Can one endow the unit interval $[0,1]$ with a group operation to make it a topological group under its natural Euclidean topology?

  • $\begingroup$ But one can endow $(0,1)$ with such a topology... $\endgroup$ – N. S. Dec 18 '12 at 20:07
  • $\begingroup$ @N.S.: With a topology or an operation? $\endgroup$ – Asaf Karagila Dec 18 '12 at 20:09
  • $\begingroup$ @AsafKaragila Ty, operation... I need my coffee :) $\endgroup$ – N. S. Dec 18 '12 at 20:12
  • $\begingroup$ Couldn't you define an operation on $Y=(-1,1)$ by the homeomorphism $f:\mathbb R\to Y,\ x\mapsto x/(|x|+1)$? It is possible to translate the operation on $\mathbb R$ onto $Y$, right? $\endgroup$ – Stefan Hamcke Dec 18 '12 at 20:19
  • $\begingroup$ Any continuous bijection between $\mathbb R$ and $(0,1)$ should do the trick.. But there is probably something subtle happening, I think: the topology would be the Euclidian topology, but the uniformity which defines this topology is not the same.... $\endgroup$ – N. S. Dec 18 '12 at 20:27

No. A topological group is homogeneous, and $[0,1]$ is not, since it has the two endpoints. (An open neighborhood of one of the endpoints, like $[0,1/2)$, is not homeomorphic to any open neighborhood of an interior point via a homeomorphism mapping $0$ to the interior point.)

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