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

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    But one can endow $(0,1)$ with such a topology...2012-12-18
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    @N.S.: With a topology or an operation?2012-12-18
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    @AsafKaragila Ty, operation... I need my coffee :)2012-12-18
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    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?2012-12-18
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    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....2012-12-18
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    A continuous homomorphism is uniformly continuous and an isomorphism between topological groups is an isomorphism between uniform spaces. So if we equip $(-1,1)$ with the algebraic structure of $\mathbb R$, then the uniform structure of the open interval induced by the new operation is finer from the usual uniformity inherited from $\mathbb R$, since $f^{-1}$ is not uniformly continuous.2012-12-18

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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.)