(Recall that a set is well-ordered when every nonempty subset has a least element. That is, S is well-ordered when every nonempty subset S' of S has an element x smaller than every other element of S'. Every finite set is well-ordered. The classic example of an infinite well-ordered set is {1,2,3,...}, which is infinite but of course only countable.)

**A**
No.

We shall show that any well-ordered set S of real numbers
is countable by writing S as a countable union of subsets
S_{1},S_{2},S_{3},..., each of which is countable.
(NB we allow finite, even empty, sets to be “countable”.)

For any x in S, define s(x) to be the successor of x, i.e., the smallest
element of S greater than x. By our hypothesis on S, this s(x) exists
unless x happens to be the largest element of S, in which case we set
s(x)=x+1.
Now define d(x) = s(x) - x, a positive real number.
Thus if we let S_{n} be the subset of S consisting of
x such that d(x)>1/n then each element of S is contained in some
(indeed in all but finitely many) of the subsets S_{n}.
Hence S is the union of these subsets. But S_{n} is clearly countable
because |x-x'|>1/n for all distinct x,x' in S_{n}.
Hence S is also countable, as claimed.

*Added July 2006:* This topic arose on sci.math and
Victor Meldrew noted an even simpler proof:
for each x in S, choose a rational number r(x) in the interval (x,s(x));
these are distinct, so we have an injection of S into the set
**Q** of rational numbers, which is countable.
[Easy exercise: as written, this proof implicitly invokes
the Axiom of Choice; explain why in fact Choice is not needed
to specify such a map r from S to **Q**.]

V.Meldrew also notes that by induction *any* totally ordered
countable set has an order-preserving imbedding
into **Q**, and thus into **R**;
in particular this is true for any well-ordered countable set
(a.k.a. countable ordinal).