Consider the set $A=\{X\in\mathcal P(\mathbb Z),X=\{k,k+2\} \}$. Show that $A$ is countable infinite.

We first define a function that maps $\{k,k+2\}$ to $k$. So consider the function $f:A\rightarrow \mathbb{Z}$ with 

$f(\{x,x+2\})=\left\{\begin{array}{lll}2x&,&x\text{ even}\\ 3x&,&x\text{ odd}\end{array}\right.$

Then we can show that a bijection exists. 

Injection: Let $\{a,a+2\},\{b,b+2\}\in A$, such that $f(\{a,a+2\})=f(\{b,b+2\})$. If $a,b$ is odd, $a+2,b+2$ is odd, so we have $3a=3b$ and $3a+6=3b+6$.

If $a,b$ is even, then $a+2,b+2$ is even, so we have $2a=2b$ and $2a+4=2b+4$. We clearly see that $a=b$, so $f$ is injective.

Surjection: Since $\mathbb{Z}$ is infinite, we know that the set $A$ is infinite. This means that for all $k\in\mathbb{Z}$, there is an $a\_k$ such that $f(\{a\_k,a\_k+2\})=k$. Thus, $f$ is surjective, so it is bijective. 

Thus, $A$ is countably infinite.