Controllables and Observables, Revisited

scott-garrabrant

Controllables and Observables, Revisited

post by Scott Garrabrant · 2020-10-29T16:38:14.878Z · LW · GW · 5 comments

  1. Controllables
    1.1. Ensurables and Preventables
    1.2. Controllables
  2. Observables
  3. Controllables and Observables Are Still Disjoint
None
5 comments

This is the fourth post in the Cartesian frames sequence. Read the first post here [LW · GW].

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 $S \subseteq W$ such that there exists a morphism $ϕ : 1_{S} \to C$ .

As an example, let $C_{0} = (A, E, \cdot)$ . Recall that $1_{S} = ({b}, S, ⋆)$ , where $b ⋆ s = s$ for all $s \in S$ . E.g., if $S_{0} = {w_{0}, w_{3}, w_{4}}$ , then

$1_{S_{0}} = \begin{matrix} \begin{matrix} w_{0} & w_{3} & w_{4} \end{matrix} \begin{matrix} b \end{matrix} & (\begin{matrix} w_{0} & w_{3} & w_{4} \end{matrix}) \end{matrix}$ .

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 $h : E \to S_{0}$ .

That function $h : E \to S_{0}$ is exactly like the function you get by looking at that row, so a morphism $(g, h) : 1_{S_{0}} \to C_{0}$ 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 $1_{S_{0}} \to C_{0}$ picking out different $a \in A$ .

In "Biextensional Equivalence [LW · GW]," 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 $1_{S} \to C$ means that there's an interface that allows a powerless bystander who has been promised $S$ to play $C$ 's game. Since $Agent (1_{S})$ 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 [LW · GW]": $Ensure (C) = {S \subseteq W | \exists a \in A, \forall e \in E, a \cdot e \in S}$ .

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

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 $(g, h) : 1_{S} \to C$ given by $g (b) = a$ , and $h (e) = a \cdot e$ . This is a morphism because

\begin{matrix} g (b) \cdot e & = a \cdot e = h (e) = b ⋆ h (e) \end{matrix}

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

Definition: $Prevent (C)$ is the set of all $S \subseteq W$ such that there exists a morphism $ϕ_{1} : 1_{W ∖ S} \to C$ .

Claim: This definition is equivalent to the one in "Introduction to Cartesian Frames [LW · GW]": $Prevent (C) = {S \subseteq W | \exists a \in A, \forall e \in E, a \cdot e \notin S}$ .

Proof: This follows from the proof for $Ensure (C)$ , 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 $ϕ : C \to D$ , then $Ensure (C) \subseteq Ensure (D)$ .

Proof: If $S \in Ensure (C)$ , there exists a morphism $ψ : 1_{S} \to C$ , so we have $ϕ \circ ψ : 1_{S} \to D,$ so $S \in Ensure (D)$ . $□$

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 $C \oplus D$ 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: $Ensure (C & D) = Ensure (C) \cap Ensure (D)$ .

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 $Ensure (C & D) \supseteq Ensure (C) \cap Ensure (D)$ . Conversely, since $&$ is a categorical product, there exist projection morphisms from $C & D$ to $C$ and from $C & D$ to $D$ , so $Ensure (C & D) \subseteq Ensure (C) \cap Ensure (D)$ . $□$

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

Proof: Let $C = (A, E, \cdot)$ , 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 $2^{W}$ . $C \oplus D$ would also have empty environment and nonempty agent, and so $Ensure (C \oplus D)$ 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 $Ensure (C \oplus D) \supseteq Ensure (C) \cup Ensure (D)$ 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 $S \in Ensure (C \oplus D)$ , then there is a morphism $(g, h) : 1_{S} \to C \oplus D$ . Since $g : Agent (1_{S}) \to A ⊔ B$ , and $Agent (1_{S}) = {b^{'}}$ 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 $(g^{'}, h^{'}) : 1_{S} \to C$ given by $g^{'} (b^{'}) = g (b^{'})$ , and $h^{'} (e) = h (e, f)$ . This is a morphism because

\begin{matrix} b^{'} ⋄ h^{'} (e) & = b^{'} ⋄ h (e, f) = g (b^{'}) ∙ (e, f) = g^{'} (b^{'}) \cdot e, \end{matrix}

where $⋄ = Eval (1_{S})$ and $∙ = Eval (C \oplus D)$ . Thus, $S \in Ensure (C)$ . $□$

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 $C = ({a}, {e}, \cdot)$ , where $a \cdot e = w$ , and let $D = null$ . $Ensure (C) = ⟨ {w} ⟩_{\supseteq}$ and $Ensure (D) = {}$ . However, $C \oplus D = ({a}, {}, ⋆)$ , so $Ensure (C \oplus D) = 2^{W}$ , 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 [LW · GW], 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: $Ensure (C) \cap Prevent (C^{*}) = {}$ .

