Enquiry on union of countable set is countable I was reading principle of mathematical analysis by Walter Rudin.  And saw the proof of union of countable set is also countable. According that proof they have written elements of the sets in a sequence in the following way
$x_{11};x_{12},x_{21};x_{13},x_{22},x_{31};...... $. Where $x_{mn} $ is the $m $th set's $n$the element
And giving the argument if two sets have same elements they will  appear twice and that's why we can enumerate them. My question is how can we assign different integer value to same element. To prove one-one correspondence between two elements. 
 A: This proof gives a surjection from the natural numbers to your union $U$ (i.e. a mapping which covers every element of $U$ at least once). This is actually enough to conclude that $U$ is countable, so a lot of proofs will just do this -- it is often simpler than finding a one-to-one correspondence (bijection) explicitly.
The reason is that, provided $U$ is infinite, you can use a surjection $f:\mathbb N\to U$ to construct a bijection $g:\mathbb N\to U$ as follows. For each $u\in U$ there is a value $\min\{n\in\mathbb N:f(n)=u\}$ (call this $h(u)$). These values are all different. Now for each $u\in U$ we can compute $g(u)=|\{v\in U:h(v)<h(u)\}|$ (this is finite since it is at most $h(u)$). If $u\neq v$ then $h(u)\neq h(v)$, so one of the corresponding sets is strictly contained in the other, so $g(u)\neq g(v)$. It is also easy to check that every natural number* is $g(u)$ for some $u$ (provided $U$ is infinite; if it is finite then this will give a bijection to the first $n$ natural numbers).
Basically what this is doing is taking the sequence you started with but only keeping the first occurrence of each element, then renumbering.
*I'm taking $\mathbb n=\{0,1,...\}$; if you want to start at $1$ replace the $<$ by $\leq$ in the definition of $g$.
