Standard Discrete Metric induces Discrete Topology

Theorem

Let $M = \struct {A, d}$ be the (standard) discrete metric space on $A$.

Then $d$ induces the discrete topology on $A$.

Thus the discrete topology is metrizable.

Proof

Let $a \in A$.

From Subset of Standard Discrete Metric Space is Open, a set $U \subseteq A$ is open in $M$.

So, in particular, $\set a$ is open in $\struct {A, d}$.

This holds for all $a \in A$.

From Metric Induces Topology it follows that $\set a$ is an open set in $\struct {A, \tau_{A, d} }$.

The result follows from Basis for Discrete Topology.

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Exercise $2$
• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $6$