# Controllables and Observables, Revisited

This is the fourth post in the Cartesian frames sequence. Read the first post [here](https://www.lesswrong.com/posts/BSpdshJWGAW6TuNzZ/introduction-to-cartesian-frames).

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:** $\mbox{Ensure}(C)$ is the set of all $S\subseteq W$ such that there exists a morphism $\phi:1_S\rightarrow C$.

As an example, let $C_0=(A,E,\cdot)$. Recall that $1_S=(\{b\},S,\star)$, where $b \star s = s$ for all $s\in S$. E.g., if $S_0=\{w_0,w_3,w_4\}$, then

$1_{S_0}=\begin{array}{cc} &\begin{array}{cc} {w_0}   &{w_3}   &{w_4} \end{array}\ \begin{array}{c} b \end{array} & \begin{pmatrix} w_0  &  w_3  &  w_4 \ \end{pmatrix} \end{array}$.

If there is a morphism $(g,h)$ from $1_{S_0}$ to $C_0$, this means that:

*   There is a function $g$ from $1_{S_0}$'s agent $\{b\}$ to $C_0$'s agent $A$, i.e., a function that always outputs a specific $a \in A$.
*   There is a function $h$ from $C_0$'s environment $E$ to $1_{S_0}$'s environment $S_0$.
*   The specific $a \in A$ picked out by $g$ exactly implements that function $h : E\rightarrow S_0$.

That function $h : E\rightarrow S_0$ is exactly like the function you get by looking at that row, so a morphism $(g,h):1_{S_0}\rightarrow C_0$ is like a row in $C_0$ that is entirely contained in $S_0$. If there are multiple such rows, then there will be multiple distinct morphisms $1_{S_0}\rightarrow C_0$ picking out different $a \in A$.

In "[Biextensional Equivalence](https://www.lesswrong.com/posts/pWruFSY7494vnucCE/biextensional-equivalence#4__Some_Small_Cartesian_Frames)," we noted that $1_S$ is like a passive observer who has a promise from the environment that the world will be in $S$. The existence of a morphism $1_S\rightarrow C$ means that there's an interface that allows a powerless bystander who has been promised $S$ to play $C$'s game. Since $\mbox{Agent}(1_S)$ 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](https://www.lesswrong.com/posts/BSpdshJWGAW6TuNzZ/introduction-to-cartesian-frames#3__Controllables)": $\mbox{Ensure}(C)=\{S⊆W  | \ ∃a∈A, ∀e∈E, a⋅e∈S\}$.

**Proof:** Let $C=(A,E,\cdot)$ and let $1_S=(\{b\},S,\star)$, where $b \star s = s$ for all $s\in S$. First, assume there exists a morphism $(g,h):1_S\rightarrow C$. Here, $g:\{b\}\rightarrow A$ and $h:E\rightarrow S$. Consider the element $g(b)\in A$. It suffices to show that $g(b)\cdot e\in S$ for all $e\in E$. Indeed, $g(b)\cdot e=b\star h(e)\in S$.

Conversely, assume that there exists an $a\in A$, such that $a\cdot e\in S$ for all $e\in E$. Then, there is a morphism $(g,h):1_S\rightarrow C$ given by $g(b)=a$, and $h(e)=a\cdot e$. This is a morphism because

$$
\begin{equation} \begin{split} g(b)\cdot e &=a\cdot e \ &=h(e) \ &=b\star h(e) \end{split}\end{equation}
$$

for all $b\in\{b\}$ and $e\in E$. $\square$

**Definition: **$\mbox{Prevent}(C)$is the set of all $S\subseteq W$ such that there exists a morphism $\phi_1: 1_{W\backslash S}\rightarrow C$.

**Claim:** This definition is equivalent to the one in "[Introduction to Cartesian Frames](https://www.lesswrong.com/posts/BSpdshJWGAW6TuNzZ/introduction-to-cartesian-frames#3__Controllables)": $\mbox{Prevent}(C)=\{S\subseteq W  |  \exists a\in A, \forall e\in E, a\cdot e\notin S\}$.

**Proof:** This follows from the proof for $\mbox{Ensure}(C)$, substituting $W\backslash S$ for $S$. $\square$

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 $\phi:C\rightarrow D$, then $\mbox{Ensure}(C)\subseteq \mbox{Ensure}(D)$.

