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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{{Stub page|grade=A|msg=Taking a break now, flesh out in the future}} | {{Stub page|grade=A|msg=Taking a break now, flesh out in the future}} | ||
__TOC__ | __TOC__ | ||
| − | == | + | =={{subpage|Statement}}== |
| − | + | {{/Statement}} | |
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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<br/><center><gallery> |
| + | File:ConnectedSubsetCoveredProof.JPG|Most of proof is done here | ||
| + | </gallery></center> | ||
| + | |easy=true}} | ||
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==Leads to== | ==Leads to== | ||
* [[A space is a disconnected subset of itself if and only if the space itself is disconnected]] | * [[A space is a disconnected subset of itself if and only if the space itself is disconnected]] | ||
Latest revision as of 09:39, 2 October 2016
Stub grade: A
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
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 (the very definition of disconnected subset)
- [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
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
This proof has been marked as an page requiring an easy proof
The message provided is:
This is actually rather important and ought to be done
Most of proof is done here
This proof has been marked as an page requiring an easy proof
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
