Difference between revisions of "Connected (topology)"
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==Definition== | ==Definition== | ||
| − | A [[Topological space|topological space]] <math>(X,\mathcal{J})</math> is connected if there is no separation of <math>X</math> | + | 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> | |
| − | + | ||
| − | A separation of | + | |
| + | If there is no such separation then the space is ''connected''<ref name="Analysis">Analysis - Part 1: Elements - Krzysztof Maurin</ref> | ||
==Equivalent definition== | ==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}} | {{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}} | {{Begin Proof}} | ||
'''Connected<math>\implies</math>only sets both open and closed are <math>X,\emptyset</math>''' | '''Connected<math>\implies</math>only sets both open and closed are <math>X,\emptyset</math>''' | ||
| Line 27: | Line 29: | ||
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''': | 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>\exists U,V\in\mathcal{J}</math> such that: | ||
| − | ** <math> | + | ** <math>Y\subseteq U\cup V</math> and |
| − | ** <math>U\cap V\subseteq C( | + | ** <math>U\cap V\subseteq C(Y)</math> and |
| − | ** