Sierpinski's theorem Sierpinski's theorem:
Every countable metric space without isolated points is homeomorphic to $\ {Q}$
Use it to prove:
a) Every dense and countable subspace of $ (\ {R}, τ_u) $ is homeomorphic to $\ {Q}$
b) With the usual topology, prove that $\ {Q}$ and $\ {Q^2}$ are homeomorphic
No idea to prove it. Any ideas?

Given Sierpiński's theorem, just show that if $A$ is dense  in $(\Bbb R, \tau_u)$ (the upper topology? with base all sets $[a,b), a < b$?) $A$ does not have any isolated points. Suppose $\{a\} = O \cap A$ for some open set $O \in \tau_u$ and $a \in A$, but then $A$ does not intersect the open set $O\setminus \{a\}$ contradiction. This only uses that $\tau_u$ must be $T_1$ and not have isolated points! So it works for the Euclidean, upper and lower topology.
So in fact if $A$ is dense in any space $X$ that is $T_1$ without isolated points then $A$ has no isolated points in its subspace topology either.
And if $A$ is moreover countable, Sierpiński tells us $A \simeq \Bbb Q$.
(b) is a special case for $X=\Bbb R^2$.
