Equivalence of metrics

In the study of metric spaces in mathematics, there are various notions of two metrics on the same underlying space being "the same", or equivalent.

In the following, $M$ will denote a non-empty set and $d\_\{1\}$ and $d\_\{2\}$ will denote two metrics on $M$.

Topological equivalence

The two metrics $d\_\{1\}$ and $d\_\{2\}$ are said to be topologically equivalent if they generate the same topology on $M$. There are many equivalent ways of expressing this condition:
* a subset $A\; subseteq\; M$ is $d\_\{1\}$-open if and only if it is $d\_\{2\}$-open;
* the open balls "nest": for any point $x\; in\; M$ and any radius $r\; >\; 0$, there exist radii $r\text{'},\; r"\; >\; 0$ such that:$B\_\{r\}\; (x;\; d\_\{1\})\; subseteq\; B\_\{r\text{'}\}\; (x;\; d\_\{2\})$ and $B\_\{r\}\; (x;\; d\_\{2\})\; subseteq\; B\_\{r"\}\; (x;\; d\_\{1\}).$
* the identity function $I\; :\; M\; o\; M$ is both $(d\_\{1\},\; d\_\{2\})$-continuous and $(d\_\{2\},\; d\_\{1\})$-continuous.

The following is a sufficient but not necessary condition for topological equivalence:
* for each $x\; in\; M$, there exist constants $c\_\{1\},\; c\_\{2\}\; >\; 0$ such that, for every point $y\; in\; M$,:$c\_\{1\}\; d\_\{1\}\; (x,\; y)\; leq\; d\_\{2\}\; (x,\; y)\; leq\; c\_\{2\}\; d\_\{1\}\; (x,\; y).$