Continuous Mapping on Topological Space Induced by Metric

Theorem

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

Let $\vartheta_{\left({A_1, d_1}\right)}$ and $\vartheta_{\left({A_2, d_2}\right)}$ be the topologies induced by the metrics $d_1$ and $d_2$.

Let $T_1 = \left({A_1, \vartheta_{\left({A_1, d_1}\right)}}\right)$ and $T_2 = \left({A_2, \vartheta_{\left({A_2, d_2}\right)}}\right)$ be the resulting topological spaces.

Let $f: A_1 \to A_2$ be $\left({d_1, d_2}\right)$-continuous.

Then $f$ is also $\left({\vartheta_{\left({A_1, d_1}\right)}, \vartheta_{\left({A_2, d_2}\right)}}\right)$-continuous.

Proof

Follows directly from:

• the open set definition of continuity on a metric space
• the definition of continuity on a topological space.

$\blacksquare$