**Proof:** If $S\in \mbox{Ensure}(C)$, there exists a morphism $\psi:1_S\rightarrow C$, so we have $\phi\circ\psi:1_S\rightarrow D,$ so $S\in \mbox{Ensure}(D)$. $\square$

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 $C \oplus D$ 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 $C \oplus D$ 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:** $\mbox{Ensure}(C\& D)= \mbox{Ensure}(C)\cap\mbox{Ensure}(D)$.

**Proof:** Since $\&$ is a categorical product, if there exists a morphism from $1_S$ to $C$ and a morphism from $1_S$ to $D$, there must exist a morphism from $1_S$ to $C\&D$. Thus $\mbox{Ensure}(C\& D)\supseteq \mbox{Ensure}(C)\cap\mbox{Ensure}(D)$. Conversely, since $\&$ is a categorical product, there exist projection morphisms from $C\&D$ to $C$ and from $C\&D$ to $D$, so $\mbox{Ensure}(C\& D)\subseteq \mbox{Ensure}(C)\cap\mbox{Ensure}(D)$. $\square$

**Claim:** If $C\neq\mbox{null}$ and $D\neq\mbox{null}$, then $\mbox{Ensure}(C\oplus D)=\mbox{Ensure}(C)\cup\mbox{Ensure}(D)$.

**Proof:** Let $C=(A,E,\cdot)$, and let $D=(B,F,\star)$. Since $C$ is not $\mbox{null}$, if $E$ were empty, then $A$ would have to be nonempty, and $\mbox{Ensure}(C)$ would be the full set $2^W$. $C\oplus D$ would also have empty environment and nonempty agent, and so $\mbox{Ensure}(C\oplus D)$ would also be the full set $2^W$. 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 that$\mbox{Ensure}(C\oplus D)\supseteq \mbox{Ensure}(C)\cup\mbox{Ensure}(D)$ because $C\oplus D$ is a coproduct, so there are canonical injection morphisms from $C$ to $C\oplus D$ and from $D$ to $C\oplus D$.

Conversely, if $S\in\mbox{Ensure}(C\oplus D)$, then there is a morphism $(g,h):1_S\rightarrow C\oplus D$. Since $g:\mbox{Agent}(1_S)\rightarrow A\sqcup B$, and $\mbox{Agent}(1_S)=\{b^\prime\}$ 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 $(g^\prime, h^\prime):1_S\rightarrow C$ given by $g^\prime(b^\prime)=g(b^\prime)$, and $h^\prime(e)=h(e,f)$. This is a morphism because

$$
\begin{equation} \begin{split}b^\prime\diamond h^\prime(e)& =b^\prime\diamond h(e,f) \& =g(b^\prime)\bullet(e,f) \& =g^\prime(b^\prime)\cdot e,\end{split} \end{equation}
$$

where $\diamond=\mbox{Eval}(1_S)$ and $\bullet=\mbox{Eval}(C\oplus D)$. Thus, $S\in \mbox{Ensure}(C)$. $\square$

The condition that $C$ and $D$ are not $\mbox{null}$ is annoying. It is also not very informative. It is reasonable to just think of $\mbox{null}$ as not a real Cartesian frame, and not worry about it. To see what concretely goes wrong, let $C=(\{a\},\{e\},\cdot)$, where $a\cdot e=w$, and let $D=\mbox{null}$. $\mbox{Ensure}(C)=\langle \{w\}\rangle_\supseteq$ and $\mbox{Ensure}(D)=\{\}$. However, $C\oplus D=(\{a\},\{\},\star)$, so $\mbox{Ensure}(C\oplus D)=2^W$, since the $a$ ensures everything. The $\mbox{null}$ 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 $C\simeq D$, $\mbox{Ensure}(C)=\mbox{Ensure}(D)$.

**Proof:** From the [homotopy equivalence definition](https://www.lesswrong.com/posts/pWruFSY7494vnucCE/biextensional-equivalence#2_2__Homotopy_Equivalence), we have that if $C\simeq D$, there exist morphisms from $C$ to $D$ and vice versa. $\square$

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:** $\mbox{Ensure}(C)\cap \mbox{Prevent}(C^*)=\{\}$.

**Proof:** Trivial. $\square$

**1.2. Controllables**

Controllables also have a simple categorical definition.

