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

- All Categories 2.2K
- Applied Category Theory Course 353
- Applied Category Theory Seminar 4
- Exercises 149
- Discussion Groups 49
- How to Use MathJax 15
- Chat 480
- Azimuth Code Project 108
- News and Information 145
- Azimuth Blog 149
- Azimuth Forum 29
- Azimuth Project 189
- - Strategy 108
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 711
- - Latest Changes 701
- - - Action 14
- - - Biodiversity 8
- - - Books 2
- - - Carbon 9
- - - Computational methods 38
- - - Climate 53
- - - Earth science 23
- - - Ecology 43
- - - Energy 29
- - - Experiments 30
- - - Geoengineering 0
- - - Mathematical methods 69
- - - Meta 9
- - - Methodology 16
- - - Natural resources 7
- - - Oceans 4
- - - Organizations 34
- - - People 6
- - - Publishing 4
- - - Reports 3
- - - Software 21
- - - Statistical methods 2
- - - Sustainability 4
- - - Things to do 2
- - - Visualisation 1
- General 39

Options

Let \(S=\{1,2,3\}\).

- Come up with any preorder relation \(\leq\) on \(S\), and let \(L\subseteq S\times S\) be the set \(\{(s_1,s_2)|s_1\leq s_2\}\), i.e. \(L\) is the image of \(\leq\) under the inclusion \(\mathbf{Pos}(S)\rightarrow\mathbf{Rel}(S)\).
- Come up with two binary relations \(Q\subseteq S\times S\) and \(Q'\subseteq S\times S\) such that \(L\leq Q\) but \(L\nleq Q'\).
- Show that \( \leq\trianglelefteq \text{Cl}(Q) \).
- Show that \( \leq\ntrianglelefteq \text{Cl}(Q')\).

The following is my answer:

- \( L=\{(1,1),(1,2),(2,2),(3,3)\}\)
- \( Q=\{(1,1),(1,2),(2,2),(3,3),(2,3)\} \) and \( Q'=\{(1,1),(1,2),(2,3),(3,3)\} \)

For question 2, I interpreted \( \leq L \) to mean 'subset of \( L \)'. While \( Q'\) contains some elements of \( L\), one of its elements is not in \( L\), i.e. (2,3), which means \( L\) is not a subset of \( Q'\). We can check that \( \text{Cl}(Q) =\text{Cl}(Q')\) and \( L\) is a subset of both \(\text{Cl}(Q)\) and \(\text{Cl}(Q')\). Thus,

- \(\leq\trianglelefteq\text{Cl}(Q)\)
- \(\leq\trianglelefteq\text{Cl}(Q')\) which contradicts what question 4 is asking.

The statement of question 4 seems to imply some general statement on relationship between the sets containing \( L \) and the order of their closures. So, am I misinterpreting some definition in section 1.5.5? Or have I missed out some other details?

## Comments

I think this is a typo in the book - I submitted it to the mistakes list a couple days ago.

The preceding paragraph should say "[The inclusion \(\mathbf{Pos}(S)\rightarrow\mathbf{Rel}(S)\)] is actually the right adjoint of a Galois connection. Its

leftadjoint is a monotone map \(\text{Cl}:\mathbf{Rel}(S)\rightarrow\mathbf{Pos}(S)\)."and the exercise should ask you to pick \(Q\leq L\) and \(Q'\nleq L\), and then show that \(\text{Cl}(Q)\trianglelefteq\leq\) and \(\text{Cl}(Q')\ntrianglelefteq\leq\).

`I think this is a typo in the book - I submitted it to the mistakes list a couple days ago. The preceding paragraph should say "[The inclusion \\(\mathbf{Pos}(S)\rightarrow\mathbf{Rel}(S)\\)] is actually the right adjoint of a Galois connection. Its **left** adjoint is a monotone map \\(\text{Cl}:\mathbf{Rel}(S)\rightarrow\mathbf{Pos}(S)\\)." and the exercise should ask you to pick \\(Q\leq L\\) and \\(Q'\nleq L\\), and then show that \\(\text{Cl}(Q)\trianglelefteq\leq\\) and \\(\text{Cl}(Q')\ntrianglelefteq\leq\\).`

Yes, that makes more sense. I thought there was something weird about that paragraph but could not put my finger on it. Thanks, Thomas.

