Order types of positive reals Suppose one has a set $S$ of positive real numbers, such that the usual numerical ordering on $S$ is a well-ordering. Is it possible for $S$ to have any countable ordinal as its order type, or are the order types that can be formed in this way more restricted than that?
 A: To complete the picture (the obvious remaining part). If ${S\subset\mathbb R}$ is well ordered, then it is countable: indeed it has countable cofinality. Thus well-ordered subsets of R are exactly countable ordinals.
A: Yes, one can have any countable ordering. Indeed any countable totally
ordered set can be embedded in $\mathbb{Q}$. Write your ordered set as
$ \lbrace a_1,a_2,\ldots \rbrace $
and define the embedding recursively: once you have placed $a_1,\ldots,a_{n-1}$
there will always be an interval to slot $a_n$ into.
A: You can get any order type. Let's assume you can get all order types up to but not including alpha, using subsets of (0,1]. If alpha=beta + 1 then squash your representation of beta and add an extra point. If alpha is a limit ordinal, choose a sequence of ordinals that converges to alpha and put the first one into (0,1/2], the second into (1/2,3/4] etc. and the result will have order type alpha.
A: Using wellorderings of positive reals is actually the standard way to construct an Aronszajn tree.