**Definition:** Let $2_S$ denote the Cartesian frame $1_S\oplus 1_{W\backslash S}$. 

Again, $C \oplus D$ 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 $1_S$ has one possible agent, $2_S$ has two possible agents, representing the choice between $S$ and $W\backslash S$.

$2_S$'s environments are all possible pairs of exactly one $s \in S$ and exactly one $s \in W\backslash S$. For example, if $S_0=\{w_0,w_3,w_4\}$ (as in our earlier example of a $1_S$) and $W=\{w_0,w_1,w_2,w_3,w_4\}$, then

$2_{S_0}=\begin{array}{cc} &\begin{array}{cccc} w_0w_1 &w_0w_2 &w_3w_1 &w_3w_2 &w_4w_1 &w_4w_2 \end{array}\ \begin{array}{c} b_{S_0} \ b_{W\backslash S_0} \end{array} & \begin{pmatrix}  w_0   & w_0   & w_3   & w_3   & w_4   & w_4 \  w_1   & w_2   & w_1   & w_2   & w_1   & w_2 \end{pmatrix} \end{array}$.

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: **$\mbox{Ctrl}(C)$ is the set of all $S\subseteq W$ such that there exists a morphism $\phi:2_S\rightarrow C$.

**Claim:** This definition is equivalent to the one in "[Introduction to Cartesian Frames](https://www.lesswrong.com/posts/BSpdshJWGAW6TuNzZ/introduction-to-cartesian-frames#3__Controllables)": $\mbox{Ctrl}(C)=\mbox{Ensure}(C)\cap \mbox{Prevent}(C)$.

**Proof:** Since $\oplus$ is the categorical coproduct, there exists a morphism from $2_S=1_S\oplus 1_{W\backslash S}$ to $C$ if and only if there exist a pair of morphisms $\phi_0:1_S\rightarrow C$ and $\phi_1: 1_{W\backslash S}\rightarrow C$, which by the above definitions of ensurables and preventables is true if and only if $S$ is in both $\mbox{Ensure}(C)$ and $\mbox{Prevent}(C)$. $\square$

Since $\mbox{Ctrl}(C)$ is the set of all $S\subseteq W$ with both $S$ and $W\backslash S$ in $\mbox{Ensure}(C)$, we immediately have that the following closure properties on ensurables also apply to controllables.

**Claim:** If there exists a morphism $\phi:C\rightarrow D$, then $\mbox{Ctrl(C)}\subseteq \mbox{Ctrl}(D)$.

**Proof:** Trivial. $\square$

**Claim:** $\mbox{Ctrl}(C\& D)= \mbox{Ctrl}(C)\cap\mbox{Ctrl}(D)$.

**Proof:** Trivial. $\square$

**Claim:** If $C\simeq D$, then $\mbox{Ctrl}(C)=\mbox{Ctrl}(D)$.

**Proof:** Trivial. $\square$

Note that although $\mbox{Ensure}(C\oplus D)=\mbox{Ensure}(C)\cup\mbox{Ensure}(D)$ 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:** $\mbox{Obs}(C)$ is the set of all $S\subseteq W$ such that there exist $C_0$ and $C_1$ with $\mbox{Image}(C_0)\subseteq S$ and $\mbox{Image}(C_1)\subseteq W\backslash S$ such that $C\simeq C_0\&C_1$.

**Claim:** This definition is equivalent to the one in "[Introduction to Cartesian Frames](https://www.lesswrong.com/posts/BSpdshJWGAW6TuNzZ/introduction-to-cartesian-frames#4__Observables)": $\mbox{Obs}(C)=\{S\subseteq W  |  \forall a_0,a_1\in A, \exists a\in A, a\in\mbox{if}(S,a_0,a_1)\}$.

**Proof:** Throughout the proof, we will let $\mbox{Obs}_{\mathrm{old}}(-)$ refer to observables as they were originally defined, and let $\mbox{Obs}(-)$ refer to observables under our categorical definition.

