Can one endow the unit interval $[0,1]$ with a group operation to make it a topological group under its natural Euclidean topology?
Is $[0,1]$ a topological group?
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$\begingroup$
general-topology
topological-groups
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0But one can endow $(0,1)$ with such a topology... – 2012-12-18
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0@N.S.: With a topology or an operation? – 2012-12-18
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0@AsafKaragila Ty, operation... I need my coffee :) – 2012-12-18
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0Couldn'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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0Any 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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0A 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
1 Answers
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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.)