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 $\struct {A, \tau_1}$
- $U \subseteq A$ 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.
$\blacksquare$