Metric Induces a Topology

Theorem

Consider a metric space $M = \left({S, d}\right)$ where $S \ne \varnothing$ is some nonempty set and $d: S \times S \to \R_{\ge 0}$ is a metric.

Then $\left({S, d}\right)$ gives rise to a topological space $\left({S, \vartheta_{\left({S, d}\right)}}\right)$ whose topology $\vartheta_{\left({S, d}\right)}$ is defined (or induced) by $d$ in the following sense:

Let $B_{\left({S, d}\right)} := \left\{{N_\epsilon \left({a}\right): \epsilon \in \R, a \in S, N_\epsilon \left({a}\right) \subseteq S}\right\}$, where:

$N_\epsilon \left({a}\right) := \left\{{x \in S: d \left({x, a}\right) < \epsilon}\right\}$

That is, $B_{\left({S, d}\right)}$ is the set of all open $\epsilon$-ball neighborhoods of all points of $S$.

It can be seen that $B_{\left({S, d}\right)}$ forms a (synthetic) topology basis; see Set of Neighborhoods is Topology Basis for a proof.

Now $\vartheta_{\left({S, d}\right)}$ is the topology associated to $B_{\left({S, d}\right)}$ in the sense of Union from Synthetic Basis is a Topology.

Any topological space which is homeomorphic to such a $\left({S, \vartheta_{\left({S, d}\right)}}\right)$ is defined as metrizable.

Proof

Let $\vartheta_{\left({S, d}\right)}$ be the set of all $X \subseteq S$ which are open in the sense that:

$\forall y \in X: \exists \epsilon \left({y}\right) > 0: N_{\epsilon \left({y}\right)} \left({y}\right) \subseteq X$

where $N_{\epsilon \left({y}\right)} \left({y}\right)$ is the open $\epsilon \left({y}\right)$-ball neighborhood of $y$.

Equivalently:

$\forall x \in X: \exists \epsilon \in \R_+: \forall y \in S: d \left({x, y}\right) < \epsilon \implies y \in X$

We need to show that $\vartheta_{\left({S, d}\right)}$ forms a topology on $S$.

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

$(1): \quad$ From Open Sets in Metric Space $\varnothing \in \vartheta_{\left({S, d}\right)}$ and $S \in \vartheta_{\left({S, d}\right)}$.

$(2): \quad$ From Union of Open Subsets, the union of any collection of open subsets of a metric space is also open.

$(3): \quad$ From Intersection of Open Subsets, the intersection of a finite number of open subsets is open.

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 reason behind the definition of open sets in topology.

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 5$