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Ancient page, needs an update, linking to theorems, so forth

There are many ways to state connectedness, and one can just as well start with disconnected and then define "connected" as "not disconnected". I have attempted to pick one and mention the others, do not be put off if you have found another definition! I have started with the most intuitive definition

Definition

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space. We say [ilmath]X[/ilmath] is connected if^[1]:

it is not disconnected^{[Note 1]} (expand the yellow box below for a reminder of this definition)

There are equivalent definitions, some are given below. Note also, that by this convention the [ilmath]\emptyset[/ilmath] is connected.

Recall the definition of a topological space being disconnected

A topological space, [ilmath](X,\mathcal{ J })[/ilmath], is said to be disconnected if^[1]:

[ilmath]\exists U,V\in\mathcal{J}[U\ne\emptyset\wedge V\neq\emptyset\wedge V\cap U=\emptyset\wedge U\cup V=X][/ilmath], in words "if there exists a pair of disjoint and non-empty open sets, [ilmath]U[/ilmath] and [ilmath]V[/ilmath], such that their union is [ilmath]X[/ilmath]"

In this case, [ilmath]U[/ilmath] and [ilmath]V[/ilmath] are said to disconnect [ilmath]X[/ilmath]^[1] and are sometimes called a separation of [ilmath]X[/ilmath].

Of a subset

Let [ilmath]A\in\mathcal{P}(X)[/ilmath] be an arbitrary subset of [ilmath]X[/ilmath], then:

we say [ilmath]A[/ilmath] is connected if it is connected when considered as a topological subspace of [ilmath](X,\mathcal{ J })[/ilmath]^[1]^[2].

There are equivalent definitions, some are given below.

Equivalent conditions

To a topological space [ilmath](X,\mathcal{ J })[/ilmath] being connected:

To an arbitrary subset, [ilmath]A\in\mathcal{P}(X)[/ilmath], being connected:

Obviously, the only sets being both relatively open and relatively closed in [ilmath]A[/ilmath] are [ilmath]\emptyset[/ilmath] and [ilmath]A[/ilmath] itself. (This comes directly from the subspace definition above)
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
- Then apply the definition above, a subset is considered connected if it is not disconnected
A subset of a topological space is connected if and only if and only if the only two subsets that are both relatively open and relatively closed with respect to the subset are the empty-set and the subset itself

Notes

↑
We could write this as:
- [ilmath]\neg(\exists U,V\in\mathcal{J}[U\ne\emptyset\wedge V\neq\emptyset\wedge U\cap V=\emptyset\wedge U\cup V=X])[/ilmath]
Which is:
- [ilmath]\forall U,V\in\mathcal{J}[U=\emptyset\vee V=\emptyset\vee U\cap V\ne\emptyset\vee U\cup V\ne X][/ilmath]
but, whilst completely "true", this is difficult to read and far less intuitive.

References

↑ ^1.0 ^1.1 ^1.2 ^1.3 Introduction to Topological Manifolds - John M. Lee
↑ ^2.0 ^2.1 Introduction to Topology - Bert Mendelson

OLD PAGE

Definition

A topological space [math](X,\mathcal{J})[/math] is connected if there is no separation of [math]X[/math]^[1] A separation of [ilmath]X[/ilmath] is:

A pair of non-empty open sets in [ilmath]X[/ilmath], which we'll denote as [math]U,\ V[/math] where:
1. [math]U\cap V=\emptyset[/math] and
2. [math]U\cup V=X[/math]

If there is no such separation then the space is connected^[2]

Equivalent definition

This definition is equivalent (true if and only if) the only empty sets that are both open in [ilmath]X[/ilmath] are:

[ilmath]\emptyset[/ilmath] and
[ilmath]X[/ilmath] itself.

I will prove this claim now:

Claim: A topological space [math](X,\mathcal{J})[/math] is connected if and only if the sets [math]X,\emptyset[/math] are the only two sets that are both open and closed.

Connected[math]\implies[/math]only sets both open and closed are [math]X,\emptyset[/math]

Suppose [math]X[/math] is connected and there exists a set [math]A[/math] that is not empty and not all of [math]X[/math] which is both open and closed. Then as :this is closed, [math]X-A[/math] is open. Thus [math]A,X-A[/math] is a separation, contradicting that [math]X[/math] is connected.

