Topologically Equivalent Metrics induce Equal Topologies

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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 if and only if:

$U \subseteq A$ is $d_1$-open if and only if $U \subseteq A$ 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 $\struct {A, \tau_1}$

if and only if:

$U \subseteq A$ 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.

$\blacksquare$