Metric Induces Topology

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Theorem

Let $M = \left({A, d}\right)$ be a metric space.


Then the topology $\tau$ induced by the metric $d$ is a topology on $M$.


Proof

We examine each of the criteria for being a topology separately.

$(1): \quad$ By Union of Open Sets of Metric Space is Open, the union of any collection of open sets of a metric space is open.
$(2): \quad$ By Finite Intersection of Open Sets of Metric Space is Open, a finite intersection of [Definition:Open Set of Metric Space|open sets]] of a metric space is open.
$(3): \quad$ By Open Sets in Metric Space, $\varnothing \in \tau$ and $A \in \tau$.

Hence the result.

$\blacksquare$


Note

Thus it can be seen that the concept of an open set as applied to a metric space is directly equivalent to that of an open set as applied to a topological space.

This is the motivation behind the definition of open sets in topology.


Sources