The proof will be broken into three parts:

*   First, we will show that $\mbox{Obs}_{\mathrm{old}}(-)$ is closed under biextensional equivalence.
*   Second, we will show that if $\mbox{Image}(C_0)\subseteq S$ and $\mbox{Image}(C_1)\subseteq W\backslash S$, then $S\in \mbox{Obs}_{\mathrm{old}}(C_0\&C_1)$. In combination with the first part, this implies that $\mbox{Obs}(C)\subseteq\mbox{Obs}_{\mathrm{old}}(C)$.
*   Third, we will show that if $S\in \mbox{Obs}_{\mathrm{old}}(A,E,\cdot)$, then $(A,E,\cdot)\simeq (A,E_S,\cdot)\&(A,E_{W\backslash S},\cdot)$.

This gives us that $\mbox{Obs}_{\mathrm{old}}(C)\subseteq \mbox{Obs}(C)$, by taking $C_0=(A,E_S,\cdot)$ and $C_1=(A,E_{W\backslash S},\cdot)$.

Part 1.

We want to show that $\mbox{Obs}_{\mathrm{old}}(-)$ is closed under biextensional equivalence.

Let $C=(A,E,\cdot)$, let $D=(B,F,\star)$, and let $C\simeq D$. We will use the homotopy equivalence definition, so let $(g_0,h_0):C\rightarrow D$ and $(g_1,h_1):D\rightarrow C$ be such that $(g_0,h_0)\circ(g_1,h_1)$ and $(g_1,h_1)\circ (g_0,h_0)$ are both homotopic to the identity. Given $S\in \mbox{Obs}_{\mathrm{old}}(C)$, we want to show that $S\in \mbox{Obs}_{\mathrm{old}}(D)$.

Given $b_0,b_1\in B$, we want to show that there exists a $b\in B$ such that for all $f\in F$, if $b\star f\in S$, then $b\star f=b_0\star f$, and otherwise $b\star f=b_1\star f$. Letting $a_i=g_1(b_i)$, the fact that $S\in\mbox{Obs}_{\mathrm{old}}(C)$ gives that there exists an $a\in A$, such that for all $e\in E$, if $a\cdot e\in S$, then $a\cdot e=a_0\cdot e$, and otherwise, $a\cdot e=a_1\cdot e$. We will take $b=g_0(a)$.

For all $f\in F$, we have that

$$
\begin{equation} \begin{split} b\star f &=g_0(a)\star f \ &=a\cdot h_0(f). \end{split}\end{equation}
$$

Further, since $(g_0,h_0)\circ(g_1,h_1)$ is homotopic to the identity, we also have that for all $f\in F$,

$$
\begin{equation} \begin{split} b_i\star f & =g_0(g_1(b_i))\star f \ & =g_1(b_i)\cdot h_0(f) \ & =a_i\cdot h_0(f). \end{split} \end{equation}
$$

Together these give that if $b\star f\in S$, then $a\cdot h_0(f)\in S$, so

$$
\begin{equation} \begin{split} b\star f & =a\cdot h_0(f) \ & =a_0\cdot h_0(f) \ & =b_0\star f, \end{split} \end{equation}
$$

 and if $b\star f\notin S$, then $a\cdot h_0(f)\notin S$, so

$$
\begin{equation} \begin{split} b\star f & =a\cdot h_0(f) \ & =a_1\cdot h_0(f) \ & =b_1\star f. \end{split} \end{equation}
$$

Thus $S\in \mbox{Obs}_{\mathrm{old}}(D)$, so $\mbox{Obs}_{\mathrm{old}}(-)$ is closed under biextensional equivalence.

Part 2.

We want to show that if $\mbox{Image}(C_0)\subseteq S$ and $\mbox{Image}(C_1)\subseteq W\backslash S$, then $S\in\mbox{Obs}_{\mathrm{old}}(C_0\&C_1)$.

Let $C_0=(A,E,\cdot)$, let $C_1=(B,F,\star)$, and let $C_0\&C_1=(A\times B,E\sqcup F,\diamond)$. Given $(a_0,b_0), (a_1,b_1)\in A\times B$, we want to show that there exists an $(a,b)\in A\times B$ such that if $(a,b)\diamond e\in S$, then $(a,b)\diamond e=(a_0,b_0)\diamond e$, and otherwise $(a,b)\diamond e=(a_1,b_1)\diamond e$. We will take $(a,b)=(a_0,b_1)$.

