Is there a bijection $f: X \to X$ such that $f(a) \succ a$ for some $a \in X$ and $f(x) \succeq x$ for all $x \in X?$

Question

$X$ is a prewellordered set with respect to $\succeq$. The prewellordering is in general not antisymmetric relation in contrast to a well-order, i.e. $x \succeq y$ and $y \succeq x$ doesn't imply $x=y.$ The map $f$ doesn't descend to the well-order induced by the prewellordering. The question looks simple at least for the case when prewellordering is a well-order but what are formal arguments?

Answer 1

$X=\omega\sqcup\omega=\left\{ \left(n,0\right)\mid n=0,1,\dots\right\} \cup\left\{ \left(n,1\right)\mid n=0,1,\dots\right\} $ ordered by:

$\left(n,0\right)\leq\left(m,0\right)$ , $\left(n,0\right)\leq\left(m,1\right)$ and $\left(n,1\right)\leq\left(m,1\right)$ for all $n,m\in\omega$.

Then $f:X\rightarrow X$ prescribed by $\left(0,0\right)\mapsto\left(0,1\right)$, $\left(n+1,0\right)\mapsto\left(n,0\right)$ and $\left(n,1\right)\mapsto\left(n+1,1\right)$ is a bijection.

This with $\left(0,0\right)<\left(0,1\right)=f\left(0,0\right)$ and $\left(n,m\right)\leq f\left(n,m\right)$ for all $n,m\in\omega$.