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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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    What is not clear? I know my English is very bad, but I'm working on it :)2012-12-17
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    Could you tell us what $X$ is?2012-12-17
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    I meant, the statement of the problem is clear, but you say you got problem on this exercise. My question what: which king of problems? Where are you stuck?2012-12-17
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    The statement of the problem is not clear. Or rather, it is clearly wrong, if the required conclusion $(X,d_{\phi})=(X,d)$ is read as $X=X$ (obviosly true) and $d_{\phi}=d$ (obviously false). My guess: The requirement is to show that $d_\phi$ and $d$ are *equivalent* metrics.2012-12-17
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    $X$ is a set. I know that i have to prove that $(X,d_\phi) \subset (X,d)$ and $(X,d) \subset (X,d_\phi)$, i.e. every open set in $(X,d_\phi)$ is open in $(X,d)$. @Harald Hanche-Olsen, yes, you're right, sorry for that :)2012-12-17
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    And my guess is confirmed, thanks! What a terrible abuse of the equality and inclusions signs. But all is clear now.2012-12-17
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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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