For all $f\in F$, $(a_0,b_1)\diamond f=b_1\star f\notin S$, since $\mbox{Image}(C_1)\subseteq W\backslash S$. Thus, if $(a_0,b_1)\diamond e\in S$, then $e\in E$, so

$$
\begin{equation} \begin{split} (a_0,b_1)\diamond e &= a_0\cdot e \ &=(a_0,b_0)\diamond e. \end{split}\end{equation}
$$

Similarly, if $(a_0,b_1)\diamond e\not\in S$, then $e\in F$, so

$$
\begin{equation} \begin{split} (a_0,b_1)\diamond e &= b_1\cdot e \ &=(a_1,b_1)\diamond e. \end{split}\end{equation}
$$

Thus, $S\in \mbox{Obs}_{\mathrm{old}}(C_0\&C_1)$.

Part 3.

We want to show that if $S\in \mbox{Obs}_{\mathrm{old}}(A,E,\cdot)$, then $(A,E,\cdot)\simeq (A,E_S,\cdot)\&(A,E_{W\backslash S},\cdot)$.

Let $C=(A,E,\cdot)$, let $C_0=(A,E_S,\cdot)$, and let $C_1=(A,E_{W\backslash S},\cdot)$. (Here the $\cdot$ in $C_0$ and $C_1$ is the restriction of $\cdot$ in $C$ to the respective domain.) Let $\star=\mbox{Eval}(C_0\&C_1)$.

First, let's quickly deal with the degenerate case where $A$ is empty. In this case $E_S=E_{W\backslash S}=E$. If $E$ is also empty, then $C\simeq \mbox{null}\simeq \mbox{null}\&\mbox{null}\simeq C_0\&C_1$. If $E$ is nonempty, then $C\simeq 0\simeq 0\&0\simeq C_0\&C_1$. Thus, we can restrict our attention to the case where $A$ is nonempty. Note that in this case, $E_S$ and $E_{W\backslash S}$ are disjoint, and as we saw before they cover $E$, so $E=E_S\sqcup E_{W\backslash S}$.

We need to construct a $(g_0,h_0): C\rightarrow C_0\&C_1$ and a $(g_1,h_1):C_0\&C_1\rightarrow C$, which compose to something homotopic to the identity in both orders. Since $E=E_S\sqcup E_{W\backslash S}$, we can just take $h_0$ and $h_1$ to be the identity on $E$. We will take $g_0:A\rightarrow A\times A$ to be the diagonal given by $g_0(a)=(a,a)$. Finally, for $g_1:A\times A\rightarrow A$, we will use the fact that $S\in\mbox{Obs}_{\mathrm{old}}(C)$. We will let $g_1(a_0,a_1)$ be chosen such that $g_1(a_0,a_1)\cdot e=a_0\cdot e$ if $g_1(a_0,a_1)\cdot e\in S$, and $g_1(a_0,a_1)\cdot e=a_1\cdot e$ otherwise. We can always choose such a $g_1(a_0,a_1)$, by the definition of $\mbox{Obs}_{\mathrm{old}}(C)$.

To see that $(g_0,h_0)$ is a morphism, observe that for all $a\in A$ and $e\in E_S\sqcup E_{W\backslash S}$, we have

$$
\begin{equation} \begin{split} g_0(a)\star e & =(a,a)\star e \ & =a\cdot e \ & =a\cdot h_0(e), \end{split} \end{equation}
$$

regardless of which half $e$ is in.

To see that $(g_1,h_1)$ is a morphism, observe that for all $(a_0,a_1)\in A\times A$ and $e\in E$, if $e\in E_S$, then

$$
\begin{equation} \begin{split} g_1(a_0,a_1)\cdot e & =a_0\cdot e \ & =(a_0,a_1)\star e \ & =(a_0,a_1)\star h_1(e), \end{split} \end{equation}
$$

 while if $e\in E_{W\backslash S}$, then 

$$
\begin{equation} \begin{split} g_1(a_0,a_1)\cdot e & =a_1\cdot e \ & =(a_0,a_1)\star e \ & =(a_0,a_1)\star h_1(e). \end{split} \end{equation}
$$

Finally, the fact that $(g_0,h_0)$ and $(g_1,h_1)$ compose to something homotopic to the identity in both orders is trivial, since $h_0\circ h_1$ and $h_1\circ h_0$ are both the identity, so trivially $a\cdot e=a\cdot h_0(h_1(e))$, and $(a_0,a_1)\star e=(a_0,a_1)\star h_1(h_0(e))$. (Technically, this is verifying that the identity is homotopic each composition, but since being homotopic is symmetric, this is fine.)

