Topologically Equivalent Metrics induce Equal Topologies

Theorem
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.

Let $d_1$ and $d_2$ be topologically equivalent.

Let $\tau_1$ and $\tau_2$ denote the topologies on $A$ induced by $d_1$ and $d_2$ respectively.

Then $\tau_1$ and $\tau_2$ are equal.

Proof
Let $d_1$ and $d_2$ be topologically equivalent by hypothesis.

By definition of topological equivalence:

$d_1$ and $d_2$ are topologically equivalent :


 * $U \subseteq A$ is $d_1$-open $U \subseteq A$ is $d_2$-open.

By definition of the induced topology:


 * The topology on the metric space $M = \struct {A, d}$ induced by (the metric) $d$ is defined as the set $\tau$ of all open sets of $M$.

Hence it follows that:


 * $U \subseteq A$ is open in the topological space $\struct {A, \tau_1}$


 * $U \subseteq A$ is open in the topological space $\struct {A, \tau_2}$
 * $U \subseteq A$ is open in the topological space $\struct {A, \tau_2}$

That is:
 * $U \in \tau_1 \iff U \in \tau_2$

and the result follows.