Let $ \;(E,\leq) \;$ be a well ordered set, and $ f \, : \, E \to E \; $ a strictly increasing bijection. Show that $ \; f=Id_E$.

Question

I thought of considering a sequence $(a_n)$ defined as follows : \begin{cases} a_0=\mathrm{min}(E) \\ a_{n+1}=\mathrm{min}(E \: \backslash\:\{a_0,\dots,a_n\}) \: , \: \forall{n}\in\mathbb{N} \end{cases} and showing that each element of $ \, E \, $ is an element of the sequence, and then use induction. But I think there can be a simpler way.

Answer 1

Hint: If $f$ is not the identity, then $A=\lbrace e\in E\ | \ f(e)\neq e\rbrace$ is not empty. Then, consider the minimum element $a$ of $A$. Show that $a$ cannot be attained by $f$ at any point but $a$.