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 Preventables**

The categorical definition of ensurables is very simple.

**Definition:** is the set of all such that there exists a morphism .

As an example, let . Recall that , where for all . E.g., if , then

.

If there is a morphism from to , this means that:

- There is a function from 's agent to 's agent , i.e., a function that always outputs a specific .
- There is a function from 's environment to 's environment .
- The specific picked out by exactly implements that function .

That function is exactly like the function you get by looking at that row, so a morphism is like a row in that is entirely contained in . If there are multiple such rows, then there will be multiple distinct morphisms picking out different .

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

**Claim:** This definition is equivalent to the one in "Introduction to Cartesian Frames": .

**Proof: **Let and let , where for all . First, assume there exists a morphism . Here, and . Consider the element . It suffices to show that for all . Indeed, .

Conversely, assume that there exists an , such that for all . Then, there is a morphism given by , and . This is a morphism because

for all and .

**Definition: **** **is the set of all such that there exists a morphism .

**Claim: **This definition is equivalent to the one in "Introduction to Cartesian Frames": .

**Proof: **This follows from the proof for , substituting for .

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 , then .

**Proof:** If , there exists a morphism , so we have so .

This fact justifies our interpretation of the existence of a morphism from to as saying that " is at least as strong as ."

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

**Claim:** .

**Proof: **Since is a categorical product, if there exists a morphism from to and a morphism from to , there must exist a morphism from to . Thus . Conversely, since is a categorical product, there exist projection morphisms from to and from to , so .

**Claim:** If and , then .

**Proof: **Let , and let . Since is not , if were empty, then would have to be nonempty, and would be the full set . would also have empty environment and nonempty agent, and so would also be the full set . Thus, we are done in the case where is empty. Similarly, we are done in the case where is empty. Assume and are nonempty.

It is clear that** ** because is a coproduct, so there are canonical injection morphisms from to and from to .

Conversely, if , then there is a morphism . Since , and is a singleton, the image of must be entirely in or in . Without loss of generality, assume it is in . Then, let be any element of . There is a morphism given by , and . This is a morphism because

where and . Thus, .

The condition that and are not is annoying. It is also not very informative. It is reasonable to just think of as not a real Cartesian frame, and not worry about it. To see what concretely goes wrong, let , where , and let . and . However, , so , since the ensures everything. The 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 .

We also have that ensurables are preserved by biextensional equivalence.

**Claim:** If , .

**Proof:** From the homotopy equivalence definition, we have that if , there exist morphisms from to 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:** .

**Proof:** Trivial.

**1.2. Controllables**

Controllables also have a simple categorical definition.

**Definition:** Let denote the Cartesian frame .

Again, represents a game where the agent chooses whether to play or , and the environment must be able to respond in either case. While has one possible agent, has two possible agents, representing the choice between and .

's environments are all possible pairs of exactly one and exactly one . For example, if (as in our earlier example of a ) and , then

.

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

**Definition: ** is the set of all such that there exists a morphism .

**Claim:** This definition is equivalent to the one in "Introduction to Cartesian Frames": .

**Proof:** Since is the categorical coproduct, there exists a morphism from to if and only if there exist a pair of morphisms and , which by the above definitions of ensurables and preventables is true if and only if is in both and .

Since is the set of all with both and in , we immediately have that the following closure properties on ensurables also apply to controllables.

**Claim:** If there exists a morphism , then .

**Proof:** Trivial.

**Claim:** .

**Proof:** Trivial.

**Claim:** If , then .

**Proof:** Trivial.

Note that although 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:** is the set of all such that there exist and with and such that .

**Claim:** This definition is equivalent to the one in "Introduction to Cartesian Frames": .

**Proof:** Throughout the proof, we will let refer to observables as they were originally defined, and let refer to observables under our categorical definition.

The proof will be broken into three parts:

- First, we will show that is closed under biextensional equivalence.
- Second, we will show that if and , then . In combination with the first part, this implies that .
- Third, we will show that if , then .

This gives us that , by taking and .

__Part 1.__

We want to show that is closed under biextensional equivalence.

Let , let , and let . We will use the homotopy equivalence definition, so let and be such that and are both homotopic to the identity. Given , we want to show that .

