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Suppose $d_1,d_2$ are topologically equivalent metrics on a set $X$. Suppose also that $d_1$ is bounded, that is there exists $K>0$ such that $d_1(x,y) \leq K$ for all $x,y\in X$.

Does this mean that $d_2$ is bounded?

My attempt:

The statement above is false, consider $X=\mathbb{C} -\{0+0i\}$ with $d_1$ as the discrete metric and $d_2(z,w) = 0$ if $z=w$ and $d_2(z,w) = |z|+|w|$ otherwise.

$d_1$ is equivalent to $d_2$ since only eventually constant sequences converge, and $d_1$ is bounded by $1$ but $d_2$ is not bounded.

Is this correct?

bof
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fosho
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  • The answer might depend on what you mean by "equivalent metrics". Are you talking about topological equivalence, or strong equivalence, or something else? – bof Apr 25 '16 at 08:18
  • Topological equivalence (i.e. they induce the same topology on $X$). – fosho Apr 25 '16 at 08:20
  • Isn't every metric topologically equivalent to a bounded metric? – bof Apr 25 '16 at 08:22
  • @bof I am not sure... – fosho Apr 25 '16 at 08:24
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    If $d(x,y)$ is a metric, doesn't the metric $d_1(x,y)=\min{1,d(x,y)}$ induce the same topology as $d(x,y)$? – bof Apr 25 '16 at 08:26
  • But this doesn't necessarily show that if two metrics are equivalent and one is bounded, then the other is bounded, does it? – fosho Apr 25 '16 at 08:28
  • Oh of course, thanks a lot for this. Did you look over my counter example? – fosho Apr 25 '16 at 08:31

1 Answers1

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Your counterexample is fine. Here's how you can get more counterexamples, which might be considered more natural.

Plainly, "metrizable by a bounded metric" is a topological property; if two topological spaces are homeomorphic, and if one is metrizable by a bounded metric, so is the other.

The usual metric on $\mathbb R$ is unbounded. Since a finite open interval $(a,b)$ is homeomorphic to $\mathbb R,$ and since the usual metric on $(a,b)$ is bounded, the usual topology on $\mathbb R$ is also induced by a bounded metric.

For example, $x\mapsto\arctan x$ is a homeomorphism from $\mathbb R$ to $(-\frac\pi2,\frac\pi2),$ and correspondingly $d(x,y)=|\arctan x-\arctan y|$ is a bounded metric for $\mathbb R.$

To get even more counterexamples, note that any metric $d(x,y)$ is topologically equivalent to a bounded metric, e.g., $d_1(x,y)=\min\{1,d(x,y)\}.$

bof
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