# Suppose $d_1$ and $d_2$ are equivalent metrics and $d_1$ is bounded, is $d_2$ bounded?

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?

• 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 8:18
• Topological equivalence (i.e. they induce the same topology on $X$). – fosho Apr 25 '16 at 8:20
• Isn't every metric topologically equivalent to a bounded metric? – bof Apr 25 '16 at 8:22
• @bof I am not sure... – fosho Apr 25 '16 at 8:24
• 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 8:26

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., $d_1(x,y)=\min\{1,d(x,y)\}.$