Difference between revisions of "Open set"
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| + | {{Refactor notice}} | ||
| + | ==Definition== | ||
| + | ===Topological space=== | ||
| + | In a [[topological space]] {{M|(X,\mathcal{J})}} we have: | ||
| + | * {{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}} | ||
| + | ===Metric space=== | ||
| + | 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}}. | ||
| + | : '''I claim that the following definitions are equivalent:''' | ||
| + | ====Definition 1==== | ||
| + | * 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: | ||
| + | ** {{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 | ||
| + | ** For all {{M|x}} in {{M|U}} there is an [[open ball]] centred at {{M|x}} entirely contained within {{M|U}} | ||
| + | ====Definition 2==== | ||
| + | * A set {{M|U\subseteq X}} is ''open'' in {{M|(X,d)}} (or just {{M|X}} if the metric is implicit) if{{rITTGG}}: | ||
| + | ** {{M|1=\text{Int}(U)=U}} - (recall that {{M|\text{Int}(U)}} denotes the [[interior]] of {{M|U}}), that is to say | ||
| + | **: (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]}}) | ||
| + | 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: | ||
| + | * {{M|\emptyset}} is open ''(claim 2)'' | ||
| + | * {{M|X}} itself is open ''(claim 3)'' | ||
| + | * {{M|\text{Int}(U)}} is open | ||
| + | *: (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)}}) | ||
| + | ==Proof of claims== | ||
| + | {{Begin Inline Theorem}} | ||
| + | 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}} | ||
| + | {{Begin Inline Proof}} | ||
| + | {{Todo|Trivial proof, be bothered to do it}} | ||
| + | {{End Proof}}{{End Theorem}} | ||
| + | {{Begin Inline Theorem}} | ||
| + | Claim 2: {{M|\emptyset}} is open | ||
| + | {{Begin Inline Proof}} | ||
| + | {{Todo|Trivial proof, be bothered to do it}} | ||
| + | {{End Proof}}{{End Theorem}} | ||
| + | {{Begin Inline Theorem}} | ||
| + | Claim 3: {{M|X}} itself is open | ||
| + | {{Begin Inline Proof}} | ||
| + | {{Todo|Trivial proof, be bothered to do it}} | ||
| + | {{End Proof}}{{End Theorem}} | ||
| + | ==See also== | ||
| + | * [[Open ball]] | ||
| + | * [[Closed set]] | ||
| + | * [[Neighbourhood]] | ||
| + | * [[Topological space]] | ||
| + | * [[Metric space]] | ||
| + | ==References== | ||
| + | <references/> | ||
| + | |||
| + | {{Definition|Metric Space|Topology|Functional Analysis}} | ||
| + | |||
| + | |||
| + | =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> | ||
Latest revision as of 02:26, 29 November 2015
Contents
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]:
- [ilmath]\text{Int}(U)=U[/ilmath] - (recall that [ilmath]\text{Int}(U)[/ilmath] denotes the interior of [ilmath]U[/ilmath]), that is to say
- (recall that [ilmath]\ x\text{ is interior to }U\} [/ilmath] and that a point, [ilmath]x[/ilmath] is interior to [ilmath]U[/ilmath] if [ilmath]\exists\delta>0[B_\delta(x)\subseteq U][/ilmath])
- [ilmath]\text{Int}(U)=U[/ilmath] - (recall that [ilmath]\text{Int}(U)[/ilmath] denotes the interior of [ilmath]U[/ilmath]), that is to say
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
- ↑ Introduction to Topology - Bert Mendelson
- ↑ 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]
