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.

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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.

 is wrong. You can see it has Ensurables that  does not have.

You can also duplicate rows in , and then add columns, so you can get things like . There are infinitely many biextensional Cartesian frames over  with morphism to , with arbitrarily large dimensions.