Proving that no natural number is equinumerous to a proper subset of itself. I have to prove that no natural number is equinumerous to a proper subset of itself. 
Let $f$ be a mapping from $k^+$ into $k^+$, where $k.k^+\in \Bbb{N}$. My book says 

If the set $k$ is not closed under $f$, then $f(p)=k$ for some number $p$ less than $k$.

I wonder why $p$ has to be less than $k$. If $k$ is not closed under $f$, it just means that the image of at lest one element in $k$ does not belong to $k$. For that to remain true, the inverse image of $k$ need not be less than $k$! It can also be $k$.
Where am I going wrong?

It is always confusing to consider numbers as sets, but I suppose that here $k=\{\, p\in\Bbb N\mid p<k\,\}$ and $k^+=k+1=k\cup\{k\}$. Then $f$ being a map to $k^+$, every image under $f$ is either an element of $k$ or equal to $k$. Closure of $k$ under $f$ means $\forall p\in k:f(p)\in k$, and so non-closure of $k$ under $f$ means $\exists p\in k:f(p)\notin k$. But that just means $\exists p\in k:f(p)=k$, as stated. There is no consideration involved of the inverse image of $\{k\}$ (which indeed might contain the element $k$).
