Transitivity?

Question

" If R and S are tansitive, then R U S is tansitive."
The above statement is FALSE, but I couldn't understand why.
Can someone show me it's not true by showing me a counterexample?

Thanks :)

Answer 1

Here's a minimal counterexample: R = {(0, 1)} and S = {(1, 0)}. Both R and S are vacuously transitive, but R∪S = {(0, 1), (1, 0)} is not transitive, because (0, 1)∈R∪S and (1, 0)∈R∪S, but (0, 0)∉R∪S.

Note that not all counterexamples are trivial. For instance, consider R'={(x, y)∈ℝ: x
The reason why transitivity is not preserved under unions is that transitivity of R and S means only that (xRy ∧ yRz) ⇒ xRz and (xSy ∧ ySz) ⇒ ySz, whereas transitivity of R∪S would require in addition that (xRy ∧ ySz) ⇒ (xRz ∨ xSz) and (xSy ∧ yRz) ⇒ (xRz ∨ xSz), and neither of the statements regarding transitivity of R and S says anything about the case where (xRy ∧ ySz) or (xSy ∧ yRz), so there is no more reason to expect that R∪S is transitive than to suspect any other relation is transitive.