Munkres Topology , minimal uncountable well ordered set This question is from Munkres Topology, page 67:

Let $S_{\Omega}$ be the minimal uncountable well ordered set. Since there is no largest element in $S_{\Omega}$, every element in $S_{\Omega}$ has immediate successor. Let $s(b)$ denotes immediate successor of $s\in S_{\Omega}$. So, if $a\in S_{\Omega}$, then $A=\{s(a),s(s(a)),s(s(s(a))),\dots\}$ is countable. But $B=\{x\in S_{\Omega}|a<x\}$ is uncountable.

Why isn't $A\ne B$?
 A: I can see how this may be confusing to someone not familiar with ordinals, or more precisely $\omega_1$. Let me try to explain:
Let $\prec$ be the well-order on $S_\Omega$. First let us prove that every element in $S_\Omega$ actually has an immediate successor:
Fix $a \in S_\Omega$. Then $a_\downarrow :=\{x \in S_\Omega \mid x \prec a\}$ is well-ordered by $\prec \restriction a_\downarrow$. Now $a \in S_\Omega \setminus a_\downarrow$ and by the minimality of $S_\Omega$ we have that $a_\downarrow$ is at most countable. Thus $S_\Omega \setminus a_\downarrow$ is uncountable and therefore $S_\Omega \setminus ( a_\downarrow \cup \{ a \} )$ is nonenmpty. Let $b \in S_\Omega \setminus ( a_\downarrow \cup \{ a \} )$ be minimal with respect to $\prec$. Then $x \prec b$ for all $x \prec a$ and thus $a \preceq b$. Since $b \neq a$ we have that $a \prec b$ and $b$ is minimal with this property. Thus $b$ is the immediate successor of $a$.
Next, let $s \colon S_\Omega \to \Omega$ map every $x \in S_\Omega$ to its immediate succesor $s(x)$ with respect to $\prec$. Fix $a \in S_\Omega$ and let $A = \{s(a), ss(a), sss(a), \ldots \}$. Write $s^n(a)$ for $\underbrace{ss\ldots s}_{n \text{-times}}(a)$. Then $f \colon \mathbb N \to A, n \mapsto s^n(a)$ is a surjection and therefore $A$ is countable.
It remains to be seen that $a_\uparrow := \{ x \in S_\Omega \mid a \prec x \}$ is uncountable. Note that $S_\Omega = a_\downarrow \cup \{a\} \cup a_\uparrow$. We already know that $a_\downarrow \cup \{a\}$ is countable. If $a_\uparrow$ were countable, then $S_\Omega$ would be the union of two countable sets and therefore countable. But $S_\Omega$ is uncountable and therefore $a_\uparrow$ has to be uncountable.
A: Instead of $S_\Omega$ (which is rather difficult to visualize), let's look at a simpler well-ordered set: $S=\mathbb{N}\cup\{\infty\}$, ordered such that $\infty$ is greater than every element of $\mathbb{N}$.  Take $a=0$; then $A$ is the set of all elements of $S$ which can be obtained by taking successors of $a$ over and over some finite number of times.  That is, $A=\{1,2,3,4,\dots\}=\{n\in\mathbb{N}:n>0\}$.  On the other hand, $B=\{n\in\mathbb{N}:n>0\}\cup\{\infty\}$, since $\infty$ is also greater than $0$, even though it cannot be obtained by repeatedly taking successors.  So $A\neq B$ in this case.  Thus in general, it is possible that $A\neq B$, and in fact that cardinality argument given by Munkres shows that for $a\in S_\Omega$, we must always have $A\neq B$ since $A$ is countable and $B$ is uncountable.
