Equivalent relations and partitions.
Recall that ” ” is a certain subset of .
More precisely, for each , there exists unique Satisfying .
We can observe that every has a “relation” with some element in B.
Definition. Given two Sets , a relation is a subset of . i.e., .
For each , we denote .
Ex) .
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Ex) Let abe function . Then is a relation.
Definition. Let be a relation. We say that is an equivalent relation if satisfies (1), (2), (3):
(1) (Reflexive) for each . i.e, .
(2) (symmetric) If then .
i.e., if then .
(3) (Transitive) If and , then .
i.e., if and then .
Ex) . .
Ex) Given a set , consider the power set of .
Define a relation in , by iff there exists one to one, onto function .
Then is an equivalence (Exercise).
Q1. Reflexive?
For each , we have idx: mapping : this map should be 1-1 and onto.
Q2. Symmetric?
Suppose . i.e., there exists a 1-1, onto
By previous (2nd) lecture, there exists such that and .
Exercise. Given and ,
1) If is onto, then is onto.
2) If is 1-1, then is1-1.
By 1) and 2), the above must be 1-1, onto.
Thus .
Hence .
Q3. Transitive?
Suppose X and Y.
i.e., there exists , which are both 1-1 and onto.
Consider the composition .
Exercise 3) If are 1-1, then is 1-1.
4) If and are onto then is onto
Hence is 1-1 and onto. i.e.,
Definition. Given a set , consider the index set I. (i.e., for each , there exists ).
(3) .
We call the partition of .
ex). Let be set of integers.
Denote .
: partition of
Remark.
1) Given a set , consider the equivalence relation . For each , define .
Then becomes a partition of (Exercise).
2) Conversely, given a set , consider a partition of .
Define , if there exists sach that .
Then ” ” is an equivalence Relation (Exercise).
3) Consequently, there is the 1-1 correspondence between
Exercise. Let be functions.
1) If are 1-1, then is 1-1
2) If is onto, then is onto.
If is 1-1, then is 1-1.