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# Exercise 14 - Chapter 1

edited May 2018

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 $$(A_p)_{p \in P}$$.

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

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1.
edited April 2018
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.

Comment Source: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.
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2.

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

Comment Source:Jerry: your equation looks good to me. Sometimes you have to refresh the page to get the rendering to occur.
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3.

Thanks Dan, that worked!

Comment Source:Thanks Dan, that worked!
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4.
edited April 2018

And what is P in this case?

Comment Source:And what is <i> P </i> in this case?
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5.
edited April 2018

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:

$$P$$ 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 $$X$$.

Comment Source: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\$$. 
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6.

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.

Comment Source: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.