(edited)$\mathrm{Dom}(R_1 \cup R_2) \subseteq \mathrm{Dom}(R_1) \cup \mathrm{Dom}(R_2)$?

Question

For sets $A,B,C,D$, let $R_1, R_2$ be relations such as $R_1 : A \rightarrow B, R_2 : C\rightarrow D$. Then is the following sentence true? $$\mathrm{Dom}(R_1 \cup R_2) \subseteq \mathrm{Dom}(R_1) \cup \mathrm{Dom}(R_2)$$

I guess that it is true, but I don't know how to prove it by predicate logic. I can prove that $R_1 \cup R_2$ is also relation that $R_1 \cup R_2 : A\cup C \rightarrow B\cup D$. If the above statement is false, what condition is needed for the statement being true?

*edit

I read drhab's answer. I think there is a problem in setting domains. Remark that $R_1 : A \rightarrow B$ then $$x \in \mathrm{Dom}(R_1) \Leftrightarrow \exists y \in B : (x,y) \in R_1$$ In addition the existential quantifier holds distributive law with $\vee$ which means "or", and the following property. $$\exists x, p(x) \vee q(x) \Leftrightarrow \exists x, p(x) \vee \exists x, q(x)$$ $$\exists p(x), q(x) \Leftrightarrow \exists x,p(x)\wedge q(x)$$ Using the above definition and properties there is a problem. \begin{align} &a \in \mathrm{Dom}(R_1 \cup R_2 )\\ \Leftrightarrow& \exists b \in B\cup D : (a,b)\in R_1 \cup R_2\\ \Leftrightarrow& \exists b \in B\cup D : (a,b) \in R_1 \vee (a,b)\in R_2 \\ \Leftrightarrow& \exists b_1 \in B\cup D : [(a,b_1) \in R_1] \vee \exists b_2 \in B\cup D: [(a,b_2) \in R_2] \\ \Leftrightarrow& \exists b_1: [ b_1 \in B\cup D \wedge (a,b_1) \in R_1] \vee \exists b_2 : [b_2 \in B\cup D \wedge (a,b_2) \in R_2 ]\\ \Leftrightarrow& \exists b_1 : [(b_1 \in B \vee b_1\in D) \wedge (a,b_1) \in R_1 ] \vee \exists b_2 : [(b_2 \in B \vee b_2\in D) \wedge (a,b_2) \in R_2 ]\\ \Leftrightarrow& \exists b_1 : [b_1 \in B \wedge (a,b_1) \in R_1] \vee \exists b_1:[b_1 \in D \wedge (a,b_1) \in R_1] \vee \exists b_2 : [b_2 \in B \wedge (a,b_2) \in R_2] \vee \exists b_2:[b_2 \in D \wedge (a,b_2) \in R_2] \end{align} In the last statement, $\exists b_1 : [b_1 \in B \wedge (a,b_1) \in R_1] \vee \exists b_2:[b_2 \in D \wedge (a,b_2) \in R_2]$ means $a\in \mathrm{Dom}(R_1) \cup \mathrm{Dom}(R_2)$. So, $\mathrm{Dom}(R_1) \cup \mathrm{Dom}(R_2) \subseteq \mathrm{Dom}(R_1 \cup R_2)$ is proved, because \begin{align} &\exists b_1 : [b_1 \in B \wedge (a,b_1) \in R_1] \vee \exists b_2:[b_2 \in D \wedge (a,b_2) \in R_2]\\ \Rightarrow& \exists b_1 : [b_1 \in B \wedge (a,b_1) \in R_1] \vee \exists b_1:[b_1 \in D \wedge (a,b_1) \in R_1] \vee \exists b_2 : [b_2 \in B \wedge (a,b_2) \in R_2] \vee \exists b_2:[b_2 \in D \wedge (a,b_2) \in R_2] \end{align}

I think the "$\Rightarrow$" can be changed into "$\Leftrightarrow$". To do this, $\exists b_1:[b_1 \in D \wedge (a,b_1) \in R_1] \vee \exists b_2 : [b_2 \in B \wedge (a,b_2) \in R_2]$ should be false, but I think if we choose a suitable relation such that $b_1, b_2$ which are in $B\cap D$, then this statement can be also true, contrary to my purpose.

How did you prove these things?

Answer 1

For any set $U$ you can define: $$\mathsf{Dom}(U):=\{x\mid\exists y[\langle x,y\rangle\in U\}$$

Then evidently the following statements are equivalent:

• $x\in\mathsf{Dom}(U\cup V)$

• $\exists y\langle x,y\rangle\in U\cup V$

• $\exists y[\langle x,y\rangle\in U\vee\langle x,y\rangle\in V]$

• $[\exists y\langle x,y\rangle\in U]\vee[\exists y\langle x,y\rangle\in V]$
• $x\in\mathsf{Dom}(U)\vee x\in\mathsf{Dom}(V)$
• $x\in\mathsf{Dom}(U)\cup\mathsf{Dom}(V)$

So we have:$$\mathsf{Dom}(U\cup V)=\mathsf{Dom}(U)\cup\mathsf{Dom}(V)$$

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$R_1$ is a relation $A\to B$ iff $R_1\subseteq A\times B$ and $R_2$ is a relation $C\to D$ iff $R_2\subseteq C\times D$.

Then evidently $R_1\cup R_2\subseteq (A\cup C)\times (B\cup D)$ so is a relation $A\cup C\to B\cup D$, as you suspected.