Is the cardinality of this well-ordered set $\aleph_0$? Given a well ordered set $S$, any element $a\in S$ except the least element of $S$ has a unique predecessor, is the cardinal of $S$ equal to $\aleph_0$?
 A: Assuming that you want each element of $S$ to have an immediate predecessor, and that you want $S$ to be infinite, then yes. To show this, we need to construct a bijection with $\omega$; define $f:\omega \to S$ inductively by setting $f(0)$ to be the least element of $S$ and $f(n + 1)$ to be the least element of $S$ greater than $f(n)$. This is clearly injective, so it remains to show that it's surjective.
Suppose $a \in S$ is not in the range of $f$. Then we can define a new function $g:\omega \to S$ by taking $g(0) = a$ and $g(n+1)$ to be the immediate predecessor of $g(n)$ in $S$. By induction, $g(n)$ is not in the range of $f$ for any $n$, so in particular $g(n)$ is not the least element of $S$, and so the necessary predecessors exist. But then the range of $g$ is a subset of $S$ with no least element, contradicting the assumption that $S$ is well-ordered. So no such $a$ exists, so $f$ is a bijection.
As a side note, $f$ is actually an order-isomorphism - $f(n) < f(m)$ whenever $n < m$. So not only does $S$ have cardinality $\aleph_0$, it has the smallest possible order type of that cardinality.
