View source for Connected (topology)

{{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==
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}}

==Notes==
<references group="Note"/>
==References==
<references/>
{{Topology navbox|plain}}
{{Definition|Topology}}
<br/><br/><hr/><br/><hr/><br/>

=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==
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.

'''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}}