Can somebody help me out with the following:
If (X,d) is a metric space and r(x,y)=min{d(x,y},1} for all x,y in X. I proved that r is a metric and that r and d are equivalent. Now I want to prove that (X,r) is complete if and only if (X,d) is complete.
This was my idea: Suppose {xn} is a Cauchy sequence in X, then {xn} is r-convergent in X (completeness). Because of the equivalence of r and d, {xn} is also d-convergent and thus (X,d) is complete. The other way is the same. Can somebody say if this is correct?
Your idea doesn't quite work. Topological equivalence is agnostic to distances, which are used to define the notion of a Cauchy sequence. For example, we can endow $\Bbb{R\setminus Q}$ with a complete metric which is equivalent to the standard metric, which is far from being complete.
HINT: Recall that in the definition of a Cauchy sequence we care more about small $\varepsilon$ than about large $\varepsilon$.
In particular, what happens when we consider $\varepsilon<1$? How do the metrics $r$ and $d$ differ in that case?