Assume $S$ and $R$ are two relations from the set $A$ to the set $B$, then $R∪S$ is a relation from $A$ to $B$. prove:
$Dom(R∪S)=Dom(R) ∪ Dom(S)$
$Im(R∪S)=Im(R) ∪ Im(S)$
for every $X⊆A$,$(R∪S)(X)=(R)(X) ∪ (S)(X)$
I tried using definition but I could not reach the desired result, any help or a little hint appreciated.
Heavy sketch for the first one, the rest is left for you. $$\operatorname{Dom}(R\cup S)=\operatorname{Dom}(R)\cup\operatorname{Dom}(S).$$ Since these are sets, use a standard set-equality proof:
Let $a\in \operatorname{Dom}(R\cup S)$. Then, there is some $b$ such that $(a,b)\in R\cup S$. By the definition of union, $(a,b)\in R$ or $(a,b)\in S$. Therefore, $a\in\operatorname{Dom}(R)$ or $a\in\operatorname{Dom}(S)$ by cases.
Let $a\in\operatorname{Dom}(R)\cup\operatorname{Dom}(S)$. Then $a\in\operatorname{Dom}(R)$ or $a\in\operatorname{Dom}(S)$. Wlog, we'll consider the first case. Then there is some $b$ such that $(a,b)\in R$. Since $R\subseteq R\cup S$, $(a,b)\in R\cup S$. Hence $a\in \operatorname{Dom}(R\cup S)$.