Only sets both open and closed are [math]X,\emptyset\implies[/math]connected

TODO:

Connected subset

A subset [ilmath]A[/ilmath] of a Topological space [ilmath](X,\mathcal{J})[/ilmath] is connected if (when considered with the Subspace topology) the only two Relatively open and Relatively closed (in A) sets are [ilmath]A[/ilmath] and [ilmath]\emptyset[/ilmath]^[3]

Useful lemma

Given a topological subspace [ilmath]Y[/ilmath] of a space [ilmath](X,\mathcal{J})[/ilmath] we say that [ilmath]Y[/ilmath] is disconnected if and only if:

[math]\exists U,V\in\mathcal{J}[/math] such that:
- [math]Y\subseteq U\cup V[/math] and
- [math]U\cap V\subseteq C(Y)[/math] and
- Both [math]U\cap Y\ne\emptyset[/math] and [math]V\cap Y\ne\emptyset[/math]

This is basically says there has to be a separation of [ilmath]Y[/ilmath] that isn't just [ilmath]Y[/ilmath] and the [ilmath]\emptyset[/ilmath] for [ilmath]Y[/ilmath] to be disconnected, but the sets may overlap outside of {{M|Y}

Proof of lemma:

TODO:

Results

Theorem:Given a topological subspace [ilmath]Y[/ilmath] of a space [ilmath](X,\mathcal{J})[/ilmath] we say that [ilmath]Y[/ilmath] is disconnected if and only if [math]\exists U,V\in\mathcal{J}[/math] such that: [math]A\subseteq U\cup V[/math], [math]U\cap V\subseteq C(A)[/math], [math]U\cap A\ne\emptyset[/math] and [math]V\cap A\ne\emptyset[/math]

TODO: Mendelson p115

Theorem: The image of a connected set is connected under a continuous map

TODO: Mendelson p116

References

↑ Topology - James R. Munkres - 2nd edition
↑ Analysis - Part 1: Elements - Krzysztof Maurin
↑ Introduction to topology - Mendelson - third edition

[2] We could write this as:
[ilmath]\neg(\exists U,V\in\mathcal{J}[U\ne\emptyset\wedge V\neq\emptyset\wedge U\cap V=\emptyset\wedge U\cup V=X])[/ilmath]
Which is:
[ilmath]\forall U,V\in\mathcal{J}[U=\emptyset\vee V=\emptyset\vee U\cap V\ne\emptyset\vee U\cup V\ne X][/ilmath]
but, whilst completely "true", this is difficult to read and far less intuitive.

[2] [ilmath]\neg(\exists U,V\in\mathcal{J}[U\ne\emptyset\wedge V\neq\emptyset\wedge U\cap V=\emptyset\wedge U\cup V=X])[/ilmath]

[3] [ilmath]\forall U,V\in\mathcal{J}[U=\emptyset\vee V=\emptyset\vee U\cap V\ne\emptyset\vee U\cup V\ne X][/ilmath]

[ITTMJML-1] 1.0 ^1.1 ^1.2 ^1.3 Introduction to Topological Manifolds - John M. Lee

[ITTBM-3] 2.0 ^2.1 Introduction to Topology - Bert Mendelson

[Topology-4] Topology - James R. Munkres - 2nd edition

[Analysis-5] Analysis - Part 1: Elements - Krzysztof Maurin

[6] Introduction to topology - Mendelson - third edition

[1]

[Note 1]

[2]

[1]

[2]

[3]

