Metrics are Topologically Equivalent iff Continuity Preserved

Theorem
Let $A$ be a set upon which there are two metrics imposed: $d_1$ and $d_2$.

The two definitions of topological equivalence of $d_1$ and $d_2$ are themselves equivalent:


 * $(1): \quad U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_2$-open.


 * $(2): \quad$ Let $\left({B, d}\right)$ and $\left({C, d\,'}\right)$ be any metric spaces.

Let $f: B \to A$ and $g: A \to C$ be any mappings such that:


 * $f$ is $\left({d, d_1}\right)$-continuous iff $f$ is $\left({d, d_2}\right)$-continuous;
 * $g$ is $\left({d_1, d\,'}\right)$-continuous iff $g$ is $\left({d_2, d\,'}\right)$-continuous.

Proof

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

From the definition of open set continuity, we have that:
 * $f$ is $\left({d, d_1}\right)$continuous iff for every set $U \subseteq A$ which is open in $\left({A, d_1}\right)$, $f^{-1} \left({U}\right)$ is open in $\left({B, d}\right)$;
 * $f$ is $\left({d, d_2}\right)$continuous iff for every set $U \subseteq A$ which is open in $\left({A, d_2}\right)$, $f^{-1} \left({U}\right)$ is open in $\left({B, d}\right)$.

Hence $f$ is $\left({d, d_1}\right)$continuous iff $f$ is $\left({d, d_2}\right)$continuous.

Similarly:
 * $g$ is $\left({d_1, d\,'}\right)$continuous iff for every set $U \subseteq C$ which is open in $\left({C, d\,'}\right)$, $g^{-1} \left({U}\right)$ is open in $\left({A, d_1}\right)$;
 * $g$ is $\left({d_2, d\,'}\right)$continuous iff for every set $U \subseteq C$ which is open in $\left({C, d\,'}\right)$, $g^{-1} \left({U}\right)$ is open in $\left({A, d_2}\right)$.

Hence $g$ is $\left({d_1, d\,'}\right)$continuous iff $g$ is $\left({d_2, d\,'}\right)$continuous.


 * Now suppose that $g$ is $\left({d_1, d\,'}\right)$-continuous iff $g$ is $\left({d_2, d\,'}\right)$-continuous.

Suppose $U \subseteq A$ is $d_2$-open.

Let $h: A \to A$ be the identity mapping.

That is, $\forall a \in A: g \left({a}\right) = a$.

This is clearly $\left({d_2, d_2}\right)$-continuous.

Hence because $g$ is $\left({d_1, d\,'}\right)$-continuous iff $g$ is $\left({d_2, d\,'}\right)$-continuous, $h$ is $\left({d_1, d_2}\right)$-continuous.

By the definition of open set continuity, it implies that $h^{-1} \left({U}\right)$ is $d_1$-open.

But $h^{-1} \left({U}\right) = U$, so $U$ is $d_1$-open.

A similar argument, still starting with the same proposition that $g$ is $\left({d_1, d\,'}\right)$-continuous iff $g$ is $\left({d_2, d\,'}\right)$-continuous, shows that if $U$ is $d_1$-open then it is $d_2$-open.

Note
Note that from the proposition that $g$ is $\left({d_1, d\,'}\right)$-continuous iff $g$ is $\left({d_2, d\,'}\right)$-continuous, we show that $U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_2$-open.

From that, we show that both:
 * $f$ is $\left({d, d_1}\right)$-continuous iff $f$ is $\left({d, d_2}\right)$-continuous;
 * $g$ is $\left({d_1, d\,'}\right)$-continuous iff $g$ is $\left({d_2, d\,'}\right)$-continuous.

Hence the proposition that $f$ is $\left({d, d_1}\right)$-continuous iff $f$ is $\left({d, d_2}\right)$-continuous is superfluous, as it follows directly from the proposition concerning $g$.