It looks like you're new here. If you want to get involved, click one of these buttons!

- All Categories 2.4K
- Chat 505
- Study Groups 21
- Petri Nets 9
- Epidemiology 4
- Leaf Modeling 2
- Review Sections 9
- MIT 2020: Programming with Categories 51
- MIT 2020: Lectures 20
- MIT 2020: Exercises 25
- Baez ACT 2019: Online Course 339
- Baez ACT 2019: Lectures 79
- Baez ACT 2019: Exercises 149
- Baez ACT 2019: Chat 50
- UCR ACT Seminar 4
- General 75
- Azimuth Code Project 111
- Statistical methods 4
- Drafts 10
- Math Syntax Demos 15
- Wiki - Latest Changes 3
- Strategy 113
- Azimuth Project 1.1K
- - Spam 1
- News and Information 148
- Azimuth Blog 149
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 718

Options

Since \( \textbf{Cost} \) is a symmetric monoidal preorder, Proposition 2.35 says that \( \textbf{Cost}^{op} \) is too. $$ \textbf{Cost} = ( [ 0, \infty ], \ge, 0, + ) $$

- What is \( \textbf{Cost}^{op} \) as a preorder?
- What is its monoidal unit?
- What is its monoidal product?

## Comments

Proposition 2.20 implies that $$ \textbf{Cost}^{op} = ( [ 0, \infty ], \le, 0, + ) $$ is a perfectly good symmetric monoidal preorder.

Using \( 0 \) for the monoidal unit, and \( + \) as the monoidal product seem fine.

Thinking of cost as a negative value \( [ -\infty, 0 ] \) gives an alternate \(op\) mapping that I will call \(neg\).

An alternate

Proposition : Preorder NegationSuppose \( \mathcal{X} = (X, \le) \) is a preorder and \( \mathcal{X}^{neg} = (X^{neg}, \le) \) is an opposite. That is to say, all the arrows are reversed but the meaning of order function is retained. If \( (X, ≤, I, \otimes) \) is a symmetric monoidal preorder then so is its negation, \( (X^{neg}, \le, I, \otimes) \). As an example, consider the following. $$ \textbf{Cost}_{neg} = ( [ -\infty, 0 ], \ge, 0, + ) $$$$ \textbf{X} = \textbf{Cost}^{op} = ( [ 0, \infty ], \le, 0, + ) $$ $$ \textbf{X}^{op} = \textbf{Cost} = ( [ 0, \infty ], \ge, 0, + ) $$ Using the negative real numbers augmented with \( - \infty \).

$$ \textbf{X}^{neg} = \textbf{Cost}^{neg^{op}} = ( [ - \infty, 0 ], \ge, 0, + ) $$ $$ \textbf{X}^{neg^{op}} = \textbf{Cost}^{neg} = ( [ - \infty, 0 ], \le, 0, + ) $$ Does the \(neg\) already have a name?

`Proposition 2.20 implies that $$ \textbf{Cost}^{op} = ( [ 0, \infty ], \le, 0, + ) $$ is a perfectly good symmetric monoidal preorder. Using \\( 0 \\) for the monoidal unit, and \\( + \\) as the monoidal product seem fine. Thinking of cost as a negative value \\( [ -\infty, 0 ] \\) gives an alternate \\(op\\) mapping that I will call \\(neg\\). An alternate **Proposition : Preorder Negation** Suppose \\( \mathcal{X} = (X, \le) \\) is a preorder and \\( \mathcal{X}^{neg} = (X^{neg}, \le) \\) is an opposite. That is to say, all the arrows are reversed but the meaning of order function is retained. If \\( (X, ≤, I, \otimes) \\) is a symmetric monoidal preorder then so is its negation, \\( (X^{neg}, \le, I, \otimes) \\). As an example, consider the following. $$ \textbf{Cost}_{neg} = ( [ -\infty, 0 ], \ge, 0, + ) $$ ![Does this generalize?](https://docs.google.com/drawings/d/e/2PACX-1vSaZS_iF5odXxpSMFNiPtH58VMEAgYuLuXV5JMT4dOwc01ZplZW1rU0oZ95wLFTlhxXDU8l_nqPR2V4/pub?w=327&h=204) $$ \textbf{X} = \textbf{Cost}^{op} = ( [ 0, \infty ], \le, 0, + ) $$ $$ \textbf{X}^{op} = \textbf{Cost} = ( [ 0, \infty ], \ge, 0, + ) $$ Using the negative real numbers augmented with \\( - \infty \\). $$ \textbf{X}^{neg} = \textbf{Cost}^{neg^{op}} = ( [ - \infty, 0 ], \ge, 0, + ) $$ $$ \textbf{X}^{neg^{op}} = \textbf{Cost}^{neg} = ( [ - \infty, 0 ], \le, 0, + ) $$ Does the \\(neg\\) already have a name?`

@FredrikEisele : why do you need the negative numbers? Also, the first question ask “what is this as a preorder?”; I understood this to require an interpretation of what \(Cost^{op}\) is, what it can represent - and I cannot figure this out: to me it seems just the same thing as \(Cost\) except that it tells you which things are cheaper rather than which ones are more expensive. Any hint? Thanks!

`@FredrikEisele : why do you need the negative numbers? Also, the first question ask “what is this as a preorder?”; I understood this to require an interpretation of what \\(Cost^{op}\\) is, what it can represent - and I cannot figure this out: to me it seems just the same thing as \\(Cost\\) except that it tells you which things are cheaper rather than which ones are more expensive. Any hint? Thanks!`

@ValterSorana You do not need negative numbers; I wanted to point out a concept related to an alternative opposite. The alternative invokes the 'cost' set as a negative value. I believe your interpretation of it as a simple change in perspective between 'cheaper' vs. 'more expensive' is a correct interpretation.

`@ValterSorana You do not need negative numbers; I wanted to point out a concept related to an alternative opposite. The alternative invokes the 'cost' set as a negative value. I believe your interpretation of it as a simple change in perspective between 'cheaper' vs. 'more expensive' is a correct interpretation.`

Got it. Thanks @FredrickEisele !

`Got it. Thanks @FredrickEisele !`