# Metrizable in infinite number of ways

The question states that if a topological space is metrizable it is metrizable in infinite number of ways.

Of course scaling the distances by any positive number will do the trick. But i want to know whether any concave and strictly increasing transform applied to distances will also result in a metric.

Edit: Earlier i got confused and asked convex transforms. Thanks @Murthy and Santos for correcting me. The author of this post has proven relatoin between concavity and sub additivity

• The claim is not true. A one-point space is metrizable in only one way – Hagen von Eitzen May 14 at 6:24
• @Murthy and Santos: Thanks for answering. I wanted to understand relation between subadditivity and convexity and confused. I realized that they should indeed be concave, because then smaller distance will grow faster and maintain the triangular inequality. – Curious May 14 at 7:07

Partial answer. As pointed out above you cannot have infinite number of different metrics in general. But I will answer the other question you have asked: suppose $$f:[0,\infty) \to \mathbb R$$ is convex strictly increasing and $$f(0)=0$$. Can we say that $$f(d(x,y))$$ is a metric? The answer is no. What you need is not convexity but sub-additivity. For example if $$f(x)=x^{2}$$ then $$f(d(x,y))$$ is not a metric since triangle in equality is not satisfied (for example when $$d$$ is the usual metric on $$\mathbb R$$).
In general, no, as you have been told. But if you have a metric space in which the metric $$d$$ takes infinitley many values, then you can consider the family of metrics $$\min(d,a)$$, with $$a>0$$. This will give you infinitely many metrics which are equivalent to the original one.