Proof: Trivial. $□$

1.2. Controllables

Controllables also have a simple categorical definition.

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

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 $W ∖ S$ .

$2_{S}$ 's environments are all possible pairs of exactly one $s \in S$ and exactly one $s \in W ∖ S$ . 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

$2_{S_{0}} = \begin{matrix} \begin{matrix} w_{0} w_{1} & w_{0} w_{2} & w_{3} w_{1} & w_{3} w_{2} & w_{4} w_{1} & w_{4} w_{2} \end{matrix} \begin{matrix} b_{S_{0}} b_{W ∖ S_{0}} \end{matrix} & (\begin{matrix} w_{0} & w_{0} & w_{3} & w_{3} & w_{4} & w_{4} w_{1} & w_{2} & w_{1} & w_{2} & w_{1} & w_{2} \end{matrix}) \end{matrix}$ .

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 $S \subseteq W$ such that there exists a morphism $ϕ : 2_{S} \to C$ .

Claim: This definition is equivalent to the one in "Introduction to Cartesian Frames [LW · GW]": $Ctrl (C) = Ensure (C) \cap Prevent (C)$ .

Proof: Since $\oplus$ is the categorical coproduct, there exists a morphism from $2_{S} = 1_{S} \oplus 1_{W ∖ S}$ to $C$ if and only if there exist a pair of morphisms $ϕ_{0} : 1_{S} \to C$ and $ϕ_{1} : 1_{W ∖ S} \to C$ , 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 $S \subseteq W$ 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 $ϕ : C \to D$ , then $Ctrl(C) \subseteq Ctrl (D)$ .

Proof: Trivial. $□$

Claim: $Ctrl (C & D) = Ctrl (C) \cap Ctrl (D)$ .

Proof: Trivial. $□$

Claim: If $C ≃ D$ , then $Ctrl (C) = Ctrl (D)$ .

Proof: Trivial. $□$

Note that although $Ensure (C \oplus D) = Ensure (C) \cup Ensure (D)$ 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 $S \subseteq W$ such that there exist $C_{0}$ and $C_{1}$ with $Image (C_{0}) \subseteq S$ and $Image (C_{1}) \subseteq W ∖ S$ such that $C ≃ C_{0} & C_{1}$ .

Claim: This definition is equivalent to the one in "Introduction to Cartesian Frames [LW · GW]": $Obs (C) = {S \subseteq W | \forall a_{0}, a_{1} \in A, \exists a \in A, a \in if (S, a_{0}, a_{1})}$ .

Proof: Throughout the proof, we will let ${Obs}_{o l d} (-)$ 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:

First, we will show that ${Obs}_{o l d} (-)$ is closed under biextensional equivalence.
Second, we will show that if $Image (C_{0}) \subseteq S$ and $Image (C_{1}) \subseteq W ∖ S$ , then $S \in {Obs}_{o l d} (C_{0} & C_{1})$ . In combination with the first part, this implies that $Obs (C) \subseteq {Obs}_{o l d} (C)$ .
Third, we will show that if $S \in {Obs}_{o l d} (A, E, \cdot)$ , then $(A, E, \cdot) ≃ (A, E_{S}, \cdot) & (A, E_{W ∖ S}, \cdot)$ .

This gives us that ${Obs}_{o l d} (C) \subseteq Obs (C)$ , by taking $C_{0} = (A, E_{S}, \cdot)$ and $C_{1} = (A, E_{W ∖ S}, \cdot)$ .

Part 1.

We want to show that ${Obs}_{o l d} (-)$ is closed under biextensional equivalence.

