John Baez in litteris <email@example.com> scripsit: > It's easy to map the ordinal omega^2 into the real numbers > in a one-to-one and order-preserving way. Here's an artist's > conception, which uses the second dimension to make things > easier to see: > > http://math.ucr.edu/home/baez/omega_squared.png
If you're asking which ordinals are order-isomorphic to a subset of the real numbers, the answer is simple (at least, assuming the axiom of choice): exactly the countable ordinals. First, any well-ordered subset of the reals is countable because between any element of the subset and the next ("the next" makes sense, of course, since the set is well-ordered) there is a rational. Conversely, we can prove by transfinite induction that any countable ordinal can be embedded in the reals: it is true of 0, if it is true of alpha it is true of alpha+1, and if it is true of every ordinal alpha<delta for a limit ordinal delta, choose an increasing sequence alpha_n leading up to delta, embed each alpha_n between (n-1)/n and n/(n+1) and put them together... the details are left to the reader (it may be necessary to remove some elements since we took the sum of the alpha_n rather than the limit, but it can also be arranged so that the two coincide).
In fact, every countable ordinal can be embedded in the reals as a closed set (then the embedding is a homeomorphism from the ordinal with the order topology to the subset of the reals with the induced topology), or as a discrete set, but obviously not both (except up to omega).
[I won't bet on what happens in the absence of Choice, but it wouldn't at all surprise me it were consistent that some large countable ordinals can't be embedded in the real line.]
I had produced a graphical representation of epsilon_0, once, but it's actually entirely uninteresting to look at, it's just a mess.