Can we define a metric $d$ on $(0,1)$ such that the topology induced by this metric is same as that of the usual Euclidean metric on this set ?
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We'll you certainly can't do it in a way that would make $(0,1)$ complete, since then $(0,1)$ would be complete with the usual metric. – Tim Raczkowski Mar 03 '15 at 14:51
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2@ Tim Raczkowski : Why ???!! completeness is not a topological property .... – Mar 03 '15 at 14:51
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The problem with using the usual metric is that a sequence could be approaching $0$ or $1$, but it would not converge in $(0,1)$. So you need to define a metric that makes $0$ or $1$ "infinitely far away" (loosely speaking) from any point in $(0,1)$, but that preserves open sets. You can do this by finding a continuous map from $(0,1)$ onto the reals, and applying the standard metric on the mapped values.
