Standard Discrete Metric is not Topologically Equivalent to p-Product Metrics

Theorem

For $n \in \N$, let $\R^n$ be an Euclidean space.

Let $p \in \R_{\ge 1}$.

Let $d_p$ be the $p$-product metric on $\R^n$.

Let $d_0$ be the standard discrete metric on $\R^n$.

Then $d_p$ and $d_0$ are not topologically equivalent.

Proof

From Open Ball in Standard Discrete Metric Space it is seen that singletons are open sets in $\struct {\R^n, d_0}$.

However, this is not the case in the $\struct {\R^n, d_p}$.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics: Example $2.4.5$