`Yes, that makes more sense. I thought there was something weird about that paragraph but could not put my finger on it. Thanks, Thomas.`

In comment 1 Aqilah proposes a solution:

I want to point out to the casual reader that the \( \le \) relation here defined is not its conventional meaning over \( \mathbb{N} \), it can be defined as any set of pairs from \( S \times S \). That conventional meaning would define the \( \le \) relation as \( \{(1,1),(1,2),(2,2),(2,3),(3,3)\} \) [incidentally chosen to be \( Q \)]. This is essential to the exercise, to see that there are many ways which the preorder relation may be defined and that they form a preorder structure \( \trianglelefteq \) .

`In [comment 1](https://forum.azimuthproject.org/discussion/1892/exercise-1-97-chapter-1#1) Aqilah proposes a solution: 1. \\( L=\\{(1,1),(1,2),(2,2),(3,3)\\}\\) I want to point out to the casual reader that the \\( \le \\) relation here defined is not its conventional meaning over \\( \mathbb{N} \\), it can be defined as any set of pairs from \\( S \times S \\). That conventional meaning would define the \\( \le \\) relation as \\( \\{(1,1),(1,2),(2,2),(2,3),(3,3)\\} \\) [incidentally chosen to be \\( Q \\)]. This is essential to the exercise, to see that there are many ways which the preorder relation may be defined and that they form a preorder structure \\( \trianglelefteq \\) .`

We show this by recognizing inclusion. We find \(Cl(q)\) by adding to \(q\) the reflective (adding \((s,s)\) for every \(s\) ) and transitive (adding \((s,u)\) when \((s,t)\) and \((t,u)\) ) closures.

Thus we can see

That this must be true takes a bit more work.

`1. \\( L=\\{(1,2),(3,3)\\} \\) 2. \\( Q=\\{(1,2),(2,3),(3,3)\\} \\) and \\( Q'=\\{(1,1),(2,3),(3,3)\\} \\) 3. Show \\( \le \trianglelefteq Cl(Q) \\) 4. Show \\( \le \ntrianglelefteq Cl(Q') \\) We show this by recognizing inclusion. We find \\(Cl(q)\\) by adding to \\(q\\) the reflective (adding \\((s,s)\\) for every \\(s\\) ) and transitive (adding \\((s,u)\\) when \\((s,t)\\) and \\((t,u)\\) ) closures. * \\( L=\\{(1,2),(3,3)\\} \\) * \\( Cl(Q) = \\{(1,2),(2,3),(3,3)\\} \cup \\{(1,1),(2,2)\\} \cup \\{ (1,3) \\} = \\{(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)\\} \\) * \\( Cl(Q') = \\{(1,1),(2,3),(3,3)\\} \cup \\{(2,2)\\} \cup \\{ \\} = \\{(1,1),(2,2),(2,3),(3,3)\\} \\) Thus we can see * \\( \\{(1,2),(3,3)\\} \subseteq \\{(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)\\} \\) * \\( \\{(1,2),(3,3)\\} \nsubseteq \\{(1,1),(2,2),(2,3),(3,3)\\} \\) That this must be true takes a bit more work. * \\( L \subseteq Q \subseteq Cl(Q) \\) is true by the transitive property of subsets. * It I had chosen a \\( Q'' = \\{(1,2),(2,3)\\} \\) then \\( Cl(Q'') = \\{(1,2),(2,3)\\} \cup \\{(1,1),(2,2),(3,3)\\} \cup \\{ (1,3) \\} = \\{(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)\\} \\) and the condition would not have been met.`

"This is actually the right

leftadjoint of a Galois connection." The description of \( Cl(q) \) sounds like a right adjoint.`"This is actually the <s>right</s> **left** adjoint of a Galois connection." The description of \\( Cl(q) \\) sounds like a right adjoint.`