It is also not very informative. It is reasonable to just think of null as not a real Cartesian frame, and not worry about it. To see what concretely goes wrong, let C=({a},{e},⋅), where a⋅e=w, and let D=null. Ensure(C)=⟨{w}⟩⊇ and Ensure(D)={}. However, C⊕D=({a},{},⋆), so Ensure(C⊕D)=2W, since the a ensures everything. The null brought us into the degenerate case where there are no possible environments, but since it had no possible agents, it had no ensurables until it was combined with C.

We also have that ensurables are preserved by biextensional equivalence.

Claim:If C≃D, Ensure(C)=Ensure(D).Proof:From the homotopy equivalence definition, we have that if C≃D, there exist morphisms from C to D and vice versa. □Finally, we have that there is a tradeoff between a Cartesian frame's ability to ensure things and its dual's ability to prevent things.

Claim:Ensure(C)∩Prevent(C∗)={}.Proof:Trivial. □1.2. ControllablesControllables also have a simple categorical definition.

Definition:Let 2S denote the Cartesian frame 1S⊕1W∖S.Again, C⊕D represents a game where the agent chooses whether to play C or D, and the environment must be able to respond in either case. While 1S has one possible agent, 2S has two possible agents, representing the choice between S and W∖S.

2S's environments are all possible pairs of exactly one s∈S and exactly one s∈W∖S. For example, if S0={w0,w3,w4} (as in our earlier example of a 1S) and W={w0,w1,w2,w3,w4}, then

2S0=w0w1w0w2w3w1w3w2w4w1w4w2bS0bW∖S0( w0 w0 w3 w3 w4 w4 w1 w2 w1 w2 w1 w2 ).

So the agent decides whether we're in S, and the environment picks a strategy for S and another for the complement of S.

Definition:Ctrl(C) is the set of all S⊆W such that there exists a morphism ϕ:2S→C.Claim:This definition is equivalent to the one in "Introduction to Cartesian Frames": Ctrl(C)=Ensure(C)∩Prevent(C).Proof:Since ⊕ is the categorical coproduct, there exists a morphism from 2S=1S⊕1W∖S to C if and only if there exist a pair of morphisms ϕ0:1S→C and ϕ1:1W∖S→C, which by the above definitions of ensurables and preventables is true if and only if S is in both Ensure(C) and Prevent(C). □Since Ctrl(C) is the set of all S⊆W with both S and W∖S in Ensure(C), we immediately have that the following closure properties on ensurables also apply to controllables.

Claim:If there exists a morphism ϕ:C→D, then Ctrl(C)⊆Ctrl(D).Proof:Trivial. □Claim:Ctrl(C&D)=Ctrl(C)∩Ctrl(D).Proof:Trivial. □Claim:If C≃D, then Ctrl(C)=Ctrl(D).Proof:Trivial. □Note that although Ensure(C⊕D)=Ensure(C)∪Ensure(D) is usually true, there isn't a corresponding result for controllables.

## 2. Observables

We also have a new definition of observables, but it is not nearly as trivial as the definition of controllables.

Definition:Obs(C) is the set of all S⊆W such that there exist C0 and C1 with Image(C0)⊆S and Image(C1)⊆W∖S such that C≃C0&C1.Claim:This definition is equivalent to the one in "Introduction to Cartesian Frames": Obs(C)={S⊆W | ∀a0,a1∈A, ∃a∈A, a∈if(S,a0,a1)}.Proof:Throughout the proof, we will let Obsold(−) refer to observables as they were originally defined, and let Obs(−) refer to observables under our categorical definition.The proof will be broken into three parts:

This gives us that Obsold(C)⊆Obs(C), by taking C0=(A,ES,⋅) and C1=(A,EW∖S,⋅).

Part 1.We want to show that Obsold(−) is closed under biextensional equivalence.

Let C=(A,E,⋅), let D=(B,F,⋆), and let C≃D. We will use the homotopy equivalence definition, so let (g0,h0):C→D and (g1,h1):D→C be such that (g0,h0)∘(g1,h1) and (g1,h1)∘(g0,h0) are both homotopic to the identity. Given S∈Obsold(C), we want to show that S∈Obsold(D).

