Is $\Bbb Q$ homeomorphic to $\Bbb Q^2$?

Question

It's an easy excercise in set theory to exhibit a bijection $\Bbb Q \cong \Bbb Q\times \Bbb Q$. However, none of the bijections I'm aware of respect the topologies on $\Bbb Q$ and $\Bbb Q^2$, inherited from their respective embeddings into $\Bbb R$ and $\Bbb R^2$.

Therefore, I'm asking whether there exists a homeomorphism $\phi: \Bbb Q^2 \to \Bbb Q$.

I don't believe that there is any such map, but since the standard techniques of algebraic topology don't enable one to discern between $\Bbb Q$ and a discrete space, I wasn't able to prove it. Maybe Cech cohomology provides a means to attack this problem, but I haven't even got the slightest Iiea how to calculate $H^1(\Bbb Q,\Bbb Z)$.

Can't you just take the usual product topology from the embedded topology on Q? — Mathemagician1234, Feb 09 '15 at 21:55
Here's the answers you're looking for:
http://mathoverflow.net/questions/26001/are-the-rationals-homeomorphic-to-any-power-of-the-rationals — Mathemagician1234, Feb 09 '15 at 22:02
There might be better duplicates. This was the first one I found. — Asaf Karagila, Feb 09 '15 at 22:14

score 3 · Accepted Answer · answered Feb 09 '15 at 21:56

3

The set of the rational numbers with its usual topology is the unique countable metrizable space without isolated points.

Can you use that property to conclude?

answered Feb 09 '15 at 21:56

Mariano Suárez-Álvarez

135,076

2

Of course, you need to prove that characterization of $\mathbb Q$ :-) – Mariano Suárez-Álvarez Feb 09 '15 at 21:58
Thanks. However it would be also nice to see an explicit construction. – Dominik Feb 09 '15 at 21:59
Well,you were considering to use Cech cohomology... – Mariano Suárez-Álvarez Feb 09 '15 at 22:00

Is $\Bbb Q$ homeomorphic to $\Bbb Q^2$?

1 Answers1