The complement of a finite set in a countable set is countable Let $X$ be a countable set. Let  $A  \subset  X$.
How can I prove that if $A$ is finite then $X  \setminus  A$ is countable?
I started off by supposing that $X  \setminus  A$ is finite and then need to show that this leads to a contradition.
The result that $X  \setminus  A$ is countable will then follow because all subsets of $X$ are countable or finite
 A: Suppose for the sake of contradiction that $X- A$ is finite. What does that imply about $(X-A) \cup A$?
A: 
Let $X$ be a countable set. 

Therefore, there exists an injective function $f:X\to \mathbb{N}$
By definition of injective, we have $\forall a,b\in X:[f(a)=f(b)\implies a=b] $

Let  $A  \subset  X$.

Let $g$ be a function $g:X \setminus A \to \mathbb{N}$ such that $\forall a\in X\setminus A: g(a)=f(a)$


*

*Suppose $x,y\in X\setminus A$

*By definition of $\setminus$, we have $x,y \in X$

*Suppose $g(x)=g(y)$

*By definition of $g$ ,we have $g(x)=f(x)$

*By definition of $g$ ,we have $g(y)=f(y)$

*$f(x)=f(y)$ (substituting 4 and 5 into 3)

*$x=y$ (from 2 and since $f$ is injective)

*$g(x)=g(y)\implies x=y$ (conclusion from 3 and 7)

*$\forall a,b \in X\setminus A: [g(a)=g(b) \implies a=b]$ (conclusion from 1 and 8)

*$g$ is an injective function $g:X\setminus A\to \mathbb{N}$ (from 9 and definition of $g$)

*By definition, $X\setminus A$ is countable  (regardless of whether $A$ is finite)
