Trying to show that the set of all $2$-element subsets of a denumerable set is denumerable Suppose $A$ is denumerable and put $X = \{ B : B \subset A, \; \; |B| = 2 \} $. I want to show that $X$ is denumerable as well.
My try: Let $f$ be bijection from $\mathbb{N}$ to $A$. 
We know any $B \in X$ is of the form $B = \{a,b \} $ for unique $a,b \in A $. We know there exist elements $n,m \in \mathbb{N}$ such that $a = f(n) $ and $b = f(m) $. 
We define $F: X \to \mathbb{N} $ by $F( \{ f(n), f(m) \} ) = 2^{f(n)}3^{f(m)} $. 
To show this is injective it is enough to show that if $2^k3^r = 1$, then $k=r=1$ 
But I am stuck here. I mean it is obvious but how can we prove this rigorously ?
 A: If $2^k 3^r = 1$ then $k = 0$ since otherwise $2 \mid 1$ and $r = 0$ since otherwise $3 \mid 1$.
A: If you can use the fact
that the rationals are denumerable,
use the numerators and denominators
as the indices into the set.
A: As @YuvalFilmus mentioned,$2^k 3^r=1$ if and only if $k= r=0$.
Given $A$ is denumerable, and hence so is $A\times A$, since Cartesian product of denumerable sets is denumerable.
Now, if you notice, $X=\{\{x,y\} : x,y\in A\}$ is equivalent to a subset (say Y) of $A\times A$, removing from $A\times A$


*

*$(y,x)$ if $(x,y)\in Y \forall (x,y)\in A\times A$.

*$(x,x) \forall x\in A$
Subset of a denumerable set is denumerable. Hence X is denumerable.

Edit: continuing what you tried, to show that $F$ is injective, suppose
$F(f(n),f(m))=F(f(p),f(q))$
$\implies 2^{f(n)}3^{f(m)}$=$2^{f(p)}3^{f(q)}$.
Then $f(n)=f(p)$ and $f(m)=f(q)$ since prime factorization is unique.
Hence it is injective.
A: Instead, simply define
$$F(\{f(n),f(m)\})=2^n+2^m.$$
Suppose $$F(\{f(n),f(m)\})=F(\{f(k),f(r)\}),$$
i.e., $$2^n+2^m=2^k+2^r.$$
We may assume that $n\lt m,\ $ $k\lt r,\ $ and $m\le r.$
Assuming that $m\lt r,$ we have
$$2^n+2^m\le2^{r-2}+2^{r-1}\lt2^r\lt2^k+2^r,$$
contradicting the assumption that $2^n+2^m=2^k+2^r.$ Hence $m=r$ and $n=k.$
