Verify it is equiv. rel.
For the set X={m,n,p,q,r,s}, let R be the relation on P(X) (power set of X) given by A R B iff A and B have the same number of elements.
Now, list all elements in {m}/R; in{m,n,p,q,r}/R. How many elements are in X/R? How many in P(X)/R?
I know I need to show it is symmetric, transitive, and reflexive. I know how to this when we have like equations and things, but this one is confusing for me. Please help. Thank you.
Hello zhupolongjoeAs you have written the question, some of it doesn't make sense. I'll explain why, but it will take a little time, so stay with me.Originally Posted by zhupolongjoe
, the power set of
, is the set of all the subsets of
Look carefully at the words above in italics:
Nowis a relation on
is related to subset
if and only if
because they both have 2 elements
because they both have 3 elements
... and so on.
Next, you need to understand thenotation. If a relation
(called the quotient set of
For example, ifand
(in other words,
is divisible by
), then
into two equivalence classes: {even integers} and {odd integers}, and these two sets are therefore the elements of
So, in the question you're asking, whereas it makes sense to talk about, it makes no sense to talk about
or
, because
, but on sets.
OK. So let's look at the question that does make sense: How many elements are there in- are equivalent. So they form one equivalence class. (And that, I suspect, is what the question means when it says 'list all the elements in
', although that is incorrect use of the notation.)
All the subsets of
The proof that(reflexive) and
(symmetric) and
(transitive).
Grandad
I just have on further question:
"All the subsets ofcontaining 2 elements form another equivalence class; all the subsets containing 0 elements form another, and so on. Since you can form subsets containing 0, 1, 2, 3, 4, 5 or 6 elements, there are therefore 7 equivalence classes, and therefore 7 elements in
."
Why wouldn't it be the 1+6+.....+6+1..i.e. add all the elements..like there is one subset with 0 elements, 6 subsets with 1 element...6 subsets with 5 elements and so on...?
Hello zhupolongjoe[/quote]Because this is the number of elements in"...there are therefore 7 equivalence classes, and therefore 7 elements in
Why wouldn't it be the 1+6+.....+6+1..i.e. add all the elements..like there is one subset with 0 elements, 6 subsets with 1 element...6 subsets with 5 elements and so on...?. (Perhaps you can work out why?) On the other hand
In the other illustration I gave you (therelation on
Grandad
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