problem in the demo of an equinumerosity I'm trying to understand the proof of a theorem about equinumerosity of 2 sets, but I am facing to a problem. Here is the summary of my issue: > Let $h$ be a bijective function from $E$ to $h(E)$ with the particularity that $h(E)\subset E$, $h^0(E)=E$, > > Let $h^{n+1}(E) = h(h^n(E))$ be the sequence of sets given by the image of $E$ by $h$ with the particularity that $h^{n+1}(E)\subset h^n(E)$. > > Finally, let $(A_n)$ be a sequence defined by $A_0=E \setminus h(E)$, $A_{n+1}=h(A_n)=h^n(E)\setminus h^{n+1}(E)$. The proof continues, but I don't understand why $h(A_n)=h^n(E)\setminus h^{n+1}(E)$. To reduce the problem, and because a recurrence is possible, let's consider only the first iteration of the equality: $$ h(E\setminus h(E))=h(E)\setminus h^2(E) $$ Alone, I would not have produced this equality. Why is it true? What are the requirements concerning h? thank you, lowley

If we think about the set $E \setminus h(E)$, this is the set $\\{ x : x \in E \mbox{ and } x \
ot\in h(E) \\}$. So if we apply $h$ to this set, we see that for such an $x$, $h(x) \in h(E)$ (obviously), but we cannot have $h(x) \in h^{2}(x)$ (since $x$ was not in $h(E)$). This uses the fact that $h$ is one-to-one; by definition, an element $y$ is in $h^{2}(E)$ if there exists some $x$ in $h(E)$ with $h(x) = y$; since $h$ is one-to-one, you are guaranteed that there is not an $x^{\prime} \
eq x$ (and so possibly not in $h(E)$) that also has $h(x^{\prime}) = y \in h^{2}(E)$.

This shows that $h(E \setminus h(E)) \subset h(E) \setminus h^{2}(E)$. Now you just need to show the reverse inclusion to finish the proof.