How to prove that 2 sets are equinumerous? We suppose that we have 2 sets $\!A$ and $\!B$, such as that $\!A \smallsetminus \!B$ is infinite and $\!B$ is either finite or countably infinite. 
How do we prove that $\!A \smallsetminus \!B \sim \!A$? Intuitionally it does make sense? but it I cant grasp the idea on how to prove it.
 A: Assume $B$ is countably infinite, let $(b_n)_{n\in\mathbb N}$ be an enumeration of it.
Let $C\subseteq A\setminus B$ a countably infinite subset, and $(c_n)_{n\in\mathbb N}$ be an enumeration of it.
Then the function
$$f:A\setminus B \rightarrow A,f(x)=\left\{\begin{array}[rl] x x && \mbox{if }x\notin C\\
b_n && \mbox{if }x=c_{2n+1}\mbox{ for some }n\in\mathbb N \\
c_n && \mbox{if }x=c_{2n}\mbox{ for some }n\in\mathbb N 
\end{array}\right.$$
is bijective.
The basic idea was that we add the missing elements in like a zip.
If $B$ is finite, it gets a bit easier, we just put all the elements of $B$ in "front" of $C$, with $(c_n)_{n\in\mathbb N}$ being an enumeration of the countably infinite subset again. Assume $B=\{b_1,\ldots,b_m\}$.
$$f:A\setminus B \rightarrow A,f(x)=\left\{\begin{array}[rl] x x && \mbox{if }x\notin C\\
b_n && \mbox{if }x=c_{n}\mbox{ for some }n\in\mathbb N, n\leq m \\
c_n && \mbox{if }x=c_{n+m}\mbox{ for some }n\in\mathbb N 
\end{array}\right.$$
A: Hint: for any sets $A$ and $B$, $A = (A \setminus B) \cup (A \cap B)$. In your case, you are given that $A \setminus B$ is infinite and that $B$ is finite or countable infinite. Can taking the union of an infinite set $X$ and a set $Y$ that is at most countably infinite result in a set of greater cardinality than $X$?
