Keith wrote in comment #1:

> \\(\texttt{false}\\) maps to everything less than $500

I wrote:

> [...] please clarify what you mean by saying

> > \\(\texttt{false}\\) maps to everything less than $500

> You're making it sound like a feasibility relation is a 'multi-valued function' where \\(\texttt{false}\\) can map to lots of different things. That's a really cool way of talking, which we may be able to make sense of if we work a little - but that's not what we defined a feasibility relation to be.

Christopher wrote approximately:

> a relation is the same thing as a multivalued function (where 'multi' includes zero).

Right: that's the clarification I was wanting from Keith. We can think of relations in at least 3 ways:

1. A relation \\(R: X \nrightarrow Y \\) between sets is a subset \\( R \subseteq X \times Y \\).

2. A relation \\(R: X \nrightarrow Y \\) between sets is a function

\\( R \colon X \times Y \to \textbf{Bool} \\).

3. A relation \\(R: X \nrightarrow Y \\) between sets is a function

\\( R \colon X \to P(Y) \\) where \\(P(Y)\\) is the power set of \\(Y\\).

We've been using the first two outlooks a lot, but the third lets us think of a relation as a multivalued function from \\(X\\) to \\(Y\\), i.e. a function that maps each element of \\(X\\) to a _subset_ of \\(Y \\). This is the outlook that Keith was taking!

We can get from outlook 2 to outlook 3 as follows.

Functions \\(f: A \times B \to C\\) correspond in a one-to-one way with functions \\(\hat{f} : A \to C^B\\), where \\(C^B\\) is the set of functions from \\(B\\) to \\(C\\). Computer scientists call this correspondence **currying**. It goes like this:

\[ \hat{f}(a)(b) = f(a,b) .\]

So, functions

\[ f: X \times Y \to \textbf{Bool} \]

correspond in a one-to-one way with functions

\[ \hat{f} : X \to \textbf{Bool}^Y .\]

But \\(\textbf{Bool}^Y\\) is just the power set of \\(Y \\)!

It's interesting to see how this plays out when we think about _feasibility_ relations.

**Puzzle.** Suppose \\(X\\) and \\(Y\\) are preorders. Can we reinterpret a feasibility relation, namely a monotone function

\[ \Phi : X^{\text{op}} \times Y \to \textbf{Bool} , \]

as a monotone function

\[ \hat{\Phi} : X^{\text{op}} \to \textbf{Bool}^{\textbf{X}} \]

or perhaps slightly better

\[ \tilde{\Phi} : Y \to \textbf{Bool}^{\textbf{X}^{\text{op}}} ? \]

The first problem here is figuring out what's an exponential of preorders!

> \\(\texttt{false}\\) maps to everything less than $500

I wrote:

> [...] please clarify what you mean by saying

> > \\(\texttt{false}\\) maps to everything less than $500

> You're making it sound like a feasibility relation is a 'multi-valued function' where \\(\texttt{false}\\) can map to lots of different things. That's a really cool way of talking, which we may be able to make sense of if we work a little - but that's not what we defined a feasibility relation to be.

Christopher wrote approximately:

> a relation is the same thing as a multivalued function (where 'multi' includes zero).

Right: that's the clarification I was wanting from Keith. We can think of relations in at least 3 ways:

1. A relation \\(R: X \nrightarrow Y \\) between sets is a subset \\( R \subseteq X \times Y \\).

2. A relation \\(R: X \nrightarrow Y \\) between sets is a function

\\( R \colon X \times Y \to \textbf{Bool} \\).

3. A relation \\(R: X \nrightarrow Y \\) between sets is a function

\\( R \colon X \to P(Y) \\) where \\(P(Y)\\) is the power set of \\(Y\\).

We've been using the first two outlooks a lot, but the third lets us think of a relation as a multivalued function from \\(X\\) to \\(Y\\), i.e. a function that maps each element of \\(X\\) to a _subset_ of \\(Y \\). This is the outlook that Keith was taking!

We can get from outlook 2 to outlook 3 as follows.

Functions \\(f: A \times B \to C\\) correspond in a one-to-one way with functions \\(\hat{f} : A \to C^B\\), where \\(C^B\\) is the set of functions from \\(B\\) to \\(C\\). Computer scientists call this correspondence **currying**. It goes like this:

\[ \hat{f}(a)(b) = f(a,b) .\]

So, functions

\[ f: X \times Y \to \textbf{Bool} \]

correspond in a one-to-one way with functions

\[ \hat{f} : X \to \textbf{Bool}^Y .\]

But \\(\textbf{Bool}^Y\\) is just the power set of \\(Y \\)!

It's interesting to see how this plays out when we think about _feasibility_ relations.

**Puzzle.** Suppose \\(X\\) and \\(Y\\) are preorders. Can we reinterpret a feasibility relation, namely a monotone function

\[ \Phi : X^{\text{op}} \times Y \to \textbf{Bool} , \]

as a monotone function

\[ \hat{\Phi} : X^{\text{op}} \to \textbf{Bool}^{\textbf{X}} \]

or perhaps slightly better

\[ \tilde{\Phi} : Y \to \textbf{Bool}^{\textbf{X}^{\text{op}}} ? \]

The first problem here is figuring out what's an exponential of preorders!