Standard Discrete Metric induces Discrete Topology
Jump to navigation
Jump to search
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: $6$