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It is well-known that every metric space $(X,d)$ is homeomorphic to a bounded metric space $(X,\bar{d})$ where: $$ \bar{d}:X^2\ni (x_1,x_2)\mapsto \frac{d(x_1,x_2)}{1+d(x_1,x_2)} \in [0,1]; $$ (see this post for instance). Is the homeomorphism $$ h:(X,d)\ni x \mapsto x \in (X,\bar{d}), $$ ever uniformly continuous? and, if so, how does its modulus of continuity relate to $X$'s diameter?

I anticipate that the map is only uniformly continuous if the diameter of $(X,d)$; by which I mean, $\operatorname{diam}(X,d):=\sup_{x_1,x_2\in X}\, d(x_1,x_2)$), is finite.

ABIM
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  • Diameter depends on the metric you consider, and "the diameter of $X$" is not well-defined as it is stated – Didier Dec 05 '21 at 15:34
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    @Didier He means (I think) the diameter wrt the original $d$ I think, as the new metric almost surely has diameter $1$ now. And by the homeomorphism he means the identity function $h(x)=x$ I suppose. – Henno Brandsma Dec 05 '21 at 16:56
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    @HennoBrandsma Exactly; but I'll make the edits to clarify. (Also *she ;)) – ABIM Dec 05 '21 at 17:16
  • I just assumed “Tom” is a he by default. Thomas etc. – Henno Brandsma Dec 05 '21 at 17:21
  • @HennoBrandsma Oh no worries; now I see why totally fair :) Sorry about the mix up on my end :) – ABIM Dec 05 '21 at 17:30

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What you mean is that both metrics induce the same topology on $X$. In other words, as Henno Brandsma comments, that $id : (X,d) \to (X,\bar d)$ is a homeomorphism. Of course $id$ is uniformly continuous. We have $$\bar d(x_1,x_2) \le d(x_1,x_2)$$ because $1 + d(x_1,x_2) \ge 1$.

However, we cannot expect that $id : (X,\bar d) \to (X, d)$ is always uniformly continuous. But it is if $\delta = \text{diam}_d(X) < \infty$. In that case we have

$$d(x_1,x_2) = (1 +\delta) \frac{d(x_1,x_2)}{1 + \delta} \le (1 +\delta) \frac{d(x_1,x_2)}{1 + d(x_1,x_2)} = (1 +\delta)\bar d(x_1,x_2) .$$

Paul Frost
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