Equivalence Relations and Classes.?

Question

Let A = {1, 2, 3, 4, 5, 6, 7, 8}.


Consider the following families of subsets of A:
F = [A_1={1}, A_2={2,3,4}, A_3={5,6}, A_4={7,8}]
and
F = [A_1={1,2}, A_2={2,3,4}, A_3={5,6}, A_4={7,8}]

Then consider the following relation on A: xRy iff x and y belong to the same subset of the family F.(in both cases).

Prove or disprove: R is an equivalence relation on A.
If they are equivalence relations, what are the equivalence classes?

Work stuff I've done:
I've found that they are reflexive and symmetric (though don't quote me on that) and I'm not sure if they are transitive or not. For some reason I was thinking the first was and the second one wasn't...

But anyway: If that is the case the first is an equivalence relation and the second isn't?
And how do I find the equivalence classes.

Answer 1

You are correct, the first F defines an equivalence relation on A, and the subsets are the equivalence classes. Those subsets form a partition of A, which guarantees an equivalence relation is defined by saying two things are related just in case they are in the same subset.



The second F is not an equivalence relation, because sets A_1 and A_2 overlap (i.e. their intersection is not empty), and for this reason transitivity fails, e.g. 1 is related to 2 and 2 is related to 3 but 1 is not related to 3. Also, for the same reason (that two of the subsets overlap), the collection of subsets is not a partition of A.

Answer 2

It's a little confusing to use the same letter F to refer to two different things.



First F:



This is an equivalence relation. You know this because it partitions the set into equivalence classes. The classes are just A_1, ... , A_4.



Second F:



This is not an equivalence relation. You can prove: 1R2 because they both belong to A_1, and 2R3 because they both are in A_2, but it is not true that 1R3.

Answer 3

must be incorrect, yet here is going with equivalence training u prefer to satisy the three situations, reflesivity that's aparent becuase it on addition symetric n +p = m+q transivity then some xRy, yRz then xRz so (m,n) R (p,q) and (p,q) R (a,b) so this ability, that (m,n) R(a,b) which so m+q = n+p and p+a = q+r so which you prefer to tutor that m+a = n+b

Answer 4

could be incorrect, yet right here is going with equivalence instructions u desire to satisy the three situations, reflesivity that's aparent becuase it on addition symetric n +p = m+q transivity then some xRy, yRz then xRz so (m,n) R (p,q) and (p,q) R (a,b) so this means, that (m,n) R(a,b) which so m+q = n+p and p+a = q+r so which you desire to coach that m+a = n+b