# Metric Induces Topology

Jump to navigation
Jump to search

## 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

- 1962: Bert Mendelson:
*Introduction to Topology*... (previous) ... (next): $\S 2.6$: Open Sets and Closed Sets: Theorem $6.4$ - 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 5$ - 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $3.1$: Topological Spaces: Example $3.1.5$