0

How can I check whether $\mathbb{Q}$ and $\mathbb{Z}$, with their usual topologies inherited from $\mathbb{R}$, are homeomorphic?

QED
  • 12,944

2 Answers2

5

No, since $\mathbb{Z}$ is discrete, but $\mathbb{Q}$ is not. Suppose that there exists a homeomorphism $f:\mathbb{Q}\rightarrow\mathbb{Z}$, $f^{-1}(0)$ is open since $\{0\}$ is open in $\mathbb{Z}$, any non empty open subset of $\mathbb{Q}$ contains more than one element. Contradiction.

M. Winter
  • 30,828
Tsemo Aristide
  • 89,587
2

From a more sophisticate point of view, it is possible to show that $\mathbb{Q}$ can be characterized topologically as the unique countable metrizable space without isolated points. See

W. Sierpinski: Sur une propriété topologique des ensembles dénombrables denses en soi, Fundamenta Mathematicae 1 (1920), 11-16.

In particular, $\mathbb{Q}$ cannot be homeomorphic to $\mathbb{Z}$, whose points are all isolated.

Francesco Polizzi
  • 4,520
  • Something must be missing, because $\Bbb Q^2$ has the same properties. – M. Winter Nov 15 '17 at 10:08
  • Nothing is missing. In fact, $\mathbb{Q}^2$ is homeomorphic to $\mathbb{Q}$, see https://math.stackexchange.com/questions/355779/showing-mathbbq-is-homeomorphic-to-mathbbq2 – Francesco Polizzi Nov 15 '17 at 10:11
  • Okay, this is definitely surprising. Unfortunately the only answer to this question takes this result as given and the cited proof is a dead link. Do you have another source. What about $S^1\cap\Bbb Q^2$? – M. Winter Nov 15 '17 at 10:13
  • 1
    Okay I found the proof (and updated the link). Thank you for pointing out this remarkable result. – M. Winter Nov 15 '17 at 10:17
  • It is a old theorem by Sierpinski. I added the original reference in the answer. – Francesco Polizzi Nov 15 '17 at 10:18