Amanda wrote:

Hi,
I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in P
between x and y.

I tried to prove the contrapositive of this proposition. If S has no
subset with the given property, then, for every x in S there's a y in
S such that (x,y) doesn't intersect S. This means that every x in S is
squeezed between 2 open intervals in the complement of S. From this I
concluded S had no accumulation points. Since R is separable, every
set that doesn't have accumulation points is countable, which implies
S is countable. But this conclusion is false. But then I noticed I
hadn`t actually proved the contrapositive of the given proposition,
ths thing is more complicated. {0, 1, 1/2,...1/n...} is an example of
a set that doesn`t satisfy the prpoasition and, yet, has olne
accumulation point.

Like you said, each element of S is squeezed between two
open intervals. So there is at most as many elements of S as
there are open intervals in R. (And since there exists an injective
map of (open intervals in R) -> Q ... well...)