Existence of well-order on an arbitrary infinite set $X$ such that $X$ has a largest element.

Question

One can prove that every set $X$ can be well-ordered. Can we also prove (or disprove) that for any arbitrary infinite set $X$ there exists a well-order such that there is a largest element, that is, there exists $x \in X$ such that $\forall y \in X: y \leq x$?

Answer 1

Choose an ordinal $\alpha$ of cardinality $|X|$. Since $X$ is infinite, $|X|+1=|X|$, so $|\alpha+1|=|X|$.

Then choose any bijection between $\alpha+1$ and $X$. This gives a well ordering on $X$ with a largest element. (That corresponding to the element $\alpha$ of $\alpha+1$.)

Answer 2

Say $a\in X$. Well-order $X\setminus\{a\}$, then add $a$ at the end.