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Let R$_1$ and R$_2$ be two equivalence relations on the same set A.

Not sure how to interpret this statement. Does it mean...

A = {1, 2} $\quad$#for example

AR$_1$A = {(1, 1), (1, 2), (2, 1), (2, 2)}

AR$_2$A = {(1, 1), (1, 2), (2, 1), (2, 2)}

R$_1$$\cap$R$_2$ = {(1, 1), (1, 2), (2, 1), (2, 2)}

R$_1$$\cup$R$_2$ = {(1, 1), (1, 2), (2, 1), (2, 2)}

fossdeep
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  • I'm confused a bit. An equivalence relation is a function that assigns true or false to each pair of elements of a set (true if they are equivalent). Your $R_1$ and $R_2$ above look like sets. – muaddib Jun 03 '15 at 17:44
  • Don't relations have to be a set of ordered pairs between two sets? – fossdeep Jun 03 '15 at 17:47
  • I understand now. Yes, the function I mentioned is true if the ordered pair which is its argument is in that set. – muaddib Jun 03 '15 at 17:48

1 Answers1

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When we say, for example, "let $R$ be an equivalence relation on the set $X$", we mean:

Suppose we have a subset $R\subseteq X\times X$ such that:

1) $\forall x\in X$ we have $(x,x)\in R$;

2) If $(x_1,x_2) \in R$, then $(x_2,x_1) \in R$;

3) If $(x_1,x_2),(x_2,x_3) \in R$, then $(x_1,x_3)\in R$.

So your specific statement is saying you have two such relations (or subsets, not necessarily the same) of a set $A$.

anak
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  • I have already proved that the intersection and union are also equivalent, although I'm not sure if it's correct. I will post my solution later for feedback, but I just wanted to make sure my understanding of the statement was right. So it looks like my example of intersection and union are correct? – fossdeep Jun 03 '15 at 17:54
  • Your $R_1$ and $R_2$ are indeed equivalence relations (the same one in-fact). Their intersection and union are the same (which is indeed the same equivalence relation as before). However this is just a specific case. Note that in general, the condition 3) in my answer doesn't hold for the union of two equivalence relations. – anak Jun 03 '15 at 18:05
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    You're right. I forgot about union not always being transitive. Thank you. – fossdeep Jun 03 '15 at 18:10