Question

If R and S are relations on a set A, then prove the following:
(i) R and S are symmetric $\Rightarrow R\bigcap S$ and $R\bigcup S$ are symmetric.
(ii) R is reflexive and S is any relation$\Rightarrow R\bigcup S$ is reflexive.

Answer 1

Hint: We will use the definitions of reflexive and symmetric relations to solve this question. A relation is a reflexive relation If every element of set A maps to itself. A relation in a set A is a symmetric relation if $({{a}_{1}},{{a}_{2}})\in R$ implies that $({{a}_{2}},{{a}_{1}})\in R$, for all ${{a}_{1}},{{a}_{2}}\in A$.

Complete step-by-step answer:
(i) In the first part of the question it is mentioned that R and S are symmetric. So this means that if (a, b) belongs to R then (b, a) also belongs to R. Also if (a, b) belongs to S then (b, a) also belongs to S. So using this information we get,
$(a,b)\in R$ and $(b,a)\in R$. Also $(a,b)\in S$ and $(b,a)\in S$.
So now we can write that R is a subset of $A\times A$ and similarly S is a subset of $A\times A$. So from these we can write that $R\bigcap S$ is also a subset of $A\times A$.
Let $(a,b)\in A$ such that $(a,b)\in R\bigcap S.......(1)$. So this implies that $(a,b)\in R$ and $(a,b)\in S$. And now we know that both R and S are symmetric from above so this implies $(b,a)\in R$ and $(b,a)\in S$. So now $(b,a)\in R\bigcap S........(2)$. Now from equation (1) and equation (2) implies that $R\bigcap S$ is symmetric.
Let $(a,b)\in A$ such that $(a,b)\in R\bigcup S.......(3)$. So this implies that $(a,b)\in R$ or $(a,b)\in S$. And now we know that both R and S are symmetric from above so this implies $(b,a)\in R$ or $(b,a)\in S$. So now $(b,a)\in R\bigcup S........(4)$. Now from equation (3) and equation (4) implies that $R\bigcup S$ is symmetric.
(ii) In the second part of the question it is mentioned that R is reflexive and S is any relation. Now this means that $(a,a)\in R$. Also $(a,a)\in A$ as R is a subset of $A\times A$. So if $(a,a)\in R$ then $(a,a)\in R\bigcup S$ because we know that in union every element is included. Hence $R\bigcup S$ is reflexive.

Note: Remembering the definition of relations and the types of relations is the key here. We should also keep in mind that intersection means common elements in the two sets and union means all the elements of the two sets.