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Is the following statement is true?

$(0,1)$ with the usual topology admits a metric which is complete?

My answer is "False." But, the answer given is "True". I am unable to figure out. Please help me.

Thanks.

user354408
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    Do a little lateral thinking. What complete metric space could $(0,1)$ be homeomorphic to? – hardmath Jul 16 '16 at 11:08
  • @hardmath Is completeness perserved under homeomorphisms? – Burrrrb Jul 16 '16 at 11:34
  • @Asemismaiel: Yes, if the metrics are equivalent. Note the question asks about a selection of a metric that makes $(0,1)$ complete. So the metric changes, but convergence is conserved by the homemorphism. – hardmath Jul 16 '16 at 15:59

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Hint: inspired by the following thread, try to pull back $\mathbb{R}$ to $(0,1)$ by means of a homeomorphism. For a suitable choice, you will get a complete metric on $(0,1)$ that produces the same topology as the euclidean one.

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Siminore
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  • @ Siminore : Accordinng to you, we have a metric on $(0,1)$ which induces the usual topology on $(0,1)$ which is also induced by the euclidean metric on $(0,1)$ which is not comple. so what is the appropriate answer : $(0,1)$ is complete or not with the usual topology. – Struggler Jul 17 '16 at 09:23
  • Completeness is not a topological property. You should not speak of completeness in a given topology, but with a given metric. – Siminore Jul 17 '16 at 09:37
  • @ Siminore : Please tell me in which metric space $(0,1)$ is complete and induce the usual topological spaces. – Struggler Jul 17 '16 at 10:35
  • @Struggler you may start from the fact that $(-\pi/2,\pi/2)$ is homeomorphic with $\mathbb{R}$ via $x \mapsto \arctan x$. – Siminore Jul 17 '16 at 11:14
  • Like $(0,1)$, is there is a metric defined on $[0,1]$ which induces the usual topology on $[0,1]$ and $[0,1]$ is incomplete. – Struggler Jul 18 '16 at 04:45