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## 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>`