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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.

amrsa
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  • What do you think the domain is fo (1) $R$, (2) $S$, (3) $R\cup S$? – almagest Jan 08 '20 at 17:07
  • Domain for $R$ and $S$ is defined by $\left{a∈A:\left(a,b\right)∈A\ for\ some\ b∈B\right}$ –  Jan 08 '20 at 17:13
  • So what is your difficulty with dom$(R\cup S)=$dom$(R)\cup$ dom$(S)$? – almagest Jan 08 '20 at 17:37
  • I just cannot prove that –  Jan 08 '20 at 17:37
  • Suppose $a\in\text{dom}(R\cup S)$. Then for some $b$ we have $(a,b)\in R\cup S$. So either $(a,b)\in R$ or $(a,b)\in S$ (or both). If $(a,b)\in R$, then $a\in\text{dom}(R)$; if $(a,b)\in S$, then $a\in\text{dom}(S)$. So dom$(R\cup S)\subseteq\text{dom}(A)\cup\text{dom}(B)$. Similarly, prove $\supseteq$. – almagest Jan 08 '20 at 17:45
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    What does this have to do with equivalence relations? – amrsa Jan 08 '20 at 17:58

1 Answers1

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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 $a\in \operatorname{Dom}(R\cup S)$.

Michael Burr
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