How to prove that $x\leq f(x)$ if f order isomorphic?

Question

Let X is a well ordered set and $f:X\to f(X)=Y\subseteq X$ is order isomorphic. For any $x\in X$ prove that $$x\leq f(x)$$

since X is well ordered, {x,f(x)}$\subseteq$X and $x\leq f(x) $ or $f(x)\leq x$. Assume $f(x)<x$ . since f is order isomorphic $f(x)<x \Rightarrow f^{-1}(f(x))<f^{-1}(x)$ so $x<f^{-1}(x)$

edit : since $B\subseteq X$ B has a minimum element "a". from definition of B $f(a)<a$ .and $f(f(a))<f(a)$ so f(a) is in B. since $f(a) <a$ and $f(a)\in B$ this conflicts "a" is the minumum element of B

Answer 1

HINT:

Suppose that $A=\{x\in X\mid f(x)<x\}$ is non-empty, then it has a minimal element $a$, now use the fact that $f$ is injective and order preserving to conclude that $f(x)\in A$, and derive a contradiction.