How to preserve completeness between different metrics on the same space? 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 $$(M,d) \text{ is complete}\iff (M,\Delta) \text{ is complete}?$$
 A: You simply need to show that $\langle M,d\rangle$ and $\langle M,\Delta\rangle$ have the same Cauchy sequences: it follows at once from that that one is complete iff the other is. 
To get you started, suppose that $\langle p_n:n\in\Bbb N\rangle$ is a $d$-Cauchy sequence in $M$. Let $\epsilon>0$ be given. It follows from the fact that $f({0^+})=f(0)=0$ that there is a $\delta>0$ such that $f(x)<\epsilon$ whenever $0\le x<\delta$. Because $\langle p_n:n\in\Bbb N\rangle$ is $d$-Cauchy, there is an $n_\epsilon\in\Bbb N$ such that $d(p_m,p_n)<\delta$ whenever $m,n\ge n_\epsilon$. But then for $m,n\ge n_\epsilon$ we have $$\Delta(p_m,p_n)=f\left(d(p_m,p_n)\right)<\epsilon\;,$$ and it follows that $\langle p_n:n\in\Bbb N\rangle$ is $\Delta$-Cauchy as well.
Now you want to assume that $\langle p_n:n\in\Bbb N\rangle$ is $\Delta$-Cauchy and show that it’s necessarily also $d$-Cauchy. As before, start with some $\epsilon>0$. Eventually you want to come up with some $n_\epsilon\in\Bbb N$ such that $d(p_m,p_n)<\epsilon$ whenever $m,n\ge n_\epsilon$. Since you’re assuming that the sequence is $\Delta$-Cauchy, you know that you can get $\Delta(p_m,p_n)=f\left(d(p_m,p_n)\right)$ small by taking $m$ and $n$ sufficiently large; can you see how to use what you know about $f$ to translate this into making $d(p_m,p_n)$ itself small?
