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Exercise 14 - Chapter 1
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 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.
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.
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 P in 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.