Statement regarding the restriction of a function

Question

All terminology below is related to Set Theory.

Definition: Let $f$ be a function and $n∈N$. We say that $f$ is of order $n$ if the inverse image of each element from the range has at most $n$ corresponding elements from the domain.

Note: $f$ is a functional relation if $\forall x,y,y'(\langle x,y \rangle \in f \text{ and } \langle x,y' \rangle \in f \Rightarrow y=y')$.

Problem: If $f$ is a function of order $n, n > 0,$ and $A⊆Dom(f)$ then the restriction of $f$ to $f^{-1}[f[A]]$ \ $A$ is of order $n - 1$.

*This statement is quite obvious if I set for example $n=2$. However I can't seem to find a mathematical proof that shows the validity for $∀n∈N$.

Any ideas and suggestions would be greatly appreciated.

Answer 1

I think that you’re on the right track in your long comment to Daniel, but it’s not entirely clear. First, everywhere that you’ve written $f^{-1}[A]$, you actually mean $f^{-1}[f[A]]$. Then you don’t actually need two cases: it doesn’t matter whether $f^{-1}[\{y\}]\setminus A$ is empty or not. All that matters is that it isn’t all of $f^{-1}[\{y\}]$.

I’d be inclined to organize it something like this.

Let $D=f^{-1}[f[A]]\setminus A$. Let $y\in f[A]$, and let $F=f^{-1}[\{y\}]$. By hypothesis $|F|\le n$, and we want to show that $|F\cap D|\le n-1$. But $F\cap D=F\setminus A$, and $F\cap A\ne\varnothing$, so ... ?