I am self-taught and I'm trying to learn about topology with the Book of James R. Munkres. In the reading I came across with this Lemma (Lemma 10.2 for tor those who have the book):
" There exists a well-ordered set $A$ having a largest element $\Omega$, such that the section $S_\Omega$ of $A$ by $\Omega$ is uncountable but every other section of $A$ is countable."
The book defines, for a well-ordered set $X$, a section of $X$ by $\alpha$ as the set $S_\alpha=\{x|x\in X \text{ and } x<\alpha\}$
Because the proof is an important part of my doubt, I transcribe it below word by word:
"We begin with an uncountable well-ordered set $B$. Let $C$ be the well-ordered set $\{1,\,2\}\times B$ in the dictionary order; then some section $C$ is uncountable. (Indeed, the section of $C$ by any element of the form $2\times b$ is uncountable.) Let $\Omega$ be the smallest element of $C$ for wich the section of $C$ by $\Omega$ is uncountable. Then let $A$ consist of this section along with the element $\Omega$."
So, this blow my mind and I'm trying to follow the demonstration with a speciic example. I'm thinking like there is nothing in the theorem or in the proof that forbids taking $B=[0,1]\subset\mathbb R$, with the restriction of the usual order (then $B$ is well-ordered). In this way $\Omega$ can't be of the form $(2,b)$, with $b\in[0,\,1]$, because every element of $C$ of the form $(1,\,b')$, with $b'\in[0,\,1]$, will be smaller. Then, as the proof goes, there is an element $\Omega=(1,\,b')$, for wich the section $S_{\Omega}$ is uncountable, but for $\Omega'=(1,\,b'')$, with $b''<b'$, $S_{\Omega'}$ will be countable. So, for me, it sounds like, there is a continuum interval $[0,b'')$ wich is countable, and as far as I know this is impossible. So I think I'm misunderstanding something here.
As always, I thank you in advance for any help.
$[0,1]$ is totally ordered, but not well-ordered (in the standard order). That forbids you from taking $B=[0,1]$.
The definition of well-order is that any non-empty subset has a smallest element. In other words, that any non-empty subset is bounded below and contains its own $\inf$. This is not the case for $[0,1]$.
I suggest you look up ordinals to see what well-orders look like; any well-ordered set is essentially (read: order isomorphic to) an ordinal. It's not so important for you to know and understand their (standard) construction at this stage, but see and learn how they're ordered.
The subset $$ \{(i,b)\in \{1,2\}\times B\mid S_{(i,b)}\text{ is uncountable}\}\subseteq \{1,2\}\times B $$ is non-empty, and therefore has a smallest element. It could be an element of the form $(1,b)$, or it could be $(2,0)$.