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Lecture 7 - Chapter 1: Logic

So far the only examples of posets I've talked about in the lectures are the real numbers \(\mathbb{R}\) and the natural numbers \(\mathbb{N}\) with their usual order \(\le\). Of course every natural number is a real number, so there's a function

$$ i : \mathbb{N} \to \mathbb{R} $$ sending any natural number \(x \in \mathbb{N}\) to the exact same number regarded as a real number. This function is monotone, so you now know instinctively to ask this question:

Puzzle 21. Does the monotone function \(i : \mathbb{N} \to \mathbb{R}\) have a left adjoint? Does it have a right adjoint? If so, what are they?

This is nice, but we need to look at other examples to appreciate the diversity of posets. Both \(\mathbb{N}\) and \(\mathbb{Z}\) have a very special property. They are totally ordered sets: posets such that

$$ \textrm{ for all } x, y, \textrm{ either } x \le y \textrm{ or } y \le x . $$ If you want to show off, you can call totally ordered sets tosets. They're also called linearly ordered, because you can imagine them as lines:

image

Totally ordered sets are limiting. Suppose you're trying to order foods on a restaurant menu based on how much you like them. What's better: a cheese sandwich or a pancake? There may be no answer, because you like them in different ways. To get a totally ordered set you have to ignore this and arrange all the foods in a line.

In standard economics we do try to arrange everything in a line. We measure the worth of everything in real numbers: numbers of dollars. There's even a theorem to justify this, proved by von Neumann and Morgenstern. But the assumptions of this theorem don't hold in real life. It's mainly just convenient to measure value, or "utility", in real numbers. With computer technologies we could set up cryptocurrencies based on other posets. But will we?

Luckily, human thought as a whole is not limited to total orders. A good example is logic. Logic, in its simplest form, is about statements \(P, Q, R, \dots \) and whether one statement implies another. If \(P\) implies \(Q\) we often write \(P \implies Q\). There are many kinds of logic, but every kind I know, this relation \(\implies\) makes statements into a preorder, since we have

1) reflexivity: \(P \implies P\)

2) transitivity: if \(P \implies Q\) and \( Q \implies R \) then \(P \implies R\).

Often people make this preorder into a poset by imposing this rule:

3) antisymmetry: if \(P \implies Q\) and \(Q \implies P\) then \(P = Q \).

This amounts to decreeing that we count two statements as "the same" if they both imply each other. We may not always want to do this. And we certainly don't want a linear order: it's easy to find examples of statements such that neither \( P \implies Q\) nor \(Q \implies P\), like "I am a millionaire" and "I am happy", or "I like this food for breakfast" and "I like this food for lunch".

So, to continue our study of preorders, posets, monotone functions and Galois connections, we'll turn to logic! Category-theoretic logic is an enormous wonderful field, but we'll just do a bit of logic based on the poset of subsets of a set, followed by a bit of logic based on the poset of partitions of a set. The latter underlies Fong and Spivak's discussion of "generative effects" in Chapter 1.

To read other lectures go here.

Comments

  • 1.
    edited April 4

    Regarding Puzzle 21, if we consider \(\mathbb{N} = \{0, 1, 2, \ldots\}\), then we can construct a left-adjoint function \(f : \mathbb{R}\to\mathbb{N}\) to our \(i : \mathbb{N}\to\mathbb{R}\) such that

    $$f = \begin{cases} 0 & \text{if } x\le0\\ \lceil x \rceil & \text{if } x>0 \end{cases}$$ We see that our function \(f\) is monotone since if \(a \le b\) then \(f(a) \le f(b)\). Then, we can check that if \(x \le i(y)\), then \(f(x) \le y\), which is the definition of a left-adjoint.

    Comment Source:Regarding Puzzle 21, if we consider \\(\mathbb{N} = \\{0, 1, 2, \ldots\\}\\), then we can construct a left-adjoint function \\(f : \mathbb{R}\to\mathbb{N}\\) to our \\(i : \mathbb{N}\to\mathbb{R}\\) such that $$f = \begin{cases} 0 & \text{if } x\le0\\\\ \lceil x \rceil & \text{if } x>0 \end{cases}$$ We see that our function \\(f\\) is monotone since if \\(a \le b\\) then \\(f(a) \le f(b)\\). Then, we can check that if \\(x \le i(y)\\), then \\(f(x) \le y\\), which is the definition of a left-adjoint.
  • 2.

