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 719

Options

## Comments

Is this what is intended?

`Is this what is intended? ![Hasse](https://docs.google.com/drawings/d/e/2PACX-1vTCgcacsG5ZRLBZqU1ZDLiM4Swx0ifCHDRRLS_GR8GN-jplFAjS_sAcNuyMrytKFgwqQplizbFjqHew/pub?w=746&h=375)`

Fredrick: This is my understanding of the question and how I have the answer written in my notebook. It seems like it would be the same Hasse diagram as for the powerset of a set of 4 elements. Since with \(S=\{1,2\}\), \(S\times S=\{(1,1),(1,2),(2,1),(2,2)\}\).

`Fredrick: This is my understanding of the question and how I have the answer written in my notebook. It seems like it would be the same Hasse diagram as for the powerset of a set of 4 elements. Since with \\(S=\\{1,2\\}\\), \\(S\times S=\\{(1,1),(1,2),(2,1),(2,2)\\}\\).`

@JaredSummers I guess the point was noticing that \( S \times S \) forms a set of 4 elements. I suppose it makes sense in the context of the subsequent exercise.

`@JaredSummers I guess the point was noticing that \\( S \times S \\) forms a set of 4 elements. I suppose it makes sense in the context of the subsequent exercise.`

Great picture, Fredrick! Yes, a binary relation on \(S\) is the same as a subset of \(S \times S\), so the poset of binary relations on \(S \) is just our friend the power set \( P(S \times S\). So, it looks like an \(n^2\)-dimensional cube if \(S\) has \(n\) elements. You're taking \(n = 2\) so your picture looks a lot like this:

`Great picture, Fredrick! Yes, a binary relation on \\(S\\) is the same as a subset of \\(S \times S\\), so the poset of binary relations on \\(S \\) is just our friend the power set \\( P(S \times S\\). So, it looks like an \\(n^2\\)-dimensional cube if \\(S\\) has \\(n\\) elements. You're taking \\(n = 2\\) so your picture looks a lot like this: <center><img src = "http://math.ucr.edu/home/baez/mathematical/7_sketches/P4_hasse_diagram.png"></center>`