Standard Discrete Metric induces Discrete Topology

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Theorem

Let $M = \left({A, d}\right)$ 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, $\left\{{a}\right\}$ is open in $\left({A, d}\right)$.

This holds for all $a \in A$.


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

The result follows from Basis for Discrete Topology.

$\blacksquare$


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