A complicated lemma from Munkres's Topology

Question

I don't understand the following lemma of the book Topology by J Munkres :

1- If a set $A$ with the mentioned properties exists there must be an example for it and so, it could help to understand both the lemma and the proof better? I tried some different types of uncountable subset of $\mathbb R$ but they don't work thus another question is

2- How to we know for sure (a proof) that there exists an $\Omega$ such that it is the smallest element of $C$ for which the section of $C$ by $\Omega$ is uncountable?

I would appreciate any simple clear detailed explanation.

PS. Definition. Let $X$ be a well-ordered set. Given $\alpha \in X$, let $S_{\alpha}$ denote the set $$S_{\alpha} = {\{x \ | \ x \in X \ \text{and} \ x < \alpha}\}.$$ It is called the section of $X$ by $\alpha$.

Answer 1

This lemma is demonstrating one of the more bizarre consequences of the fact that the axiom of choice lets us well-order any set. The consequence is that: There is a minimum uncountable ordinal.

Ordinal numbers are like cardinal numbers except they only apply to well-ordered sets, and give a finer distinction than cardinality for such sets. Two sets are of the same ordinal, if they are of the same cardinal and there exists an order isomorphism. That is, a one-to-one and onto function (thus showing their cardinal equivalence) that is also an order preserving map in both directions.

In this lemma, the ordinal of the set $A$ is the 2nd to least uncountable ordinal.

Answer 2

So $B$ is a well-ordered set that is uncountable. I think Munkres showed the existence of such a set before this lemma.

Define $C$ as $\{1,2\} \times B$ in the lexicographical ordering. This is also a well-order by previous results.

If $b \in B$ then for all $x\in B$, we have $(1,x) < (2,b)$ and so $S_{(2,b)}$ is uncountable.

So define $U$ to be the set of all elements $x$ of $C$ that have $S_x$ uncountable. We have just shown that $U$ is non-empty, so there is a minimum $m \in U$.

So $S_m = \{x \in C: x < m \}$ (order taken in $C$ of course) is uncountable, as $m \in U$. And any $x \in S_m$ has a countable section because otherwise it would contradict the minimality of $m$. So $A = S_m \cup \{m\}$ is the required uncountable well-ordered set (with $m$ as the max) in the inherited order from $C$.

These properties define $A$ up to order isomorphism (so it's essentially unique) and indeed it can be shown that no subset of the real line in its standard order can be isomorphic to it.