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

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    $\begingroup$ The claim is not true. A one-point space is metrizable in only one way $\endgroup$ – Hagen von Eitzen May 14 at 6:24
  • $\begingroup$ @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. $\endgroup$ – 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 $\mathbb R$).

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    $\begingroup$ Of course it is convex. What is your idea of convex function? $\endgroup$ – Kabo Murphy May 14 at 6:45
  • $\begingroup$ Deleted my comment. $\endgroup$ – Cheerful Parsnip May 14 at 20:09

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.


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