Is the equality of the cardinality of open intervals on R is an equivalence relation. Show that equality of the cardinality of open intervals on R is an equivalence relation.
I'm usure of what the relation is and how to represent it. 
I know how to show it's an equivalence relation and I know that to show the transitive property, I will need to show that a composition of bijections is a bijection.
 A: The relation - assuming you've copied the problem correctly - is as follows: an open interval $U$ is equivalent to an open interval $V$ if $U$ and $V$ have the same cardinality, that is, if there is a bijection $f:U\rightarrow V$. So, for example, $(1, 2)$ is equivalent to $(3, 4)$, since the function $f: (1, 2)\rightarrow (3, 4): x\mapsto x+2$ is a bijection.
You ask how to show that the composition of two bijections is a bijection (to show transitivity). The composition of bijections is indeed a bijection. To do this, you need to show that the composition of two injections is an injection, and that the composition of two surjections is a surjection. For example, if $i:A\rightarrow B, j: B\rightarrow C$ are injections, I want to show $j\circ i$ is an injection. How do I do this? Well, pick $a_1\not=a_2$ in $A$. Then since $i$ is injective, $i(a_1)\not=i(a_2)$. And since $j$ is injective, $j(i(a_1))\not=j(i(a_2))$. But this means $j\circ i(a_1)\not=j\circ i(a_2)$, so $j\circ i$ is indeed injective.
Do you see how to show that the composition of two surjections is surjective? And, do you see why this (together with the fact proved above) will imply that the composition of two bijections is bijective?

Incidentally, the equivalence relation you understand is a pretty boring one: any two nonempty open intervals in $\mathbb{R}$ have the same cardinality. Maybe you meant $\mathbb{Z}$?
A: Just prove that any two open intervals have the same cardinality, and then reflexivity, symmetry, and transitivity are all trivially true.
