Metrics are Topologically Equivalent iff Continuity Preserved

= Theorem =

Metric Space
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. $$U \subseteq A$$ is $d_1$-open $$\iff$$ $$U \subseteq A$$ is $d_2$-open;


 * 2. 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 =

Proof for Metric Space

 * 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 $$g$$.