Let $C = (A, E, \cdot)$ , let $D = (B, F, ⋆)$ , and let $C ≃ D$ . We will use the homotopy equivalence definition, so let $(g_{0}, h_{0}) : C \to D$ and $(g_{1}, h_{1}) : D \to C$ 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 $S \in {Obs}_{o l d} (C)$ , we want to show that $S \in {Obs}_{o l d} (D)$ .

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 $b ⋆ f \in S$ , then $b ⋆ f = b_{0} ⋆ f$ , and otherwise $b ⋆ f = b_{1} ⋆ f$ . Letting $a_{i} = g_{1} (b_{i})$ , the fact that $S \in {Obs}_{o l d} (C)$ 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{matrix} b ⋆ f & = g_{0} (a) ⋆ f = a \cdot h_{0} (f) . \end{matrix}

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{matrix} b_{i} ⋆ f & = g_{0} (g_{1} (b_{i})) ⋆ f = g_{1} (b_{i}) \cdot h_{0} (f) = a_{i} \cdot h_{0} (f) . \end{matrix}

Together these give that if $b ⋆ f \in S$ , then $a \cdot h_{0} (f) \in S$ , so

\begin{matrix} b ⋆ f & = a \cdot h_{0} (f) = a_{0} \cdot h_{0} (f) = b_{0} ⋆ f, \end{matrix}

and if $b ⋆ f \notin S$ , then $a \cdot h_{0} (f) \notin S$ , so

\begin{matrix} b ⋆ f & = a \cdot h_{0} (f) = a_{1} \cdot h_{0} (f) = b_{1} ⋆ f . \end{matrix}

Thus $S \in {Obs}_{o l d} (D)$ , so ${Obs}_{o l d} (-)$ is closed under biextensional equivalence.

Part 2.

We want to show that if $Image (C_{0}) \subseteq S$ and $Image (C_{1}) \subseteq W ∖ S$ , then $S \in {Obs}_{o l d} (C_{0} & C_{1})$ .

Let $C_{0} = (A, E, \cdot)$ , let $C_{1} = (B, F, ⋆)$ , and let $C_{0} & C_{1} = (A \times B, E ⊔ F, ⋄)$ . 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 $(a, b) ⋄ e \in S$ , then $(a, b) ⋄ e = (a_{0}, b_{0}) ⋄ e$ , and otherwise $(a, b) ⋄ e = (a_{1}, b_{1}) ⋄ e$ . We will take $(a, b) = (a_{0}, b_{1})$ .

For all $f \in F$ , $(a_{0}, b_{1}) ⋄ f = b_{1} ⋆ f \notin S$ , since $Image (C_{1}) \subseteq W ∖ S$ . Thus, if $(a_{0}, b_{1}) ⋄ e \in S$ , then $e \in E$ , so

\begin{matrix} (a_{0}, b_{1}) ⋄ e & = a_{0} \cdot e = (a_{0}, b_{0}) ⋄ e . \end{matrix}

Similarly, if $(a_{0}, b_{1}) ⋄ e \notin S$ , then $e \in F$ , so

\begin{matrix} (a_{0}, b_{1}) ⋄ e & = b_{1} \cdot e = (a_{1}, b_{1}) ⋄ e . \end{matrix}

Thus, $S \in {Obs}_{o l d} (C_{0} & C_{1})$ .

Part 3.

We want to show that if $S \in {Obs}_{o l d} (A, E, \cdot)$ , then $(A, E, \cdot) ≃ (A, E_{S}, \cdot) & (A, E_{W ∖ S}, \cdot)$ .

Let $C = (A, E, \cdot)$ , let $C_{0} = (A, E_{S}, \cdot)$ , and let $C_{1} = (A, E_{W ∖ S}, \cdot)$ . (Here the $\cdot$ in $C_{0}$ and $C_{1}$ is the restriction of $\cdot$ in $C$ to the respective domain.) Let $⋆ = Eval (C_{0} & C_{1})$ .

First, let's quickly deal with the degenerate case where $A$ is empty. In this case $E_{S} = E_{W ∖ S} = E$ . If $E$ is also empty, then $C ≃ null ≃ null & null ≃ C_{0} & C_{1}$ . If $E$ is nonempty, then $C ≃ 0 ≃ 0 & 0 ≃ C_{0} & C_{1}$ . Thus, we can restrict our attention to the case where $A$ is nonempty. Note that in this case, $E_{S}$ and $E_{W ∖ S}$ are disjoint, and as we saw before they cover $E$ , so $E = E_{S} ⊔ E_{W ∖ S}$ .

