Difference between revisions of "Open set"

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{{Refactor notice}}
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==Definition==
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===Topological space===
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In a [[topological space]] {{M|(X,\mathcal{J})}} we have:
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* {{M|\forall S\in\mathcal{J} }} that {{M|S}} is an open set. {{M|\mathcal{J} }} is by definition the set of open sets of {{M|X}}
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===Metric space===
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In a [[metric space]] {{M|(X,d)}} there are 2 definitions of open set, however it will be shown that they are equivalent. Here {{M|U}} is some arbitrary subset of {{M|X}}.
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: '''I claim that the following definitions are equivalent:'''
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====Definition 1====
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* A set {{M|U\subseteq X}} is ''open'' in {{M|(X,d)}} (or just {{M|X}} if the metric is implicit) if {{M|U}} is a [[neighbourhood]] to all of its points{{rITTBM}}, that is to say:
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** {{M|U\subseteq X}} is open if {{MM|1=\forall x\in U\exists\delta_x>0[B_{\delta_x}(X)\subseteq U]}} - (recall that {{M|B_r(x)}} denotes the [[open ball]] of radius {{M|r}} centred at {{M|x}}) or
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** For all {{M|x}} in {{M|U}} there is an [[open ball]] centred at {{M|x}} entirely contained within {{M|U}}
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====Definition 2====
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* A set {{M|U\subseteq X}} is ''open'' in {{M|(X,d)}} (or just {{M|X}} if the metric is implicit) if{{rITTGG}}:
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** {{M|1=\text{Int}(U)=U}} - (recall that {{M|\text{Int}(U)}} denotes the [[interior]] of {{M|U}}), that is to say
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**: (recall that {{M|1=\text{Int}(U):=\{x\in X|\ x\text{ is interior to }U\} }} and that a [[interior point|point, {{M|x}} is interior to {{M|U}}]] if {{M|\exists\delta>0[B_\delta(x)\subseteq U]}})
  
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It is easy to see that these definitions are very similar to each other (these are indeed equivalent is ''claim 1'')
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==Immediate results==
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It is easily seen that:
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* {{M|\emptyset}} is open ''(claim 2)''
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* {{M|X}} itself is open ''(claim 3)''
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* {{M|\text{Int}(U)}} is open
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*: (see [[interior]] for a proof of this. The claim is the same as {{M|1=\text{Int}(\text{Int}(U))=\text{Int}(U)}} as by the first claim we can use either definition of open, so we use {{M|U}} is open if {{M|1=U=\text{Int}(U)}})
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==Proof of claims==
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{{Begin Inline Theorem}}
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Claim 1: A set, {{M|U\subseteq X}} is open according to definition {{M|1}} {{M|\iff}} {{M|U}} is open according to definition {{M|2}}
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{{Begin Inline Proof}}
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{{Todo|Trivial proof, be bothered to do it}}
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{{End Proof}}{{End Theorem}}
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{{Begin Inline Theorem}}
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Claim 2: {{M|\emptyset}} is open
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{{Begin Inline Proof}}
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{{Todo|Trivial proof, be bothered to do it}}
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{{End Proof}}{{End Theorem}}
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{{Begin Inline Theorem}}
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Claim 3: {{M|X}} itself is open
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{{Begin Inline Proof}}
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{{Todo|Trivial proof, be bothered to do it}}
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{{End Proof}}{{End Theorem}}
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==See also==
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* [[Open ball]]
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* [[Closed set]]
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* [[Neighbourhood]]
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* [[Topological space]]
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* [[Metric space]]
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==References==
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<references/>
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{{Definition|Metric Space|Topology|Functional Analysis}}
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=Old page=
  
 
Here <math>(X,d)</math> denotes a metric space, and <math>B_r(x)</math> the [[Open ball|open ball]] centred at <math>x</math> of radius <math>r</math>
 
Here <math>(X,d)</math> denotes a metric space, and <math>B_r(x)</math> the [[Open ball|open ball]] centred at <math>x</math> of radius <math>r</math>
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In a [[Topological space|topological space]] the elements of the topology are defined to be open sets
 
In a [[Topological space|topological space]] the elements of the topology are defined to be open sets
  
 
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===Neighbourhood===
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A subset {{M|N}} of a [[Topological space]] {{M|(X,\mathcal{J})}} is a '''neighbourhood of {{M|p}}'''<ref>Introduction to topology - Third Edition - Mendelson</ref> if:
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* <math>\exists U\in\mathcal{J}:p\in U\wedge U\subset N</math>
 
