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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m (Alec moved page Topology induced by a metric to The set of all open balls of a metric space are able to generate a topology and are a basis for that topology without leaving a redirect: Better name (this is a theorem, not defining the 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!}} | ||
+ | __TOC__ | ||
+ | ==Statement== | ||
+ | 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: | ||
+ | * {{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]] | ||
+ | ==Proof== | ||
+ | {{Begin Notebox}}Recall the definition of a [[topology generated by a basis]]{{Begin Notebox Content}} | ||
+ | {{:Topology generated by a basis/Statement}} | ||
+ | {{End Notebox Content}}{{End Notebox}} | ||
+ | '''Proof that the requisite conditions are met:''' | ||
+ | # {{M|\forall x\in X\exists B\in\mathcal{B}[x\in B]}} | ||
+ | #* Let {{M|x\in X}} be given | ||
+ | #** '''Lemma:''' {{M|\forall p\in X\forall\epsilon>0[p\in B_\epsilon(x)]}} | ||
+ | #*** Proof: | ||
+ | #**** Let {{M|p\in X}} be given. | ||
+ | #***** Let {{M|\epsilon>0}} be given (with {{M|\epsilon\in\mathbb{R}_{>0} }} of course) | ||
+ | #****** Recall, by definition of an [[open ball]] that {{M|[u\in B_\delta(v)]\iff[d(u,v)<\delta]}} | ||
+ | #******* Thus {{M|[p\in B_\epsilon(p)]\iff[d(p,p)<\epsilon]}} | ||
+ | #****** Recall, by definition of a [[metric]] that {{M|[d(u,v)\eq 0]\iff[u\eq v]}} | ||
+ | #******* Thus {{M|d(p,p)\eq 0}} | ||
+ | #****** 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)}} | ||
+ | #***** 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) | ||
+ | #**** 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}} | ||
+ | #*** This completes the proof. | ||
+ | #** By using the lemma above we see {{M|\forall\epsilon>0[x\in B_\epsilon(x)]}} | ||
+ | #*** 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} }} | ||
+ | #** Choose {{M|B:\eq B_1(x)}} | ||
+ | #*** 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}]}} | ||
+ | #*** By our choice of {{M|B}} (and the lemma) we see {{M|x\in B}} | ||
+ | #** Our choice of {{M|B}} satisfies the requirements | ||
+ | #* Since {{M|x\in X}} was arbitrary we have shown it for all {{M|x}} - as required. This completes part 1 | ||
+ | # {{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]}} | ||
+ | #* Let {{M|U,\ V\in\mathcal{B} }} be given. | ||
+ | #** 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. | ||
+ | #** Suppose {{M|U}} and {{M|V}} have ''[[non-empty]]'' [[intersection]] | ||
+ | #*** We must now show {{M|\forall x\in U\cap V\exists B\in\mathcal{B}[x\in W\wedge W\subseteq U\cap V]}} | ||
+ | #**** Let {{M|x\in U\cap V}} 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 {{M|W}} such that {{M|x\in W}} ({{M|W}} is in fact centred at {{M|x}}) and {{M|W\subseteq U\cap V}} | ||
+ | #***** As required | ||
+ | #**** Since {{M|x\in U\cap V}} 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 {{M|U,V\in\mathcal{B} }} were arbitrary we have shown it for all open balls | ||
+ | Thus {{M|\mathcal{B} }} suitable to [[topology generated by a basis|generate a topology and be a basis for that topology]] | ||
+ | ==Discussion of result== | ||
+ | {{Requires work|grade=A*|msg=Now we can link "open" in the metric sense to "open" in the topological sense. At long last.}} | ||
+ | ==Notes== | ||
+ | <references group="Note"/> | ||
+ | ==References== | ||
+ | <references/> | ||
+ | {{Requires references|grade=C|msg=Requires them, but is very widely known and borderline implicit!}} | ||
+ | {{Theorem Of|Topology|Metric Space}} | ||
+ | |||
+ | =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]]. | ||
Latest revision as of 22:16, 16 January 2017
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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:
- [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
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]
- we have both of the following conditions:
- [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
- [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:
- [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]
- Recall, by definition of an open ball that [ilmath][u\in B_\delta(v)]\iff[d(u,v)<\delta][/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)
- Let [ilmath]\epsilon>0[/ilmath] be given (with [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]
- Let [ilmath]p\in X[/ilmath] be given.
- This completes the proof.
- 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
- Lemma: [ilmath]\forall p\in X\forall\epsilon>0[p\in B_\epsilon(x)][/ilmath]
- Since [ilmath]x\in X[/ilmath] was arbitrary we have shown it for all [ilmath]x[/ilmath] - as required. This completes part 1
- Let [ilmath]x\in X[/ilmath] be given
- [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
- 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
- Since [ilmath]x\in U\cap V[/ilmath] was arbitrary we have shown it for all
- Let [ilmath]x\in U\cap V[/ilmath] be given
- We've shown it
- 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]
- 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
- Let [ilmath]U,\ V\in\mathcal{B} [/ilmath] be given.
Thus [ilmath]\mathcal{B} [/ilmath] suitable to generate a topology and be a basis for that topology
Discussion of result
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Notes
- ↑ 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].
- We then use Union of subsets is a subset of the union (with [ilmath]B_\alpha:\eq X[/ilmath]) to see that [ilmath]\bigcup\mathcal{B}\subseteq X[/ilmath] - as required.
- ↑ 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]
- ↑ 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)
- [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]
References
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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