Domain, Range, and Relation 
Show that if R is a relation, then $\cup\cup R =(\operatorname{dom}R)\cup(\operatorname{ran}R)$.



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*First, I don't know what $\cup\cup R$ means where $R$ is a relation.

*I want to know what is the membership relation on $R$.
Please help me.
 A: When we say that $R$ is a relation then we say that $R$ is a set (or class) of ordered pairs. In the usual contexts of modern set theory, this means that the elements of $R$ are sets of the form $\{\{a\},\{a,b\}\}$, and that $a$ is in $\text{dom} R$ and $b\in\text{rng} R$.
When we write $\bigcup A$ we mean $\{u\mid\exists v\in A. u\in v\}$, that is $\bigcup_{B\in A} B$.
So $\bigcup R$ would be all the sets $\{a\}$ and $\{a,b\}$, in a big collection, and $\bigcup\bigcup R$ would be simply those $a$'s and $b$'s.
I hope this clears out the idea, and I am leaving you to write the details.
A: First of all, recall the most usual (set-theoretic) definition of a relation:

A (binary) relation $R$ on $X$ is a subset of $X \times X$

Alternative accounts do not insist both factors are the same (i.e. allow $R \subseteq X \times Y$). In either case, $R$ is a set of ordered pairs.
Now the set-theoretic definition (due to Kuratowski) of an ordered pair $(x,y)$ is the set $\{\{x\},\{x,y\}\}$. The definition of union implies that we have:
$$\cup\cup R = \cup \{z \in (x,y): (x,y) \in R\}$$
By the definition of $(x,y)$ above, we see that $z \in (x,y)$ iff $z = x$ or $z = y$. Can you take it from here?