We need to construct a $(g_{0}, h_{0}) : C \to C_{0} & C_{1}$ and a $(g_{1}, h_{1}) : C_{0} & C_{1} \to C$ , which compose to something homotopic to the identity in both orders. Since $E = E_{S} ⊔ E_{W ∖ S}$ , we can just take $h_{0}$ and $h_{1}$ to be the identity on $E$ . We will take $g_{0} : A \to A \times A$ to be the diagonal given by $g_{0} (a) = (a, a)$ . Finally, for $g_{1} : A \times A \to A$ , we will use the fact that $S \in {Obs}_{o l d} (C)$ . 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 ${Obs}_{o l d} (C)$ .

To see that $(g_{0}, h_{0})$ is a morphism, observe that for all $a \in A$ and $e \in E_{S} ⊔ E_{W ∖ S}$ , we have

\begin{matrix} g_{0} (a) ⋆ e & = (a, a) ⋆ e = a \cdot e = a \cdot h_{0} (e), \end{matrix}

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{matrix} g_{1} (a_{0}, a_{1}) \cdot e & = a_{0} \cdot e = (a_{0}, a_{1}) ⋆ e = (a_{0}, a_{1}) ⋆ h_{1} (e), \end{matrix}

while if $e \in E_{W ∖ S}$ , then

\begin{matrix} g_{1} (a_{0}, a_{1}) \cdot e & = a_{1} \cdot e = (a_{0}, a_{1}) ⋆ e = (a_{0}, a_{1}) ⋆ h_{1} (e) . \end{matrix}

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 $(a_{0}, a_{1}) ⋆ e = (a_{0}, a_{1}) ⋆ h_{1} (h_{0} (e))$ . (Technically, this is verifying that the identity is homotopic each composition, but since being homotopic is symmetric, this is fine.)

Putting it together.

If $S \in Obs (C)$ , then $C ≃ C_{0} & C_{1}$ , where $Image (C_{0}) \subseteq S$ and $Image (C_{1}) \subseteq W ∖ S$ . By part 2, $S \in {Obs}_{o l d} (C_{0} & C_{1})$ , which by part 1, means that $S \in {Obs}_{o l d} (C)$ . Conversely, if $S \in {Obs}_{o l d} (A, E, \cdot)$ , then by part 3, $(A, E, \cdot) ≃ (A, E_{S}, \cdot) & (A, E_{W ∖ S}, \cdot)$ , which since $Image (A, E_{S}, \cdot) \subseteq S$ and $Image (A, E_{W ∖ S}, \cdot) \subseteq W ∖ S$ , implies that $S \in Obs (A, E, \cdot)$ . $□$

Note that from the above proof, if $S \in Obs (C)$ , we know how to construct the $C_{0}$ and $C_{1}$ such that $C ≃ C_{0} & C_{1}$ . 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 $C ≃ D$ , 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 [LW · GW] from before. First, a lemma.

Lemma: Let $C ≃ C_{0} & C_{1}$ , with $Image (C_{0}) \subseteq S$ and $Image (C_{1}) \subseteq W ∖ S$ . If $S \in Ensure (C),$ then $C_{1} ≃ ⊤$ . If $S \in Prevent (C)$ , then $C_{0} ≃ ⊤$ .

Proof: If $S \in Ensure (C)$ , 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 $(g, h) : 1_{S} \to C_{1}$ . Let $C_{1} = (A, E, \cdot)$ , and let $1_{S} = ({b}, S, ⋆)$ . If there were an $e \in E$ , then $g (b) \cdot e = b ⋆ h (e)$ would be in both $S$ and $W ∖ S$ , a contradiction. Therefore $C_{1}$ has empty environment. Also, since $g (b) \in A$ , $C_{1}$ has nonempty agent. Therefore $C_{1} ≃ ⊤$ .

Symmetrically, if $S \in Prevent (C)$ , then $W ∖ S \in Ensure (C)$ , so $C_{0} ≃ ⊤$ . $□$

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

Theorem: If $C ≄ ⊤$ , then $Ctrl (C) \cap Obs (C) = {}$ .

