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I've got problem with following: $\phi : [0,\infty) \to \mathbb{R}$ is non-decreasing, concave function. Such that $\phi (0) =0$, and $\phi (u) >0$ for $u>0$. Prove that if $\phi$ is continuous at $0$ then $\mathcal{T} (d_{\phi}))=\mathcal{T} (d))$, where $d$ is a metric on $X$ and $d_{\phi}(x,y) = \phi (d(x,y))$.

I can prove that $d_\phi$ is metric on $X$. I've stucked with above problem.

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    I suggest you change the problem title to the more accurate “Equivalence of metrics”.2012-12-17

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You can do it by proving: For any $\varepsilon>0$ there is a $\delta>0$ so that $d(x,y)<\delta$ implies $d_\phi(x,y)<\epsilon$ – and also $d_\phi(x,y)<\delta$ implies $d(x,y)<\epsilon$. (For your sanity's sake, concentrate on one of these at a time.)

These are not necessary requirements for the metrics to be queivalent, but they are sufficient. And in this case, you can prove them without too much hazzle, I hope.

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    Yes, that should work. (Did I really write queivalent? Well, it's such a cute misprint I'll leave it as is.)2012-12-18