Here is a wkst that I have been working on and would like some more insight on. I am a bit lost and what to see what other ppl think! I have somewhat an idea but an confused at parts. Thanks

If we have a relation R on a set A that is both reflexive and transitive, we can make it into a partial order relation by identifying pairs of elements that are related under R in both directions. We define partial order relation on equivalence classes of elements of A, as shown below. Throughout, let A be a set and suppose that R is a relation on A that is reflexive and transitive. Do no auume anything about the symmetric or antisymmetric properties for R.

1. Define a second relation ~ on A by saying that a ~ b if and only if aRb and bRa are both true. Show that ~ us an equivalence relation on A.

2. For each a in A, let [a] = { x exist in A | x ~ a}, that is, the equivalence class of a under the relation defined in Question 1. Let C be the set of all distinct equivalence classes [a], as a varies over all elements A. Define a relation < on C by saying that [a] < [b] if and only if aRb in A. Show that < is well-definded, that is, if [a] =[c] and [b] = [d], then [a] < [b] implies that [c] < [d]. (hint: remember that [x] =[y] if and only if x ~ y in A. Use this to rephrase what you are asked to show)

3. With C and < defined in Question 2, show that < is a partial order relation on C. (Hint” for the antisymmetric property, you will again need to use the fact that [x] = [y] if and only if x ~ y in A.)

4. An example to illustrate the procedure outlined in Question 1-3, let B = {1,2,3,4,5} and let A = P(B), the set of all subsets of B. Define R on A by saying that SRT if and only if (S intersect {1,3,4}) subset (T intersect {1,3,4}). Show that R is both reflexive and transitive in this case, but is neither symmetric nor antisymmetric. List the distinct equivalence classes of A under ~, where ~ us defined on A as in Question 1. (that is, find the set C.) Find a lattice diagram for the partial order relation < on C that is defined in Question 2 and 3. Explain.