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Does every equivalence relation defined on a set, automatically partitions the set?
If yes, give a rigorous proof.
If no, give an example.
Proof:1. If there is x in a set, then it will have to be in a partition.(symmetric)
2.If x ~y, they are in the same partition.
=>y~x(reflexive)
3.If x~y,y~z,
Then x~z as all are in the same partition.(transitive)
We can partition the points/elements according to their equivalence like - suppose we have a set S which has 6 points/elements- {A1,A2,A3,A4,A5,A6}
So, we can get partitions from equivalence relations
PROVED
Equivalence relation -> Partition
Constrtuct partitions from the relation as follows,
1) pick an element from set S, take all y in S such that x~y and put them all in a new partition
2) repeat step 1 until all elements of S are exhausted
step 1 guarantees disjointedness and step 2 guarantees exhaustiveness
Partition -> Equivalence relation
Declare all elements in a single partition to be equivalent,
1) reflexivity: trivial (x is in the same partition as x, so x~x)
2) symmetry: if x is in the same partition as y, so y is also in the same partition. Therefore, x~y=>y~x
3) transitivity: if x is in the same partition as y, y is in the same partition as z, then x is in the same partition as z. Therefore, x~y,y~z => x~z
Yes, since making equivalence relations means that the elements are in the same set, we can say that stating elements equivalent means to break the set and form some new set (partition).