Putting it together.

If $S\in\mbox{Obs}(C)$, then $C\simeq C_0\&C_1$, where $\mbox{Image}(C_0)\subseteq S$ and $\mbox{Image}(C_1)\subseteq W\backslash S$. By part 2, $S\in\mbox{Obs}_{\mathrm{old}}(C_0\&C_1)$, which by part 1, means that $S\in \mbox{Obs}_{\mathrm{old}}(C)$. Conversely, if $S\in \mbox{Obs}_{\mathrm{old}}(A,E,\cdot)$, then by part 3, $(A,E,\cdot)\simeq (A,E_S,\cdot)\&(A,E_{W\backslash S},\cdot)$, which since $\mbox{Image}(A,E_S,\cdot)\subseteq S$ and $\mbox{Image}(A,E_{W\backslash S},\cdot)\subseteq W\backslash S$, implies that $S\in \mbox{Obs}(A,E,\cdot)$. $\square$

Note that from the above proof, if $S\in\mbox{Obs}(C)$, we know how to construct the $C_0$ and $C_1$ such that $C\simeq C_0\&C_1$. In particular, every column of $C$ must be entirely contained in $S$ or entirely outside of $S$, and $C_0$ just takes the subset of columns in $S$ while $C_1$ 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 $C\simeq D$, then $\mbox{Obs}(C)=\mbox{Obs}(D)$.

**Proof:** Trivial. $\square$

3\. Controllables and Observables Are Still Disjoint
----------------------------------------------------

To become more used to our new definitions, let us reprove the [incompatibility theorems](https://www.lesswrong.com/posts/BSpdshJWGAW6TuNzZ/introduction-to-cartesian-frames#5__Controllables_and_Observables_Are_Disjoint) from before. First, a lemma.

**Lemma:** Let $C\simeq C_0\&C_1$, with $\mbox{Image}(C_0)\subseteq S$ and $\mbox{Image}(C_1)\subseteq W\backslash S$. If $S\in \mbox{Ensure}(C),$ then $C_1\simeq\top$. If $S\in \mbox{Prevent}(C)$, then $C_0\simeq\top$.

**Proof:** If $S\in \mbox{Ensure}(C)$, there exists a morphism from $1_S$ to $C$, so there exists a morphism from $1_S$ to $C_0\&C_1$. Composing this with the canonical projection from $C_0\&C_1$ to $C_1$ gives a morphism $(g,h):1_S\rightarrow C_1$. Let $C_1 = (A,E,\cdot)$, and let $1_S=(\{b\},S,\star)$. If there were an $e\in E$, then $g(b)\cdot e=b\star h(e)$ would be in both $S$ and $W\backslash S$, a contradiction. Therefore $C_1$ has empty environment. Also, since $g(b)\in A$, $C_1$ has nonempty agent. Therefore $C_1\simeq\top$.

Symmetrically, if $S\in \mbox{Prevent}(C)$, then $W\backslash S\in \mbox{Ensure}(C)$, so $C_0\simeq\top$. $\square$

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

**Theorem:** If $C\not\simeq \top$, then $\mbox{Ctrl}(C)\cap\mbox{Obs}(C)=\{\}$.

**Proof:** We prove the contrapositive. Assume $S\in\mbox{Ctrl}(C)\cap\mbox{Obs}(C)$. Let $C\simeq C_0\&C_1$, with $\mbox{Image}(C_0)\subseteq S$ and $\mbox{Image}(C_1)\subseteq W\backslash S$. By the above lemma, both $C_0\simeq \top$ and $C_1\simeq \top$. Thus $C\simeq \top\&\top\cong\top$. $\square$

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

**Theorem:** If $S\in\mbox{Ensure}(C)\cap\mbox{Obs}(C)$, then $\mbox{Image}(C)\subseteq S$.

**Proof:** If $S\in\mbox{Ensure}(C)\cap\mbox{Obs}(C)$, then $C\simeq C_0\&C_1$, with $\mbox{Image}(C_0)\subseteq S$ and $\mbox{Image}(C_1)\subseteq W\backslash S$. By the above lemma, $C_1\simeq \top$, so $C\simeq C_0\&\top\cong C_0$, so $\mbox{Image}(C)\subseteq S.$ $\square$

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.