==See also==
 
==See also==
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* [[Relatively open|Relatively open set]]
 
* [[Closed set]]
 
* [[Closed set]]
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* [[Relatively closed|Relatively closed set]]
  
 
==References==
 
==References==
 +
<references/>
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{{Definition|Topology|Metric Space}}
 
{{Definition|Topology|Metric Space}}

Latest revision as of 02:26, 29 November 2015

This page is currently being refactored (along with many others)
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Definition

Topological space

In a topological space [ilmath](X,\mathcal{J})[/ilmath] we have:

  • [ilmath]\forall S\in\mathcal{J} [/ilmath] that [ilmath]S[/ilmath] is an open set. [ilmath]\mathcal{J} [/ilmath] is by definition the set of open sets of [ilmath]X[/ilmath]

Metric space

In a metric space [ilmath](X,d)[/ilmath] there are 2 definitions of open set, however it will be shown that they are equivalent. Here [ilmath]U[/ilmath] is some arbitrary subset of [ilmath]X[/ilmath].

I claim that the following definitions are equivalent:

Definition 1

  • A set [ilmath]U\subseteq X[/ilmath] is open in [ilmath](X,d)[/ilmath] (or just [ilmath]X[/ilmath] if the metric is implicit) if [ilmath]U[/ilmath] is a neighbourhood to all of its points[1], that is to say:
    • [ilmath]U\subseteq X[/ilmath] is open if [math]\forall x\in U\exists\delta_x>0[B_{\delta_x}(X)\subseteq U][/math] - (recall that [ilmath]B_r(x)[/ilmath] denotes the open ball of radius [ilmath]r[/ilmath] centred at [ilmath]x[/ilmath]) or
    • For all [ilmath]x[/ilmath] in [ilmath]U[/ilmath] there is an open ball centred at [ilmath]x[/ilmath] entirely contained within [ilmath]U[/ilmath]

Definition 2

  • A set [ilmath]U\subseteq X[/ilmath] is open in [ilmath](X,d)[/ilmath] (or just [ilmath]X[/ilmath] if the metric is implicit) if[2]:

It is easy to see that these definitions are very similar to each other (these are indeed equivalent is claim 1)

Immediate results

It is easily seen that:

  • [ilmath]\emptyset[/ilmath] is open (claim 2)
  • [ilmath]X[/ilmath] itself is open (claim 3)
  • [ilmath]\text{Int}(U)[/ilmath] is open
    (see interior for a proof of this. The claim is the same as [ilmath]\text{Int}(\text{Int}(U))=\text{Int}(U)[/ilmath] as by the first claim we can use either definition of open, so we use [ilmath]U[/ilmath] is open if [ilmath]U=\text{Int}(U)[/ilmath])

Proof of claims

Claim 1: A set, [ilmath]U\subseteq X[/ilmath] is open according to definition [ilmath]1[/ilmath] [ilmath]\iff[/ilmath] [ilmath]U[/ilmath] is open according to definition [ilmath]2[/ilmath]




TODO: Trivial proof, be bothered to do it


Claim 2: [ilmath]\emptyset[/ilmath] is open




TODO: Trivial proof, be bothered to do it


Claim 3: [ilmath]X[/ilmath] itself is open




TODO: Trivial proof, be bothered to do it


See also

References

  1. Introduction to Topology - Bert Mendelson
  2. Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene


Old page

Here [math](X,d)[/math] denotes a metric space, and [math]B_r(x)[/math] the open ball centred at [math]x[/math] of radius [math]r[/math]

Metric Space definition

"A set [math]U[/math] is open if it is a neighborhood to all of its points"[1] and neighborhood is as you'd expect, "a small area around".

Neighbourhood

A set [math]N[/math] is a neighborhood to [math]a\in X[/math] if [math]\exists\delta>0:B_\delta(a)\subset N[/math]

That is if we can puff up any open ball about [ilmath]x[/ilmath] that is entirely contained in [ilmath]N[/ilmath]

Topology definition

In a topological space the elements of the topology are defined to be open sets

Neighbourhood

A subset [ilmath]N[/ilmath] of a Topological space [ilmath](X,\mathcal{J})[/ilmath] is a neighbourhood of [ilmath]p[/ilmath][2] if:

  • [math]\exists U\in\mathcal{J}:p\in U\wedge U\subset N[/math]

See also

References

  1. Bert Mendelson, Introduction to Topology - definition 6.1, page 52
  2. Introduction to topology - Third Edition - Mendelson