Topology induced by a metric

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Definition

Let [ilmath](X,d)[/ilmath] be a metric space. Then there is a topology we can imbue on [ilmath]X[/ilmath], called the metric topology that can be defined in terms of the metric, [ilmath]d:X\times X\rightarrow\mathbb{R}_{\ge 0} [/ilmath].

We do this using the concept of topology generated by a basis. We claim ("Claim 1"):

• [ilmath]\mathcal{B}:\eq\left\{B_r(x)\ \vert\ x\in X\wedge r\in\mathbb{R}_{>0} \right\} [/ilmath] satisfies the conditions to generate a topology (and is a basis for that topology) - where [ilmath]B_\epsilon(p)[/ilmath] denotes the open ball of radius [ilmath]\epsilon\in\mathbb{R}_{>0} [/ilmath] centred at [ilmath]p\in X[/ilmath]

The resulting topological space, say [ilmath](X,\mathcal{ J })[/ilmath], has basis [ilmath]\mathcal{B} [/ilmath]

• Explicitly the topology is [math]\mathcal{J}:\eq\left\{\left. \bigcup_{B\in\mathcal{F} }B\ \right\vert\ \mathcal{F}\in\mathcal{P}(\mathcal{B})\right\} [/math]
  • Notice [ilmath]\bigcup_{B\in\emptyset} B\eq\emptyset[/ilmath] - hence the empty-set is open - as required.
  • Notice also that [ilmath]\bigcup{B\in\mathcal{B} }B\eq X[/ilmath] - obvious as [ilmath]\mathcal{B} [/ilmath] contains (among others) an open ball centred at each point in [ilmath]X[/ilmath] and each point is in that open ball at least.
    TODO: Copy and paste a proof from elsewhere

Consequences

• The open sets are exactly those which are unions of basis elements. i.e.

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Proof of claims

• Claim 1: [ilmath]\mathcal{B} [/ilmath] satisfies the conditions to generate a topology for which it is a basis - see The set of all open balls of a metric space are able to generate a topology and are a basis for that topology