Difference between revisions of "The image of a connected set is connected"

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(Created page with "{{Stub page|grade=A*|msg=Doing some work while I've got a bit of time}} {{Caution|This is being done RIGHT BEFORE BED - do not rely on it until I've checked it}} __TOC__ ==Sta...")
 
(Proof last night was.... wrong, added references and navbox, so forth)
 
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{{Stub page|grade=A*|msg=Doing some work while I've got a bit of time}}
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{{Stub page|grade=A*|msg=Doing some work while I've got a bit of time, majority written right before bed, so needs checking. This is also "the theorem" of connectedness}}
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{{Caution|This is being done RIGHT BEFORE BED - do not rely on it until I've checked it}}
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__TOC__
 
__TOC__
 
==Statement==
 
==Statement==
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Let {{Top.|X|J}} and {{Top.|Y|K}} be [[topological spaces]] and let {{M|f:X\rightarrow Y}} be a ''[[continuous]]'' [[map]]. Then, for any {{M|A\in\mathcal{P}(X)}}, we have:
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Let {{Top.|X|J}} and {{Top.|Y|K}} be [[topological spaces]] and let {{M|f:X\rightarrow Y}} be a ''[[continuous]]'' [[map]]. Then, for any {{M|A\in\mathcal{P}(X)}}, we have{{rITTBM}}:
 
* If {{M|A}} is a {{link|connected subset|topology}} of {{Top.|X|J}} then {{M|f(A)}} is connected subset in {{Top.|Y|K}}
 
* If {{M|A}} is a {{link|connected subset|topology}} of {{Top.|X|J}} then {{M|f(A)}} is connected subset in {{Top.|Y|K}}
 
==Proof==
 
==Proof==
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Suppose {{M|f(A)}} is disconnected, and {{M|(f(A),\mathcal{K}_{f(A)})}} is a [[topological subspace]] of {{Top.|Y|K}}.
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We do this proof by [[contrapositive]], that is noting that: {{M|1=(A\implies B)\iff((\neg B)\implies(\neg A))}}, as such we will show:
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* Then there exist {{M|U,V\in\mathcal{K}_{f(A)} }} such that {{M|U}} and {{M|V}} [[disconnected (topology)|disconnect]] {{M|f(A)}}
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* if {{M|f(A)}} is {{link|disconnected|topology}} then {{M|A}} is disconnected
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** {{M|f^{-1}(U)}} and {{M|f^{-1}(V)}} are disjoint by {{M|f}} being a function and their union contains {{M|A}} (but could be bigger than it, as we might not have {{M|1=f^{-1}(f(A))=A}} of course!)
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Let us begin:
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** We apply the right-hand part of:
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*Suppose {{M|f(A)}} is disconnected, and write {{M|(f(A),\mathcal{K}_{f(A)})}} for the [[topological subspace]] on {{M|f(A)}} of {{Top.|Y|K}}.
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*** [[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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** Then there exists {{M|U,V\in\mathcal{K}_{f(A)} }} such that {{M|U}} and {{M|V}} [[disconnected (topology)|disconnect]] {{M|f(A)}}<ref group="Note">Recall {{M|U}} and {{M|V}} are said to disconnect {{M|f(A)}} if:
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# they are [[disjoint]],
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# they are both [[non-empty]], and
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# {{M|1=U\cup V=f(A)}}</ref>
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*** Notice that {{M|f^{-1}(U)}} and {{M|f^{-1}(V)}} are [[disjoint]]
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**** PROOF HERE
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*** Notice also that {{M|A\subseteq f^{-1}(U)\cup f^{-1}(V)}} (''we may or may not have: {{M|1=A=f^{-1}(U)\cup f^{-1}(V)}}, as there might exist elements outside of {{M|A}} that map into {{M|f(A)}} notice'')
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**** PROOF HERE
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{{Caution|We do not know whether or not {{M|f^{-1}(U)}} or {{M|f^{-1}(V)}} are open in {{M|X}} (and they may form a strict superset of {{M|A}} so cannot be open in {{M|A}})}}
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{{Requires proof|grade=A|msg=Working on it}}
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===OLD WORKINGS===
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* OLD
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** WORK
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*** {{M|f^{-1}(U)}} and {{M|f^{-1}(V)}} are disjoint by {{M|f}} being a function and their union contains {{M|A}} (but could be bigger than it, as we might not have {{M|1=f^{-1}(f(A))=A}} of course!)
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*** We apply the right-hand part of:
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**** [[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]]
 