    An aside on the topic of partial orders in currency: perhaps we can take inspiration to what humans did in the past when (totally-ordered) bullion money was scarce but still needed to do commerce: we often relied on recording transactions with others through tally sticks and other forms of credit. You could even make it effectively tamper-proof by recording transactions in which two halves of a stick, forging the exact locations of notches natural grain of the split stick would have been practically impossible. Debt in this form could then be traded as currency and someone to try to collect on that debt.

    This is partially ordered because a mark on one set of sticks is not necessarily fungible or comparable with another: one person's debt might not be perceived as being able to pay back their debt, so in exchange, it might not be perceived worth the nominal value.

    Comment Source:An aside on the topic of partial orders in currency: perhaps we can take inspiration to what humans did in the past when (totally-ordered) bullion money was scarce but still needed to do commerce: we often relied on recording transactions with others through [tally sticks](https://en.wikipedia.org/wiki/Tally_stick) and other forms of credit. You could even make it effectively tamper-proof by recording transactions in which two halves of a stick, forging the exact locations of notches natural grain of the split stick would have been practically impossible. Debt in this form [could then be traded as currency](http://www.bbc.com/news/business-40189959) and someone to try to collect on that debt. This is partially ordered because a mark on one set of sticks is not necessarily fungible or comparable with another: one person's debt might not be perceived as being able to pay back their debt, so in exchange, it might not be perceived worth the nominal value.
  • 3.
    edited April 10

    Continuing with Yakov's comment.

    We can construct a left-adjoint function \( L : \mathbb{R}\to\mathbb{N} \) to our \( I : \mathbb{N}\to\mathbb{R}\) such that

    $$L = \begin{cases} 0 & \text{if } x\le0\\ \lceil x \rceil & \text{if } x>0 \end{cases}$$ Can we construct a right-adjoint function \( R : \mathbb{R}\to\mathbb{N} \) to our \( I : \mathbb{N}\to\mathbb{R}\) such that

    $$R = \begin{cases} ? & \text{if } x\lt0\\ \lfloor x \rfloor & \text{if } x \ge0 \end{cases}$$ ...what do we do when \( x \lt0 \)?

    Adjoints

    Comment Source:Continuing with Yakov's comment. We can construct a left-adjoint function \\( L : \mathbb{R}\to\mathbb{N} \\) to our \\( I : \mathbb{N}\to\mathbb{R}\\) such that $$L = \begin{cases} 0 & \text{if } x\le0\\\\ \lceil x \rceil & \text{if } x>0 \end{cases}$$ Can we construct a right-adjoint function \\( R : \mathbb{R}\to\mathbb{N} \\) to our \\( I : \mathbb{N}\to\mathbb{R}\\) such that $$R = \begin{cases} ? & \text{if } x\lt0\\\\ \lfloor x \rfloor & \text{if } x \ge0 \end{cases}$$ ...what do we do when \\( x \lt0 \\)? ![Adjoints](https://docs.google.com/drawings/d/e/2PACX-1vRD1FFfwQ4qGDkT8XVX4tjcQx3XlPewnc1_UxMpHJIQCXdzv8lneYvt5YToniHrKnD2tIMhfwQfdcCY/pub?w=754&h=188)
  • 4.

    Brian Cohen wrote:

    An aside on the topic of partial orders in currency: perhaps we can take inspiration to what humans did in the past when (totally-ordered) bullion money was scarce but still needed to do commerce: we often relied on recording transactions with others through tally sticks and other forms of credit. You could even make it effectively tamper-proof by recording transactions in which two halves of a stick, forging the exact locations of notches natural grain of the split stick would have been practically impossible. Debt in this form could then be traded as currency and someone to try to collect on that debt.

    This is partially ordered because a mark on one set of sticks is not necessarily fungible or comparable with another: one person's debt might not be perceived as being able to pay back their debt, so in exchange, it might not be perceived worth the nominal value.

