Difference between revisions of "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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{{Refactor notice|grade=A|msg=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!}}
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__TOC__
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==Statement==
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Let {{M|X}} be a [[set]], let {{M|d:X\times X\rightarrow\mathbb{R}_{\ge 0} }} be a [[metric]] on that set and let {{M|(X,d)}} be the resulting [[metric space]]. Then we claim{{rITTMJML}}:
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* {{M|\mathcal{B}:\eq\left\{ B_\epsilon(x)\ \vert\ x\in X\wedge \epsilon\in\mathbb{R}_{>0}\right\} }} satisfies the condition [[topology generated by a basis|to generate a topology, for which it is a basis]]
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==Proof==
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{{Begin Notebox}}Recall the definition of a [[topology generated by a basis]]{{Begin Notebox Content}}
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{{:Topology generated by a basis/Statement}}
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{{End Notebox Content}}{{End Notebox}}
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'''Proof that the requisite conditions are met:'''
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# {{M|\forall x\in X\exists B\in\mathcal{B}[x\in B]}}
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#* Let {{M|x\in X}} be given
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#** '''Lemma:''' {{M|\forall p\in X\forall\epsilon>0[p\in B_\epsilon(x)]}}
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#*** Proof:
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#**** Let {{M|p\in X}} be given.
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#***** Let {{M|\epsilon>0}} be given (with {{M|\epsilon\in\mathbb{R}_{>0} }} of course)
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#****** Recall, by definition of an [[open ball]] that {{M|[u\in B_\delta(v)]\iff[d(u,v)<\delta]}}
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#******* Thus {{M|[p\in B_\epsilon(p)]\iff[d(p,p)<\epsilon]}}
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#****** Recall, by definition of a [[metric]] that {{M|[d(u,v)\eq 0]\iff[u\eq v]}}
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#******* Thus {{M|d(p,p)\eq 0}}
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#****** As {{M|\epsilon > 0}} we see {{M|d(p,p)\eq 0<\epsilon}}, {{ie}} {{M|d(p,p)<\epsilon}} thus {{M|p\in B_\epsilon(p)}}
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#***** Since {{M|\epsilon > 0}} was arbitrary we have shown {{M|p\in B_\epsilon(p)}} for all {{M|\epsilon>0}} ({{M|\epsilon\in\mathbb{R}_{>0} }} of course)
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#**** Since {{M|p\in X}} was arbitrary we have shown {{M|\forall\epsilon>0[p\in B_\epsilon(p)]}} for all {{M|p\in X}}
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#*** This completes the proof.
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#** By using the lemma above we see {{M|\forall\epsilon>0[x\in B_\epsilon(x)]}}
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#*** In particular we see {{M|x\in B_1(x)}} - there is nothing special about the choice of {{M|\epsilon:\eq 1}} - we could have picked any {{M|\epsilon\in\mathbb{R}_{>0} }}
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#** Choose {{M|B:\eq B_1(x)}}
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#*** Note that {{M|B\in\mathcal{B} }} by definition of {{M|\mathcal{B} }}, explicitly: {{M|[B_r(q)\in\mathcal{B}]\iff[q\in X\wedge r\in\mathbb{R}_{>0}]}}
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#*** By our choice of {{M|B}} (and the lemma) we see {{M|x\in B}}
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#** Our choice of {{M|B}} satisfies the requirements
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#* Since {{M|x\in X}} was arbitrary we have shown it for all {{M|x}} - as required. This completes part 1
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# {{M|\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]}}
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#* Let {{M|U,\ V\in\mathcal{B} }} be given.
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#** Suppose {{M|U}} and {{M|V}} are [[disjoint]]. Then the [[logical implication]] holds regardless of the RHS, and we've shown the statement to be true in this case.
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#** Suppose
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==Notes==
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<references group="Note"/>
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==References==
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<references/>
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{{Theorem Of|Topology|Metric Space}}
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=OLD PAGE=
 
For a [[Metric space|metric space]] {{M|(X,d)}} there is a [[Topological space|topology]] which the metric induces on {{M|x}} that is the topology of all sets which are [[Open set|open in the metric sense]].
 
For a [[Metric space|metric space]] {{M|(X,d)}} there is a [[Topological space|topology]] which the metric induces on {{M|x}} that is the topology of all sets which are [[Open set|open in the metric sense]].
  

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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[1]:

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][Note 1]) 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][Note 2]
      • 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.[Note 3]

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

Notes

  1. ↑ By the implies-subset relation [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B][/ilmath] really means [ilmath]X\subseteq\bigcup\mathcal{B} [/ilmath], as we only require that all elements of [ilmath]X[/ilmath] be in the union. Not that all elements of the union are in [ilmath]X[/ilmath]. However:
    • [ilmath]\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))[/ilmath] by definition. So clearly (or after some thought) the reader should be happy that [ilmath]\mathcal{B} [/ilmath] contains only subsets of [ilmath]X[/ilmath] and he should see that we cannot as a result have an element in one of these subsets that is not in [ilmath]X[/ilmath].
    Thus [ilmath]\forall B\in\mathcal{B}[B\in\mathcal{P}(X)][/ilmath] which is the same as (by power-set and subset definitions) [ilmath]\forall B\in\mathcal{B}[B\subseteq X][/ilmath].
  2. ↑ We could of course write:
    • [ilmath]\forall U,V\in\mathcal{B}\ \forall x\in \bigcup\mathcal{B}\ \exists W\in\mathcal{B}[(x\in U\cap V)\implies(x\in W\wedge W\subseteq U\cap V)][/ilmath]
  3. ↑ Suppose that [ilmath]U,V\in\mathcal{B} [/ilmath] are given but disjoint, then there are no [ilmath]x\in U\cap V[/ilmath] to speak of, and [ilmath]x\in W[/ilmath] may be vacuously satisfied by the absence of an [ilmath]X[/ilmath], however:
    • [ilmath]x\in W\subseteq U\cap V[/ilmath] is taken to mean [ilmath]x\in W[/ilmath] and [ilmath]W\subseteq U\cap V[/ilmath], so we must still show [ilmath]\exists W\in\mathcal{B}[W\subseteq U\cap V][/ilmath]
      • This is not always possible as [ilmath]W[/ilmath] would have to be [ilmath]\emptyset[/ilmath] for this to hold! We do not require [ilmath]\emptyset\in\mathcal{B} [/ilmath] (as for example in the metric topology)

References

  1. ↑ Introduction to Topological Manifolds - John M. Lee



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.


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



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