The set of all open balls of a metric space are able to generate a topology and are a basis for that topology

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This is one of the oldest pages on the wiki, created in Feb 2015. With a slight revision in Apr 2015 when Template:Theorem was moved to Template:Theorem Of. A very old page indeed!

Statement

Let [ilmath]X[/ilmath] be a set, let [ilmath]d:X\times X\rightarrow\mathbb{R}_{\ge 0} [/ilmath] be a metric on that set and let [ilmath](X,d)[/ilmath] be the resulting metric space. Then we claim:

• [ilmath]\mathcal{B}:\eq\left\{ B_\epsilon(x)\ \vert\ x\in X\wedge \epsilon\in\mathbb{R}_{>0}\right\} [/ilmath] satisfies the condition to generate a topology, for which it is a basis

Proof

Recall the definition of a topology generated by a basis

Let [ilmath]X[/ilmath] be a set and let [ilmath]\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))[/ilmath] be any collection of subsets of [ilmath]X[/ilmath], then:

• [ilmath](X,\{\bigcup\mathcal{A}\ \vert\ \mathcal{A}\in\mathcal{P}(\mathcal{B})\})[/ilmath] is a topological space with [ilmath]\mathcal{B} [/ilmath] being a basis for the topology [ilmath]\{\bigcup\mathcal{A}\ \vert\ \mathcal{A}\in\mathcal{P}(\mathcal{B})\}[/ilmath]

if and only if

• we have both of the following conditions:
  1. [ilmath]\bigcup\mathcal{B}=X[/ilmath] (or equivalently: [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B][/ilmath]) and
  2. [ilmath]\forall U,V\in\mathcal{B}\big[U\cap V\neq\emptyset\implies \forall x\in U\cap V\exists B\in\mathcal{B}[x\in W\wedge W\subseteq U\cap V]\big][/ilmath]
     • Caveat:[ilmath]\forall U,V\in\mathcal{B}\ \forall x\in U\cap V\ \exists W\in\mathcal{B}[x\in W\subseteq U\cap V][/ilmath] is commonly said or written; however it is wrong, this is slightly beyond just abuse of notation.

Proof that the requisite conditions are met:

1. [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B][/ilmath]
   • Let [ilmath]x\in X[/ilmath] be given
     • Lemma: [ilmath]\forall p\in X\forall\epsilon>0[p\in B_\epsilon(x)][/ilmath]
       • Proof:
         • Let [ilmath]p\in X[/ilmath] be given.
           • Let [ilmath]\epsilon>0[/ilmath] be given (with [ilmath]\epsilon\in\mathbb{R}_{>0} [/ilmath] of course)
             • Recall, by definition of an open ball that [ilmath][u\in B_\delta(v)]\iff[d(u,v)<\delta][/ilmath]
               • Thus [ilmath][p\in B_\epsilon(p)]\iff[d(p,p)<\epsilon][/ilmath]
             • Recall, by definition of a metric that [ilmath][d(u,v)\eq 0]\iff[u\eq v][/ilmath]
               • Thus [ilmath]d(p,p)\eq 0[/ilmath]
             • As [ilmath]\epsilon > 0[/ilmath] we see [ilmath]d(p,p)\eq 0<\epsilon[/ilmath], i.e. [ilmath]d(p,p)<\epsilon[/ilmath] thus [ilmath]p\in B_\epsilon(p)[/ilmath]
           • Since [ilmath]\epsilon > 0[/ilmath] was arbitrary we have shown [ilmath]p\in B_\epsilon(p)[/ilmath] for all [ilmath]\epsilon>0[/ilmath] ([ilmath]\epsilon\in\mathbb{R}_{>0} [/ilmath] of course)
         • Since [ilmath]p\in X[/ilmath] was arbitrary we have shown [ilmath]\forall\epsilon>0[p\in B_\epsilon(p)][/ilmath] for all [ilmath]p\in X[/ilmath]
       • This completes the proof.
     • By using the lemma above we see [ilmath]\forall\epsilon>0[x\in B_\epsilon(x)][/ilmath]
       • In particular we see [ilmath]x\in B_1(x)[/ilmath] - there is nothing special about the choice of [ilmath]\epsilon:\eq 1[/ilmath] - we could have picked any [ilmath]\epsilon\in\mathbb{R}_{>0} [/ilmath]
     • Choose [ilmath]B:\eq B_1(x)[/ilmath]
       • Note that [ilmath]B\in\mathcal{B} [/ilmath] by definition of [ilmath]\mathcal{B} [/ilmath], explicitly: [ilmath][B_r(q)\in\mathcal{B}]\iff[q\in X\wedge r\in\mathbb{R}_{>0}][/ilmath]
       • By our choice of [ilmath]B[/ilmath] (and the lemma) we see [ilmath]x\in B[/ilmath]
     • Our choice of [ilmath]B[/ilmath] satisfies the requirements
   • Since [ilmath]x\in X[/ilmath] was arbitrary we have shown it for all [ilmath]x[/ilmath] - as required. This completes part 1
2. [ilmath]\forall U,V\in\mathcal{B}\big[U\cap V\neq\emptyset\implies \forall x\in U\cap V\exists B\in\mathcal{B}[x\in W\wedge W\subseteq U\cap V]\big][/ilmath]
   • Let [ilmath]U,\ V\in\mathcal{B} [/ilmath] be given.
     • Suppose [ilmath]U[/ilmath] and [ilmath]V[/ilmath] are disjoint. Then the logical implication holds regardless of the RHS, and we've shown the statement to be true in this case.
     • Suppose [ilmath]U[/ilmath] and [ilmath]V[/ilmath] have non-empty intersection
       • We must now show [ilmath]\forall x\in U\cap V\exists B\in\mathcal{B}[x\in W\wedge W\subseteq U\cap V][/ilmath]
         • Let [ilmath]x\in U\cap V[/ilmath] be given
           • By If the intersection of two open balls is non-empty then for every point in the intersection there is an open ball containing it in the intersection we see that
             • There exists an open ball [ilmath]W[/ilmath] such that [ilmath]x\in W[/ilmath] ([ilmath]W[/ilmath] is in fact centred at [ilmath]x[/ilmath]) and [ilmath]W\subseteq U\cap V[/ilmath]
           • As required
         • Since [ilmath]x\in U\cap V[/ilmath] was arbitrary we have shown it for all
       • We've shown it
     • Now in either case the logical implication has been shown to hold
   • Since [ilmath]U,V\in\mathcal{B} [/ilmath] were arbitrary we have shown it for all open balls

Thus [ilmath]\mathcal{B} [/ilmath] suitable to generate a topology and be a basis for that topology

Discussion of result

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Now we can link "open" in the metric sense to "open" in the topological sense. At long last.

References

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Requires them, but is very widely known and borderline implicit!

OLD PAGE

For a metric space [ilmath](X,d)[/ilmath] there is a topology which the metric induces on [ilmath]x[/ilmath] that is the topology of all sets which are open in the metric sense.

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TODO: Proof that open sets in the metric space have the topological properties

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