Initial segments of a well-founded connex relation are not necessarily finte.

Question

Good evening, any solutions or help on this one?

Let $\prec$ be a well-founded relation on a set $X$ such that the relation $\preceq$ defined by $x\preceq y$ iff $x\prec y ∨ x = y$ is a total order. Show it need not necessarily make the set $\{x\in X|x\prec y\}$ finite for all $y\in X$.

Answer 1

Take $X=\Bbb N\cup\{\infty\}$ and define $n<\infty$ for all $n\in\Bbb N$. Then $X,<$ is well-founded and $X,\leq$ is totally ordered, but $\{x\in X|x<\infty\}=\Bbb N$ is infinite.

Answer 2

Let $(W,\prec)$ be a well-ordered set. Fix $x\notin W$. Extend the definition of $\prec$ so that $w\prec x$ for every $w\in W$. Then $(W\cup\{x\},\prec)$ is a well-ordered set. Furthermore, the initial segment before $x$ is $\{\xi:\xi\prec x\}=W$.

As Fabio pointed out, by letting $W$ be any infinite set we obtain the desired result.