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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==Statement== | ==Statement== | ||
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}}: | 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}}: | ||
| − | * 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}} | + | * 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}} (the definition of {{link|disconnected subset|topology}}) |
{{iff}} | {{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}} | * {{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}}''" | ** 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!}} | {{Todo|There is a formulation similar this (see p114, Mendelson) that works in terms of closed rather than open sets, link to it!}} | ||
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==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 00:48, 2 October 2016
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Contents
[hide]Statement
Let (X,J) be a topological space and let A∈P(X) be an arbitrary subset of X, then[1]:
- The topological space (A,JA) (which is A imbued with the subspace topology inherited from (X,J)) is disconnected (the definition of disconnected subset)
- ∃U,V∈J[ U∩A≠∅ ⏟U non-empty in A∧ V∩A≠∅ ⏟V non-empty in A∧ U∩V∩A=∅ ⏟U, V disjoint in A∧ A⊆U∪V ⏟covers A] - the "disjoint in A" condition is perhaps better written as: (U∩A)∩(V∩A)=∅
- In words - "there exist two sets open in (X,J) that are disjoint in A and non-empty in A that cover 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
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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
