Let $f : ℕ → P(ℕ)$ be given by $f(n) = {k*n | k ∈ ℕ}$. (P(ℕ) is the power set of the set A.)
Is f injective?
The answer to this question goes like this:
$f(n) = (n,2n,3n,4n,5n...)$
$min({ n,2n,3n,...})$ is $n$, as $n<2n<3n<....$
Suppose $f(m)=f(n)$. Then
${ (m,2m,3m,...)}={ (n,2n,3n,...)}$
smallest element of $( m,2m,3m,...)$ = smallest element of $(n,2n,3n,...)$, $m=n$.
My question is why do we focus on the smallest element of the set?
We focus on it because it produces the contradiction we need. If you claim two sets are equal and I can show the smallest elements are different, it shows you are wrong, because I have identified and element that is in one set and not in the other. There may be other ways of showing the two sets are different, but this one works.