# equivalence relations

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• Nov 23rd 2009, 06:43 PM
g rad23
equivalence relations
say we have a set W. We also have a non-empty set Z such that each element of Z is an equivalence relation on W.

Must show that the nZ is an ER in W. (where n is intersection)
how would this be proven, like i know that u must prove that it is reflexive, transitive and symmetric but how exactly would u start it?
• Nov 23rd 2009, 07:21 PM
Drexel28
Quote:

Originally Posted by g rad23
say we have a set W. We also have a non-empty set Z such that each element of Z is an equivalence relation on W.

Must show that the nZ is an ER in W. (where n is intersection)
how would this be proven, like i know that u must prove that it is reflexive, transitive and symmetric but how exactly would u start it?

I am sorry, I don't understand what you mean.
• Nov 23rd 2009, 07:29 PM
g rad23
my question is
let W be a set.
Let Z be a non empty set such that each element of Z is an equivalence relation on W

i must show that the nZ is an equivalence relation on W. (where nZ is the intersection of Z)

how would i go about proving this is true?

i understand that i have to prove that the relation is reflexive, anti-symmetric and transitive, but how would i do that
• Nov 23rd 2009, 07:33 PM
Drexel28
Quote:

Originally Posted by g rad23
my question is
let W be a set.
Let Z be a non empty set such that each element of Z is an equivalence relation on W

i must show that the nZ is an equivalence relation on W. (where nZ is the intersection of Z)

how would i go about proving this is true?

i understand that i have to prove that the relation is reflexive, anti-symmetric and transitive, but how would i do that

I'm sorry. I don't mean to sound ignortant, but I was unsure of notation. Do you mean $nz=\bigcap_{x\in z}x$?
• Nov 23rd 2009, 07:36 PM
g rad23
dont worrry about it and yes i did mean the intersection as u stated
• Nov 23rd 2009, 07:39 PM
Drexel28
Quote:

Originally Posted by g rad23
dont worrry about it and yes i did mean the intersection as u stated

Maybe some other member can help you better than I, but I don't really undertsand what this question could possibly mean? What is the intersection of two equivalence relations? Unless, you mean the intersection of all the partitions induced by the relations?
• Nov 23rd 2009, 07:44 PM
g rad23
well im not sure myself this is the question i was given for my hmwk. well im assuming its intersection it looks exactly like the symbol for intersection but has no limit.
• Nov 23rd 2009, 07:48 PM
Drexel28
Quote:

Originally Posted by g rad23
well im not sure myself this is the question i was given for my hmwk. well im assuming its intersection it looks exactly like the symbol for intersection but has no limit.

Ok, now that I think about it I believe that I understand what this is saying. if $\sim$ is an equivalence relation on $E$ then $\sim$ is charcterized by $R=\left\{(a,b)\in E\times E:a\sim b\right\}$. Maybe that if we consider the different relations on $E$ to be charchertized by $R_1,R_2,\cdots$ (not necessarily countable..I just wrote it that way for clarity) then perhaps they mean to show that $R_1\cap R_2\cdots=R$ is an the characterization of an equivalence relation. Does that sound right? Another member may swoop in an answer this btw.
• Nov 23rd 2009, 07:51 PM
g rad23
umm it might be right im not sure i thought in order to prove it was an equivalence relation on something one would have to prove that the relation is reflexive, transitive and symmetric
• Nov 24th 2009, 02:18 AM
emakarov
The gist of the problem is to prove the following. Let $W$ be a set and let $R_1$ and $R_2$ be relations (i.e., subsets of $W\times W$) on $W$. Moreover, suppose that $R_1$ and $R_2$ are equivalence relations. Show that $R_1\cap R_2$ is an equivalence relation a well.

(This solves the problem when the set $Z$ of equivalence relations has size 2, or, in fact, any finite size. To be strict, one has to prove this also for infinite $Z$, but the proof is essentially the same.)

This should be easy to show by definition.
• Nov 24th 2009, 02:47 AM
Shanks
Yeah, i agree with emakarov!