Let $R_1$ and $R_2$ be reflexive relations on a set $S$. Prove that $R_1\cup R_2$ and $R_1 \cap R_2$ are reflexive.

Question

How do I go about proving this?

Answer 1

Basically, to prove a relation $R$ is reflexive on a set $S$, we need to prove $(s, s) \in R$ for all $s \in S$. Since we already know this is true for $R_1$ and $R_2$, it becomes very easy to prove this for their intersection and union.

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Given an element $s \in S$, we know that $(s, s) \in R_1$ and $(s, s) \in R_2$ because $R_1, R_2$ are reflexive.

Since $(s, s)$ is in both sets, it is also in the intersection of the sets, so $(s, s) \in R_1 \cap R_2$. Therefore, $R_1 \cap R_2$ is reflexive.

Since $(s, s) \in R_1$ and $R_1 \subseteq (R_1 \cup R_2)$, we can deduce that $(s, s) \in R_1 \cup R_2$. Therefore, $R_1 \cup R_1$ is reflexive.

Answer 2

For first part as Noble Mushtak said $R_1 \cap R_2$ is reflexive. Now for second part $R_1 \cup R_2=R_1 or R_2 - (R_1 \cap R_2)$ i.e. that part of $R_1$ and $R_2$ which are only one time in $R_1 \cap R_2$ which are also reflexive as we have seen . That's it