If $a\in\mathbb{R}$ how can $a$ have a successor?

Question

I'm trying to read the proof of the theorem:

If $A$ is well-ordered by $<$ and it's a subset of $\mathbb{R}$, then there is an order-embedding $f:A→ℚ$ and $A$ is countable.

If we knew that $A$ was countable since the beginning I wouldn't even ask this question, but since it's the conclusion of the theorem i don't understand. In the proof they say:

"For each $a\in A$, as long as $a$ is not the maximum element of $A$, $a$ will have a successor $a^+$ in $A$.

My question is, how can an element of a subset of $\mathbb{R}$ have a successor if $\mathbb{R}$ is uncountable? Is it because $A$ is well-ordered? but even if $A$ is well-ordered how can i know that it's also countable (without proving this theorem obviously and since the book I'm reading hasn't given any theorem for this)?

Answer 1

Yes, it's because $A$ is well-ordered.

This has nothing to do with countability, there are well-ordered sets that aren't countable, yet their elements still have successors.

Note that if $(E,<)$ is any well-ordered set (of any cardinality) and $a\in E$ is not a maximal element, then $\min\{b\in E\mid b>a\}$ is well defined (well-ordered means any nonempty subset has a minimum, here as $a$ is not maximal, this subset is nonempty), and is a successor of $a$.

Answer 2

$a$ does not have a successor in $\Bbb R$, but it does have a successor in $A$, precisely because $A$ is well-ordered.

Note that uncountably is not an obstruction to having successors; there are uncountable well-orders. It is the density of $\Bbb R$ that stops $a$ from having a successor in $\Bbb R$.

So far in the proof (as much as you have shown us), there is no assumption of countability, neither explicitly nor implicitly. My guess is that much of the proof goes to proving precisely that $A$ is countable.