Uncountable sets and interval topology

Question

Let be an uncountable well-ordered set $(X,\leq)$. We suppose that exists some $x\in X$ such that $y < x$ for uncountably many values of $y$ and define $x_1$ as the minimum $x\in X$ such that $y < x$ for uncountably many values of $y$. Let be a function $f$ such that $f(x)=0$ for $x\ne x_1$ and $f(x_1) = 1$. For the order topology on $X$ (basic open sets are open intervals), show that f is not continuous, but for every convergent sequence $u_n\to u$ in X, $f(u_n)$ converges to $f(u)$.

Answer 1

The image set is discrete, so {1} is open. The inverse image is $\{x_1\}$ that isn't open in the interval topology.

But $f$ is sequentially continuous:

The only interesting case is when $\forall n\in{\Bbb N}:\ u_n<x_1$. For a such sequence, if $\exists l=\lim_{n\to\infty} u_n$, then $l<x_1$. Namely, let be $I_n=\{x\in X\vert x\le u_n\}$ and $I=\cup_{n\in{\Bbb N}}I_n$. $I$ is countable because is a countable union of countable sets (countable choice is required here) and an interval, so $l\le\sup I<x_1$.