Equivalence of sets. 
Prove that if $M$ is an arbitrary infinite set and $A$ is countable, then $M \sim M \cup A$

$M\sim N$ are said to be equivalence if a one-to-one correspondence can be set up between their elements
This exercise comes in the book of funtional analysis to Kolmogorov.
My doubt goes around how a formal proof would be written, in this book several examples come, but, the functions are not built. I think it is clear how to create function 1-1, assigning the first elements of $M$ to $A$ and the rest of them back to $M$
 A: We may assume without loss of generality that $A \cap M = \emptyset$, otherwise replace $A$ with $A \setminus M$ (which is countable being a subset of a countable set).
Consider the case where $A$ is countably infinite. The idea is that we will take a countable subset $N = \{ n_0,n_1,n_2,\ldots\}$ of $M$, then map $N$ into itself by mapping it onto the 'even' part, i.e., $n_k \mapsto n_{2k}$. This leaves the 'odd' part free to map $A = \{a_0,a_1,a_2,\ldots\}$ onto, i.e., $a_k \mapsto n_{2k+1}$. We may fix the elements of $M\setminus N$.
Being completely formal, let $f\colon\mathbb{N} \to M$ be an injection and $g\colon \mathbb{N} \to A$ a bijection. Then $h\colon M \sqcup A \to M$ defined by
$$ h(m) = \begin{cases} f(2f^{-1}(m)) & \text{if $m\in \operatorname{im} f$,} \\ f(2g^{-1}(m)+1) & \text{if $m \in A$,} \\ m & \text{otherwise.} \end{cases}$$
The case where $A$ is finite is similar, except you only need to reserve the first $|A|$-many elements of $N$ (again, a countable subset of $M$ with some fixed enumeration) to map the elements of $A$ to, translating the elements of $N$ forward by $|A|$-many indices, and keeping the other elements of $M$ fixed.
