Metric Induces Topology

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.

$\blacksquare$


Motivation

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

  • 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.4$
  • 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.1$: 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
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