I have the following exersice:
Let $A$ be a set and $f: A \to \mathbb N$ a bijective function. Show that a set $B\subset A, B\neq A$ and a bijective function $g: A\to B$ exists.
Let $A$ be any set and $F: A\to\mathbb N$ a bijective function.
Claim: $\exists B \subset A, B\neq A$ s.t. $\exists g: B\to A$ which is bijective.
Proof: Let $A$ be an arbritary set and $f$ as above. We define $B$ to be the set that contains every second element of $A$. Since $f$ is a bijection, we have $|A|=|B|=|\mathbb N|=\infty$ (1)
let now $g$ be a function from $B$ to $A$: $g:B\to A$
Since $|A|=|B|$ it follows, that $g$ is bijective.
Does my proof hold like that? E.g. I'm not sure if $A$ could be the empty set.