Given b0,b1∈B, we want to show that there exists a b∈B such that for all f∈F, if b⋆f∈S, then b⋆f=b0⋆f, and otherwise b⋆f=b1⋆f. Letting ai=g1(bi), the fact that S∈Obsold(C) gives that there exists an a∈A, such that for all e∈E, if a⋅e∈S, then a⋅e=a0⋅e, and otherwise, a⋅e=a1⋅e. We will take b=g0(a).

For all f∈F, we have that

b⋆f=g0(a)⋆f=a⋅h0(f).Further, since (g0,h0)∘(g1,h1) is homotopic to the identity, we also have that for all f∈F,

bi⋆f=g0(g1(bi))⋆f=g1(bi)⋅h0(f)=ai⋅h0(f).Together these give that if b⋆f∈S, then a⋅h0(f)∈S, so

b⋆f=a⋅h0(f)=a0⋅h0(f)=b0⋆f,and if b⋆f∉S, then a⋅h0(f)∉S, so

b⋆f=a⋅h0(f)=a1⋅h0(f)=b1⋆f.Thus S∈Obsold(D), so Obsold(−) is closed under biextensional equivalence.

Part 2.We want to show that if Image(C0)⊆S and Image(C1)⊆W∖S, then S∈Obsold(C0&C1).

Let C0=(A,E,⋅), let C1=(B,F,⋆), and let C0&C1=(A×B,E⊔F,⋄). Given (a0,b0),(a1,b1)∈A×B, we want to show that there exists an (a,b)∈A×B such that if (a,b)⋄e∈S, then (a,b)⋄e=(a0,b0)⋄e, and otherwise (a,b)⋄e=(a1,b1)⋄e. We will take (a,b)=(a0,b1).

For all f∈F, (a0,b1)⋄f=b1⋆f∉S, since Image(C1)⊆W∖S. Thus, if (a0,b1)⋄e∈S, then e∈E, so

(a0,b1)⋄e=a0⋅e=(a0,b0)⋄e.Similarly, if (a0,b1)⋄e∉S, then e∈F, so

(a0,b1)⋄e=b1⋅e=(a1,b1)⋄e.Thus, S∈Obsold(C0&C1).

Part 3.We want to show that if S∈Obsold(A,E,⋅), then (A,E,⋅)≃(A,ES,⋅)&(A,EW∖S,⋅).

Let C=(A,E,⋅), let C0=(A,ES,⋅), and let C1=(A,EW∖S,⋅). (Here the ⋅ in C0 and C1 is the restriction of ⋅ in C to the respective domain.) Let ⋆=Eval(C0&C1).

First, let's quickly deal with the degenerate case where A is empty. In this case ES=EW∖S=E. If E is also empty, then C≃null≃null&null≃C0&C1. If E is nonempty, then C≃0≃0&0≃C0&C1. Thus, we can restrict our attention to the case where A is nonempty. Note that in this case, ES and EW∖S are disjoint, and as we saw before they cover E, so E=ES⊔EW∖S.

We need to construct a (g0,h0):C→C0&C1 and a (g1,h1):C0&C1→C, which compose to something homotopic to the identity in both orders. Since E=ES⊔EW∖S, we can just take h0 and h1 to be the identity on E. We will take g0:A→A×A to be the diagonal given by g0(a)=(a,a). Finally, for g1:A×A→A, we will use the fact that S∈Obsold(C). We will let g1(a0,a1) be chosen such that g1(a0,a1)⋅e=a0⋅e if g1(a0,a1)⋅e∈S, and g1(a0,a1)⋅e=a1⋅e otherwise. We can always choose such a g1(a0,a1), by the definition of Obsold(C).

To see that (g0,h0) is a morphism, observe that for all a∈A and e∈ES⊔EW∖S, we have

g0(a)⋆e=(a,a)⋆e=a⋅e=a⋅h0(e),regardless of which half e is in.