Proof: We prove the contrapositive. Assume $S \in Ctrl (C) \cap Obs (C)$ . Let $C ≃ C_{0} & C_{1}$ , with $Image (C_{0}) \subseteq S$ and $Image (C_{1}) \subseteq W ∖ S$ . By the above lemma, both $C_{0} ≃ ⊤$ and $C_{1} ≃ ⊤$ . Thus $C ≃ ⊤ & ⊤ ≅ ⊤$ . $□$

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

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

Proof: If $S \in Ensure (C) \cap Obs (C)$ , then $C ≃ C_{0} & C_{1}$ , with $Image (C_{0}) \subseteq S$ and $Image (C_{1}) \subseteq W ∖ S$ . By the above lemma, $C_{1} ≃ ⊤$ , so $C ≃ C_{0} & ⊤ ≅ C_{0}$ , so $Image (C) \subseteq S .$ $□$

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.

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comment by Ramana Kumar (ramana-kumar) · 2020-12-03T20:22:48.580Z · LW(p) · GW(p)

Counterexample to your claim that (with both $C$ and $D$ not $n u l l$ ):

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:

\begin{matrix} a & b a & y & x b & x & x \end{matrix}

\begin{matrix} a & b a & x & y b & y & y \end{matrix}

Then $C t r l (C) = C t r l (D) = \emptyset$ but $C t r l (C \oplus D) ∋ {x}$ .

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

Replies from: Scott Garrabrant

↑ comment by Scott Garrabrant · 2020-12-04T01:10:58.290Z · LW(p) · GW(p)

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.

comment by Rob Bensinger (RobbBB) · 2020-10-29T20:16:20.307Z · LW(p) · GW(p)

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 $C_{0}$ :

$C_{0}$

$C_{0}^{*} = \begin{matrix} \begin{matrix} e_{0} & e_{1} \end{matrix} \begin{matrix} a_{0} a_{1} \end{matrix} & (\begin{matrix} w_{0} & w_{2} w_{1} & w_{3} \end{matrix}) \end{matrix}$

$⊥_{{}} = 0 = \begin{matrix} \begin{matrix} e_{0} \end{matrix} \begin{matrix} \end{matrix} & (\begin{matrix} \end{matrix}) \end{matrix}$

$1_{{w_{0}, w_{1}}} = \begin{matrix} \begin{matrix} e_{0} & e_{1} \end{matrix} \begin{matrix} a_{0} \end{matrix} & (\begin{matrix} w_{0} & w_{1} \end{matrix}) \end{matrix}$

$1_{{w_{2}, w_{3}}} = \begin{matrix} \begin{matrix} e_{0} & e_{1} \end{matrix} \begin{matrix} a_{0} \end{matrix} & (\begin{matrix} w_{2} & w_{3} \end{matrix}) \end{matrix}$

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 $1_{S}$ frames corresponding to the other five ensurables that can be deduced from the matrix: $1_{{w_{0}, w_{1}, w_{2}}}$ , $1_{{w_{0}, w_{1}, w_{3}}}$ , $1_{{w_{0}, w_{2}, w_{3}}}$ , $1_{{w_{1}, w_{2}, w_{3}}}$ , $1_{{w_{0}, w_{1}, w_{2}, w_{3}}}$ . If $W$ has more than four elements, then there will be additional $1_{S}$ frames / additional ensurables beyond these seven.)

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

Replies from: Scott Garrabrant, Scott Garrabrant

↑ comment by Scott Garrabrant · 2020-10-29T20:31:53.087Z · LW(p) · GW(p)

is wrong. You can see it has Ensurables that $C_{0}$ does not have.

↑ comment by Scott Garrabrant · 2020-10-29T20:39:58.151Z · LW(p) · GW(p)

You can also duplicate rows in , and then add columns, so you can get things like $⎛ ⎜ ⎝ \begin{matrix} w_{0} & w_{1} & w_{2} w_{0} & w_{1} & w_{3} w_{2} & w_{3} & w_{0} \end{matrix} ⎞ ⎟ ⎠$ . There are infinitely many biextensional Cartesian frames over ${w_{0}, w_{1}, w_{2}, w_{3}}$ with morphism to $C_{0}$ , with arbitrarily large dimensions.

Controllables and Observables, Revisited

Contents

1. Controllables

2. Observables

3. Controllables and Observables Are Still Disjoint

5 comments