    What you're describing is Double-Entry Bookkeeping. David Ellerman (the same author from the Partition Logic paper) also made a paper giving a mathematical treatment of double-entry bookkeeping: On Double-Entry Bookkeeping: The Mathematical Treatment.

    Fredrick Eisele wrote:

    ...what do we do when x<0?

    Why not simply map \(x \in \mathbb{R} \) to \( 0 \in \mathbb{N} \)?

    Comment Source:Brian Cohen wrote: >An aside on the topic of partial orders in currency: perhaps we can take inspiration to what humans did in the past when (totally-ordered) bullion money was scarce but still needed to do commerce: we often relied on recording transactions with others through tally sticks and other forms of credit. You could even make it effectively tamper-proof by recording transactions in which two halves of a stick, forging the exact locations of notches natural grain of the split stick would have been practically impossible. Debt in this form could then be traded as currency and someone to try to collect on that debt. >This is partially ordered because a mark on one set of sticks is not necessarily fungible or comparable with another: one person's debt might not be perceived as being able to pay back their debt, so in exchange, it might not be perceived worth the nominal value. What you're describing is Double-Entry Bookkeeping. David Ellerman (the same author from the [Partition Logic](https://arxiv.org/abs/0902.1950) paper) also made a paper giving a mathematical treatment of double-entry bookkeeping: [On Double-Entry Bookkeeping: The Mathematical Treatment](https://arxiv.org/abs/1407.1898). Fredrick Eisele wrote: >...what do we do when x<0? Why not simply map \\(x \in \mathbb{R} \\) to \\( 0 \in \mathbb{N} \\)?
  • 5.
    edited April 11

    Fredrick wrote:

    ...what do we do when \( x \lt 0 \)?

    We have a formula for the right adjoint if it exists: we saw it near the end of Lecture 6. So, we can use this to figure out what \(R(x)\) must be if the right adjoint exists... and if the formula gives an undefined result, we know the right adjoint cannot exist.

    Another approach is to use Proposition 1.81 in Seven Sketches. Applied to our example, this implies that if \(R : \mathbb{R} \to \mathbb{N}\) is a right adjoint to \(I : \mathbb{N} \to \mathbb{R} \), we must have

    $$ I(R(x)) \le x $$ for all \(x \in \mathbb{R}\). See what this means?

    Comment Source:Fredrick wrote: > ...what do we do when \\( x \lt 0 \\)? We have a formula for the right adjoint if it exists: we saw it near the end of [Lecture 6](https://forum.azimuthproject.org/discussion/1901/lecture-6-chapter-1-computing-adjoints/p1). So, we can use this to figure out what \\(R(x)\\) must be if the right adjoint exists... and if the formula gives an undefined result, we know the right adjoint cannot exist. Another approach is to use Proposition 1.81 in _[Seven Sketches](http://math.mit.edu/~dspivak/teaching/sp18/7Sketches.pdf)_. Applied to our example, this implies that if \\(R : \mathbb{R} \to \mathbb{N}\\) is a right adjoint to \\(I : \mathbb{N} \to \mathbb{R} \\), we must have $$ I(R(x)) \le x $$ for all \\(x \in \mathbb{R}\\). See what this means?
  • 6.
    edited April 11

    \(R = 0 \text{ if } x \lt0 \) gives \( I(R(-0.2)) = 0.0 \nleq -0.2 \) meaning there is no right adjoint. Simply mapping negative values to 0 is a good mapping, it is called the left adjoint. There is a right adjoint for \( I : \mathbb{N} \rightarrow \mathbb{R}^+ \) though. Another, visual, way to think about \( I(R(x)) \le x \) is that red arrows cannot bend to the right [what I was trying to say with the picture].

    Comment Source:\\(R = 0 \text{ if } x \lt0 \\) gives \\( I(R(-0.2)) = 0.0 \nleq -0.2 \\) meaning there is no right adjoint. Simply mapping negative values to 0 is a good mapping, it is called the left adjoint. There is a right adjoint for \\( I : \mathbb{N} \rightarrow \mathbb{R}^+ \\) though. Another, visual, way to think about \\( I(R(x)) \le x \\) is that red arrows cannot bend to the right [what I was trying to say with the picture].
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