I was reading this pdf where I came across this proposition and its proof: Proposition 2: Let $S \subset \mathbb N$ be an infinite subset. Then there exists a bijection between $S$ and $\mathbb N.$ Proof: You might think this is intuitively clear. But let’s prove it carefully. We define a...

Let $B=\{n\in\Bbb N:f(n)<n\}$. If there is some $n\in\Bbb N$ such that $f(n)<n$, then $B\ne\varnothing$, so $B$ has a least element $m$. Clearly $m>0$ (why?), so $m=k+1$ for some $k\in\Bbb N$. Then $k\notin B$, so $f(k)\ge k$. But $f(m)<m$, so $f(m)\le k$, and therefore $f(m)\le f(k)$, contradict...