Proving that no natural number is equinumerous to a proper subset of itself.

Question

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?

I suppose that the mapping $f$ must be from $k$ to $k^{+}$, i.e. $k+1$. — Mauro ALLEGRANZA, Mar 27 '14 at 10:44
As per comment below (@Marc van Leeuwen) the number $k$, "modelled" as a set, is $k = { 0, 1, k-1 }$ and has $k$ elements. Thus, $k+1 = k \cup { k }$ has $k+1$ elements. We have that $k \in k+1$ but $k \notin k$. — Mauro ALLEGRANZA, Mar 27 '14 at 10:46
@MauroALLEGRANZA- Thanks! That answers my question perfectly. — algebraically_speaking, Mar 27 '14 at 10:50

score 0 · Answer 1 · answered Mar 27 '14 at 10:29

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 $k$).

Thanks! That was exactly the mistake I was making. – algebraically_speaking Mar 27 '14 at 10:50 — algebraically_speaking, Mar 27 '14 at 10:50

Proving that no natural number is equinumerous to a proper subset of itself.

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