Let R and S be partial orders on a nonempty set A prove that T = R intersection S is also a partial order on A.

See proof below

Step-by-step explanation:

We denote (x, y) ∈R as xRy, and we also use the similar notation xSy for (x, y) ∈S. Remember that R and S are reflexive, antisymmetric and transitive relations (the definition of partial order).

To prove that R∩S⊆A is a partial order, we will prove that R∩S is reflexive, antisymmetric and transitive.

Reflexive: Let a∈A. R is reflexive thus aRa. S is also reflexive, then aSa. Then (a, a) ∈R and (a, a) ∈S which implies that (a, a) ∈R∩S, that is, a (R∩S) a for all a∈A. Antisymmetric: Let a, b∈A and suppose that a (R∩S) b and b (R∩S) a hold. In particular, aRb and bRa. Since R is antisymmetric, a=b. Transitive: Let a, b, c∈A and suppose that a (R∩S) b and b (R∩S) c hold. Then aRb, bRc, aSb and bSc are true. The first two statements imply by the transitivity of R that aRc. Similarly, from the last two we have that aSc. Thus a (R∩S) c as we wanted to prove.