To see that (g1,h1) is a morphism, observe that for all (a0,a1)∈A×A and e∈E, if e∈ES, then

g1(a0,a1)⋅e=a0⋅e=(a0,a1)⋆e=(a0,a1)⋆h1(e),while if e∈EW∖S, then

g1(a0,a1)⋅e=a1⋅e=(a0,a1)⋆e=(a0,a1)⋆h1(e).Finally, the fact that (g0,h0) and (g1,h1) compose to something homotopic to the identity in both orders is trivial, since h0∘h1 and h1∘h0 are both the identity, so trivially a⋅e=a⋅h0(h1(e)), and (a0,a1)⋆e=(a0,a1)⋆h1(h0(e)). (Technically, this is verifying that the identity is homotopic each composition, but since being homotopic is symmetric, this is fine.)

Putting it together.If S∈Obs(C), then C≃C0&C1, where Image(C0)⊆S and Image(C1)⊆W∖S. By part 2, S∈Obsold(C0&C1), which by part 1, means that S∈Obsold(C). Conversely, if S∈Obsold(A,E,⋅), then by part 3, (A,E,⋅)≃(A,ES,⋅)&(A,EW∖S,⋅), which since Image(A,ES,⋅)⊆S and Image(A,EW∖S,⋅)⊆W∖S, implies that S∈Obs(A,E,⋅). □

Note that from the above proof, if S∈Obs(C), we know how to construct the C0 and C1 such that C≃C0&C1. In particular, every column of C must be entirely contained in S or entirely outside of S, and C0 just takes the subset of columns in S while C1 takes the subset of columns outside of S.

One thing to like about this new definition is that it shows that when an agent can observe S, you can actually break it up into two different agents. The first agent chooses how to behave in worlds in S and is promised that the world will in fact be in S, and the second does the same for the worlds not in S. These two agents combine using & to form the original agent.

Observables are much less well-behaved than controllables, so there is much less to say about them at this point. We do have that observability is preserved under biextensional equivalence, which is trivial under the new definition and was proven within the previous proof for the old definition.

Claim:If C≃D, then Obs(C)=Obs(D).Proof:Trivial. □## 3. Controllables and Observables Are Still Disjoint

To become more used to our new definitions, let us reprove the incompatibility theorems from before. First, a lemma.

Lemma:Let C≃C0&C1, with Image(C0)⊆S and Image(C1)⊆W∖S. If S∈Ensure(C), then C1≃⊤. If S∈Prevent(C), then C0≃⊤.Proof:If S∈Ensure(C), there exists a morphism from 1S to C, so there exists a morphism from 1S to C0&C1. Composing this with the canonical projection from C0&C1 to C1 gives a morphism (g,h):1S→C1. Let C1=(A,E,⋅), and let 1S=({b},S,⋆). If there were an e∈E, then g(b)⋅e=b⋆h(e) would be in both S and W∖S, a contradiction. Therefore C1 has empty environment. Also, since g(b)∈A, C1 has nonempty agent. Therefore C1≃⊤.Symmetrically, if S∈Prevent(C), then W∖S∈Ensure(C), so C0≃⊤. □

Now we can reprove (a slightly stronger version of) our main incompatibility theorem.

Theorem:If C≄⊤, then Ctrl(C)∩Obs(C)={}.Proof:We prove the contrapositive. Assume S∈Ctrl(C)∩Obs(C). Let C≃C0&C1, with Image(C0)⊆S and Image(C1)⊆W∖S. By the above lemma, both C0≃⊤ and C1≃⊤. Thus C≃⊤&⊤≅⊤. □We also reprove (the important direction of) the one-sided result.

Theorem:If S∈Ensure(C)∩Obs(C), then Image(C)⊆S.Proof:If S∈Ensure(C)∩Obs(C), then C≃C0&C1, with Image(C0)⊆S and Image(C1)⊆W∖S. By the above lemma, C1≃⊤, so C≃C0&⊤≅C0, so Image(C)⊆S. □In our next post, we will move to discussing Cartesian frames over different worlds, or different world models. E.g., W might be the set of all possible microphysical states of a room, while V is the smaller set of all possible arrangements of macroscopic objects in the room. We will describe how to translate between frames over W and frames over V.

In the process, we will note some surprising facts about coarser and more refined models of the world, as they relate to observables.