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

Consider the proof of Proposition 1.11. Suppose that \(\sim\) is an equivalence relation, and let \(P\) be the set of \((\sim)\)-closed and \((\sim)\)-connected subsets \((A_p)_{p \in P}\).

- Show that each \(A_p\) is nonempty.
- Show that if \(p \neq q\), i.e. if \(A_p\) and \(A_q\) are not exactly the same set, then \(A_p \cap A_q = \emptyset\).
- Show that \(A = \bigcup_{p \in P} A_p\).

## Comments

Each \(A_p\) is nonempty because it is \((\sim)\)-connected, and a \((\sim)\)-connected subset is nonempty by definition.

If \(A_p\) and \(A_q\) had a nonempty intersection, they would each contain some element \(x \in X\). But then, since they are \((\sim)\)-closed and \((\sim)\)-connected, \(A_p\) and \(A_q\) would each contain

all and onlyelements of X equivalent to x, contradicting that \(A_p\) and \(A_q\) are unequal. So \(A_p\) and \(A_q\) can’t have a nonempty intersection.We prove that A and the specified union are subsets of each other: First, every element in the union is obviously in A, because the union is a union of

subsetsof A. On the other hand, if \(x \in A\), then the set \(A_r\) of all elements \(y \in A\) with y \(\sim\) x is \((\sim)\)-closed and \((\sim)\)-connected (by transitivity and reflexivity of equivalence). So \(A_r\) is by definition one of the subsets in the specified union, and thus x (which is in \(A_r\) by reflexivity) is in that union.`1. Each \\(A_p\\) is nonempty because it is \\((\sim)\\)-connected, and a \\((\sim)\\)-connected subset is nonempty by definition. 2. If \\(A_p\\) and \\(A_q\\) had a nonempty intersection, they would each contain some element \\(x \in X\\). But then, since they are \\((\sim)\\)-closed and \\((\sim)\\)-connected, \\(A_p\\) and \\(A_q\\) would each contain *all and only* elements of X equivalent to x, contradicting that \\(A_p\\) and \\(A_q\\) are unequal. So \\(A_p\\) and \\(A_q\\) can’t have a nonempty intersection. 3. We prove that A and the specified union are subsets of each other: First, every element in the union is obviously in A, because the union is a union of *subsets* of A. On the other hand, if \\(x \in A\\), then the set \\(A_r\\) of all elements \\(y \in A\\) with y \\(\sim\\) x is \\((\sim)\\)-closed and \\((\sim)\\)-connected (by transitivity and reflexivity of equivalence). So \\(A_r\\) is by definition one of the subsets in the specified union, and thus x (which is in \\(A_r\\) by reflexivity) is in that union.`

Jerry: your equation looks good to me. Sometimes you have to refresh the page to get the rendering to occur.

`Jerry: your equation looks good to me. Sometimes you have to refresh the page to get the rendering to occur.`

Thanks Dan, that worked!

`Thanks Dan, that worked!`

And what is

Pin this case?`And what is <i> P </i> in this case?`

Paulius: the problem says

That's a bit of a mouthful, but the idea is this. If we start with an equivalence relation \(\sim\) on a set \(X\), we can use it to chop \(X\) into a bunch of parts called 'equivalence' classes:

\(P\) is the set of these parts. It's a bit ridiculous to say "the set of subsets \((A_p)_{p \in P}\)". One can equivalently just say \(P\), since this is the same thing!

If you're wondering what a " \((\sim)\)-closed and \((\sim)\)-connected subset" is, you can ask me or look it up in the book - it's in the proof of Proposition 1.11, right before this exercise.

The point of the exercise is to show that these "\((\sim)\)-closed and \((\sim)\)-connected subsets" really do form a partition of \(X\).

`Paulius: the problem says > Let \\(P\\) be the set of \\((\sim)\\)-closed and \\((\sim)\\)-connected subsets \\((A_p)_{p \in P}\\). That's a bit of a mouthful, but the idea is this. If we start with an equivalence relation \\(\sim\\) on a set \\(X\\), we can use it to chop \\(X\\) into a bunch of parts called 'equivalence' classes: <center><img width = "300" src = "http://math.ucr.edu/home/baez/mathematical/7_sketches/set_partition.png"></center> \\(P\\) is the set of these parts. It's a bit ridiculous to say "the set of subsets \\((A_p)_{p \in P}\\)". One can equivalently just say \\(P\\), since this is the same thing! If you're wondering what a " \\((\sim)\\)-closed and \\((\sim)\\)-connected subset" is, you can ask me or look it up in the book - it's in the proof of Proposition 1.11, right before this exercise. The point of the exercise is to show that these "\\((\sim)\\)-closed and \\((\sim)\\)-connected subsets" really do form a partition of \\(X\\).`

Thank you very much. Got it. I was going a bit off tangent here (the axiom of choice and all that). Was thinking how one should construct

P, when it was already given.`Thank you very much. Got it. I was going a bit off tangent here (the axiom of choice and all that). Was thinking how one should construct <i>P</i> , when it was already given.`