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Let $(M,d)$ be a metric space and $f\colon[0,\infty)\to[0,\infty)$ metric preseving map that is right continuous at $0$, i.e. $f$ has satisfies $$\forall x,y\in [0,\infty)\colon f(x+y)\le f(x)+f(y)\quad\text{and}\quad f(x)=0\iff x=0.$$ Furthermore, $f$ is non-decreasing and addition $$f(0^+)=\lim_{x\downarrow0}f(x)=f(0).$$

Then the composition $\Delta=f\circ d$ is again a metric on $M$. But how can I show $$(M,d) \text{ is complete}\iff (M,\Delta) \text{ is complete}?$$

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    The domain of $f$ is $[0,\infty)$ but the Triangle Inequality is expressed for all $x,y\in M$. Did you mean for all $x,y \ge 0$?2012-03-23
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    @bbm: Patrick was pointing out something else. There's no need to mention that $f$ is non-decreasing since this follows from the other conditions (when corrected as Patrick proposed).2012-03-23
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    @Brian M. Scott: I think that the rightcontinuity at $0$ is does the trick, since $f$ is non-negative, $f(0)=0$ and f is right continuous at $0$. $\exists \delta>0$ such that $x\in[0,\delta)\implies f(x)<\epsilon.$ So if $\Delta(x_m,x_n)<\epsilon$ is small, then so is $f(d(x_m,x_n))$. The previous comments then imply $d(x_m, x_n)$ is small. How can I write this down in a neat way?2012-03-24

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