# Metric Induces Topology

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## Theorem

Let $M = \struct {A, d}$ 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 open sets of a metric space is open.

- $(3): \quad$ By Open Sets in Metric Space, $\O \in \tau$ 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

- 1962: Bert Mendelson:
*Introduction to Topology*... (previous) ... (next): $\S 2.6$: Open Sets and Closed Sets: Theorem $6.4$ - 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $3.1$: Topological Spaces: Example $3.1.5$ - 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $5$: Metric Spaces