Difference between revisions of "A subset of a topological space is disconnected if and only if it can be covered by two non-empty-in-the-subset and disjoint-in-the-subset sets that are open in the space itself"
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m (Alec moved page A subset of a topological space is disconnected if and only if it can be covered by two non-empty disjoint open sets of the space to [[A subset of a topological space is disconnected if and only if it can be covered by two non-empty...) |
(Page now has correct title, and correct statement.) |
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==Statement== | ==Statement== | ||
− | Let {{Top.|X|J}} be a [[topological space]] and let {{M| | + | Let {{Top.|X|J}} be a [[topological space]] and let {{M|A\in\mathcal{P}(X)}} be an arbitrary [[subset of]] {{M|X}}, then{{rITTBM}}: |
− | * {{M|A}} is {{ | + | * The topological space {{M|(A,\mathcal{J}_A)}} (which is {{M|A}} imbued with the [[subspace topology]] inherited from {{Top.|X|J}}) is {{link|disconnected|topology}} |
− | + | {{iff}} | |
+ | * {{M|1=\exists U,V\in\mathcal{J}[\underbrace{\ U\cap A\ne\emptyset\ }_{U\text{ non-empty in }A}\wedge\underbrace{\ V\cap A\ne\emptyset\ }_{V\text{ non-empty in } A}\wedge\underbrace{\ U\cap V\cap A=\emptyset\ }_{U,\ V\text{ disjoint in }A}\wedge\underbrace{\ A\subseteq U\cup V\ }_{\text{covers }A}]}} - the "disjoint in {{M|A}}" condition is perhaps better written as: {{M|1=(U\cap A)\cap(V\cap A)=\emptyset}} | ||
+ | ** In words - "''there exist two sets open in {{Top.|X|J}} that are [[disjoint in]] {{M|A}} and [[non-empty in]] {{M|A}} that [[cover]] {{M|A}}''" | ||
+ | {{Todo|There is a formulation similar this (see p114, Mendelson) that works in terms of closed rather than open sets, link to it!}} | ||
==Proof== | ==Proof== | ||
{{Requires proof|grade=A|msg=This is actually rather important and ought to be done|easy=true}} | {{Requires proof|grade=A|msg=This is actually rather important and ought to be done|easy=true}} |
Revision as of 00:40, 2 October 2016
Stub grade: A
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Taking a break now, flesh out in the future
Contents
Statement
Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space and let [ilmath]A\in\mathcal{P}(X)[/ilmath] be an arbitrary subset of [ilmath]X[/ilmath], then[1]:
- The topological space [ilmath](A,\mathcal{J}_A)[/ilmath] (which is [ilmath]A[/ilmath] imbued with the subspace topology inherited from [ilmath](X,\mathcal{ J })[/ilmath]) is disconnected
- [ilmath]\exists U,V\in\mathcal{J}[\underbrace{\ U\cap A\ne\emptyset\ }_{U\text{ non-empty in }A}\wedge\underbrace{\ V\cap A\ne\emptyset\ }_{V\text{ non-empty in } A}\wedge\underbrace{\ U\cap V\cap A=\emptyset\ }_{U,\ V\text{ disjoint in }A}\wedge\underbrace{\ A\subseteq U\cup V\ }_{\text{covers }A}][/ilmath] - the "disjoint in [ilmath]A[/ilmath]" condition is perhaps better written as: [ilmath](U\cap A)\cap(V\cap A)=\emptyset[/ilmath]
- In words - "there exist two sets open in [ilmath](X,\mathcal{ J })[/ilmath] that are disjoint in [ilmath]A[/ilmath] and non-empty in [ilmath]A[/ilmath] that cover [ilmath]A[/ilmath]"
TODO: There is a formulation similar this (see p114, Mendelson) that works in terms of closed rather than open sets, link to it!
Proof
Grade: A
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This is actually rather important and ought to be done
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Leads to
- A space is a disconnected subset of itself if and only if the space itself is disconnected
- A space is a connected subset of itself if and only if the space itself is connected - one can be used to prove the other
See also
TODO: Flesh out
References