Given , we want to show that there exists a such that for all , if , then , and otherwise . Letting , the fact that gives that there exists an , such that for all , if , then , and otherwise, . We will take .

For all , we have that

Further, since is homotopic to the identity, we also have that for all ,

Together these give that if , then , so

and if , then , so

Thus , so is closed under biextensional equivalence.

__Part 2.__

We want to show that if and , then .

Let , let , and let . Given , we want to show that there exists an such that if , then , and otherwise . We will take .

For all , , since . Thus, if , then , so

Similarly, if , then , so

Thus, .

__Part 3.__

We want to show that if , then .

Let , let , and let . (Here the in and is the restriction of in to the respective domain.) Let .

First, let's quickly deal with the degenerate case where is empty. In this case . If is also empty, then . If is nonempty, then . Thus, we can restrict our attention to the case where is nonempty. Note that in this case, and are disjoint, and as we saw before they cover , so .

We need to construct a and a , which compose to something homotopic to the identity in both orders. Since , we can just take and to be the identity on . We will take to be the diagonal given by . Finally, for , we will use the fact that . We will let be chosen such that if , and otherwise. We can always choose such a , by the definition of .

To see that is a morphism, observe that for all and , we have

regardless of which half is in.

To see that is a morphism, observe that for all and , if , then

while if , then

Finally, the fact that and compose to something homotopic to the identity in both orders is trivial, since and are both the identity, so trivially , and . (Technically, this is verifying that the identity is homotopic each composition, but since being homotopic is symmetric, this is fine.)

__Putting it together.__

If , then , where and . By part 2, , which by part 1, means that . Conversely, if , then by part 3, , which since and , implies that .

Note that from the above proof, if , we know how to construct the and such that . In particular, every column of must be entirely contained in or entirely outside of , and just takes the subset of columns in while takes the subset of columns outside of .

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

**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 , with and . If then . If , then .

**Proof:** If , there exists a morphism from to , so there exists a morphism from to . Composing this with the canonical projection from to gives a morphism . Let , and let . If there were an , then would be in both and , a contradiction. Therefore has empty environment. Also, since , has nonempty agent. Therefore .

Symmetrically, if , then , so .

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

**Theorem:** If , then .

**Proof:** We prove the contrapositive. Assume . Let , with and . By the above lemma, both and . Thus .

We also reprove (the important direction of) the one-sided result.

**Theorem:** If , then .

**Proof:** If , then , with and . By the above lemma, , so , so

In our next post, we will move to discussing Cartesian frames over different worlds, or different world models. E.g., might be the set of all possible microphysical states of a room, while is the smaller set of all possible arrangements of macroscopic objects in the room. We will describe how to translate between frames over and frames over .

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

Counterexample to your claim that Ctrl(C⊕D)=Ctrl(C)∪Ctrl(D) (with both C and D not null):

Take C and D to be the following Cartesian frames over world {x,y} with agent and environment both {a,b} as the following matrices:

abayxbxxabaxybyy

Then Ctrl(C)=Ctrl(D)=∅ but Ctrl(C⊕D)∋{x}.

(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

C0= f0f1b0b1(w0w1w2w3) .

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

Frames I think have a morphism going to C0:

C0

C∗0= e0e1a0a1(w0w2w1w3)

⊥{}=0=e0( )

1{w0,w1}= e0e1a0(w0w1)

1{w2,w3}= e0e1a0(w2w3)

Every frame that looks like a frame on this list (other than 0), but with extra columns added — regardless of what's in those columns. (As a special case, this includes the five 1S frames corresponding to the other five ensurables that can be deduced from the matrix: 1{w0,w1,w2}, 1{w0,w1,w3}, 1{w0,w2,w3}, 1{w1,w2,w3}, 1{w0,w1,w2,w3}. If W has more than four elements, then there will be additional 1S frames / additional ensurables beyond these seven.)

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

C∗0 is wrong. You can see it has Ensurables that C0 does not have.

You can also duplicate rows in C0, and then add columns, so you can get things like ⎛⎜⎝w0w1w2w0w1w3w2w3w0⎞⎟⎠. There are infinitely many biextensional Cartesian frames over {w0,w1,w2,w3} with morphism to C0, with arbitrarily large dimensions.