in the theorem :
given function $f:A \to A$, with $c∈A−ran f$, and function $h:ω→A$ recursively defined as $h(0)=c$ and $h(n^+)=f(h(n))$. if $f$ is one-to-one, then $h$ is one-to-one as well
why must $c \notin ran f$ ? the formal symbolic proof is easy to follow, but i'm failling to see the intuition. any clarification as to why this is necessary is welcome
On
after thinking it through a bit more, it seems that the way the theorem presented it, is sort of "putting the cart before the horse". it seems to me more proper to say that (for this theorem) if $f$ is one-to-one and $h(0)$ is set to the val $a$ for our recursively defined $h$, then $a$ will not appear in the resultant $ran f$. for if it did, then it would not be one-to-one, since one-to-one functions "visit" each value in the range only once and this $h(0)$ denotes an already implicitly "visited" value of $f$. could anyone else corroborate this view ? i hope i'm not confusing myself even further
You can have a one-to-one function $f$ such that $f(f(x))=x$ for all $x$. Example: $A=\mathbb R$ and $f(x)=-x$. Take $c=0$ in this example. Then $h(n)=0$ for all $n$, so $h$ is not one-to-one. When $c$ is not in the range of $f$ we can use the fact that $f(c), f(f(c)),f(f(f(c))),...$ are all distinct and that is why the proof works.
[For example if $f(f(c))=f(f(f(c)))$ then $c=f(c)$ but this cannot happen if $c$ is not in the range of $f$].