@@ Line 1: / Line 1: @@
+{{Refactor notice|grade=A|msg=Ancient page, needs an update, linking to theorems, so forth}}
+: There are many ways to state connectedness, and one can just as well start with {{link|disconnected|topology}} and then define "connected" as "not disconnected". I have attempted to pick one and mention the others, do not be put off if you have found ''another'' definition! I have started with the most intuitive definition
+__TOC__
 ==Definition==
-A [[Topological space|topological space]] <math>(X,\mathcal{J})</math> is connected if there is no separation of <math>X</math>
+Let {{Top.|X|J}} be a [[topological space]]. We say {{M|X}} is ''connected'' if{{rITTMJML}}:
+* it is not {{link|disconnected|topology}}<ref group="Note">We could write this as:
+* {{M|1=\neg(\exists U,V\in\mathcal{J}[U\ne\emptyset\wedge V\neq\emptyset\wedge U\cap V=\emptyset\wedge U\cup V=X])}}
+Which is:
+* {{M|1=\forall U,V\in\mathcal{J}[U=\emptyset\vee V=\emptyset\vee U\cap V\ne\emptyset\vee U\cup V\ne X]}}
+but, whilst completely "true", this is difficult to read and far less intuitive.
+</ref> (expand the yellow box below for a reminder of this definition)
+There are equivalent definitions, some are given below. Note also, that by this convention the [[emptyset|{{M|\emptyset}}]] is connected.
+{{Begin Notebox}}
+'''Recall''' the definition of a topological space being ''{{link|disconnected|topology}}''
+{{Begin Notebox Content}}
+{{:Disconnected (topology)/Definition}}
+{{End Notebox Content}}{{End Notebox}}
+===Of a subset===
+Let {{M|A\in\mathcal{P}(X)}} be an arbitrary [[subset of]] {{M|X}}, then:
+* we say {{M|A}} is connected if it is connected when considered as a [[subspace topology|topological subspace]] of {{Top.|X|J}}{{rITTMJML}}{{rITTBM}}.
+There are equivalent definitions, some are given below.
+=={{subpage|Equivalent conditions}}==
+{{/Equivalent conditions}}
+==See also==
+* [[Disconnected (topology)|Disconnected]]
+* [[Every continuous map from a non-empty connected space to a discrete space is constant]]
+{{Todo|Flesh out, add more theorems, for example image of a connected set is connected, so forth}}
-===Separation===
+==Notes==
-This belongs on this page because a separation is only useful in this definition.
+<references group="Note"/>
+==References==
+<references/>
+{{Topology navbox|plain}}
+{{Definition|Topology}}
+<br/><br/><hr/><br/><hr/><br/>
-A separation of <math>X</math> is a pair of two non-empty [[Open set|open sets]] <math>U,V</math> where <math>U\cap V=\emptyset</math> where <math>U\cup V=X</math>
+=OLD PAGE=
+==Definition==
+A [[Topological space|topological space]] <math>(X,\mathcal{J})</math> is connected if there is no separation of <math>X</math><ref name="Topology">Topology - James R. Munkres - 2nd edition</ref> A separation of {{M|X}} is:
+* A pair of non-empty [[Open set|open sets]] in {{M|X}}, which we'll denote as <math>U,\ V</math> where:
+*# <math>U\cap V=\emptyset</math> and
+*# <math>U\cup V=X</math>
+If there is no such separation then the space is ''connected''<ref name="Analysis">Analysis - Part 1: Elements - Krzysztof Maurin</ref>
 ==Equivalent definition==
-We can also say: A topological space <math>(X,\mathcal{J})</math> is connected if and only if the sets <math>X,\emptyset</math> are the only two sets that are both open and closed.
+This definition is equivalent (true ''if and only if'') the only empty sets that are both open in {{M|X}} are:
+# {{M|\emptyset}} and
+# {{M|X}} itself.
+I will prove this claim now:
+{{Begin Theorem}}
+Claim: A topological space <math>(X,\mathcal{J})</math> is connected if and only if the sets <math>X,\emptyset</math> are the only two sets that are both open and closed.
+{{Begin Proof}}
+'''Connected<math>\implies</math>only sets both open and closed are <math>X,\emptyset</math>'''
+:Suppose <math>X</math> is connected and there exists a set <math>A</math> that is not empty and not all of <math>X</math> which is both open and closed. Then as :this is closed, <math>X-A</math> is open. Thus <math>A,X-A</math> is a separation, contradicting that <math>X</math> is connected.
-===Proof===
+'''Only sets both open and closed are <math>X,\emptyset\implies</math>connected'''
 {{Todo}}
+{{End Proof}}
+{{End Theorem}}
+==Connected subset==
+A subset {{M|A}} of a [[Topological space]] {{M|(X,\mathcal{J})}} is connected if (when considered with the [[Subspace topology]]) the only two [[Relatively open]] and [[Relatively closed]] (in A) sets are {{M|A}} and {{M|\emptyset}}<ref>Introduction to topology - Mendelson - third edition</ref>
+==Useful lemma==
+Given a [[Subspace topology|topological subspace]] {{M|Y}} of a space {{M|(X,\mathcal{J})}} we say that {{M|Y}} is disconnected '''if and only if''':
+* <math>\exists U,V\in\mathcal{J}</math> such that:
+** <math>Y\subseteq U\cup V</math> and
+** <math>U\cap V\subseteq C(Y)</math> and
+** Both <math>U\cap Y\ne\emptyset</math> and <math>V\cap Y\ne\emptyset</math>
+This is basically says there has to be a separation of {{M|Y}} that isn't just {{M|Y}} and the {{M|\emptyset}} for {{M|Y}} to be disconnected, but the sets may overlap outside of {{M|Y}
+{{Begin Theorem}}
+Proof of lemma:
+{{Begin Proof}}
+{{Todo}}
+{{End Proof}}{{End Theorem}}
+==Results==
+{{Begin Theorem}}
+Theorem:Given a [[Subspace topology|topological subspace]] {{M|Y}} of a space {{M|(X,\mathcal{J})}} we say that {{M|Y}} is disconnected '''if and only if''' <math>\exists U,V\in\mathcal{J}</math> such that: <math>A\subseteq U\cup V</math>, <math>U\cap V\subseteq C(A)</math>, <math>U\cap A\ne\emptyset</math> and <math>V\cap A\ne\emptyset</math>
+{{Begin Proof}}
+{{Todo|Mendelson p115}}
+{{End Proof}}{{End Theorem}}
+{{Begin Theorem}}
+Theorem: The image of a connected set is connected under a continuous map
+{{Begin Proof}}
+{{Todo|Mendelson p116}}
+{{End Proof}}
+{{End Theorem}}
+==References==
+<references/>
 {{Definition|Topology}}

Difference between revisions of "Connected (topology)"

Latest revision as of 01:52, 1 October 2016

Contents

Definition

Of a subset

Equivalent conditions

See also

Notes

References

OLD PAGE

Definition

Equivalent definition

Connected subset

Useful lemma

Results

References

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