This completes the proof
 
This completes the proof
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{{Caution|Good night, still to do, put page in the right place!}}
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==Notes==
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<references group="Note"/>
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==References==
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<references/>
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{{Topology navbox|plain}}
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{{Theorem Of|Topology}}

Latest revision as of 04:10, 3 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:
Doing some work while I've got a bit of time, majority written right before bed, so needs checking. This is also "the theorem" of connectedness

Statement

Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces and let [ilmath]f:X\rightarrow Y[/ilmath] be a continuous map. Then, for any [ilmath]A\in\mathcal{P}(X)[/ilmath], we have[1]:

  • If [ilmath]A[/ilmath] is a connected subset of [ilmath](X,\mathcal{ J })[/ilmath] then [ilmath]f(A)[/ilmath] is connected subset in [ilmath](Y,\mathcal{ K })[/ilmath]

Proof

We do this proof by contrapositive, that is noting that: [ilmath](A\implies B)\iff((\neg B)\implies(\neg A))[/ilmath], as such we will show:

  • if [ilmath]f(A)[/ilmath] is disconnected then [ilmath]A[/ilmath] is disconnected

Let us begin:

  • Suppose [ilmath]f(A)[/ilmath] is disconnected, and write [ilmath](f(A),\mathcal{K}_{f(A)})[/ilmath] for the topological subspace on [ilmath]f(A)[/ilmath] of [ilmath](Y,\mathcal{ K })[/ilmath].
    • Then there exists [ilmath]U,V\in\mathcal{K}_{f(A)} [/ilmath] such that [ilmath]U[/ilmath] and [ilmath]V[/ilmath] disconnect [ilmath]f(A)[/ilmath][Note 1]
      • Notice that [ilmath]f^{-1}(U)[/ilmath] and [ilmath]f^{-1}(V)[/ilmath] are disjoint
        • PROOF HERE
      • Notice also that [ilmath]A\subseteq f^{-1}(U)\cup f^{-1}(V)[/ilmath] (we may or may not have: [ilmath]A=f^{-1}(U)\cup f^{-1}(V)[/ilmath], as there might exist elements outside of [ilmath]A[/ilmath] that map into [ilmath]f(A)[/ilmath] notice)
        • PROOF HERE

Caution:We do not know whether or not [ilmath]f^{-1}(U)[/ilmath] or [ilmath]f^{-1}(V)[/ilmath] are open in [ilmath]X[/ilmath] (and they may form a strict superset of [ilmath]A[/ilmath] so cannot be open in [ilmath]A[/ilmath])

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.
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The message provided is:
Working on it

OLD WORKINGS

This completes the proof

Notes

  1. ↑ Recall [ilmath]U[/ilmath] and [ilmath]V[/ilmath] are said to disconnect [ilmath]f(A)[/ilmath] if:
    1. they are disjoint,
    2. they are both non-empty, and
    3. [ilmath]U\cup V=f(A)[/ilmath]

References

  1. ↑ Introduction to Topology - Bert Mendelson

v • d • e
 
Topology
Overview of the structures studied in topology
Primitives
Global properties of topological spaces
Local properties of topological spaces
Constructs
Subtypes of topological spaces
inner product spaces [ilmath]\subset[/ilmath] normed spaces [ilmath]\subset[/ilmath] metric spaces [ilmath]\subset[/ilmath] topological spaces
Maps between topological spaces



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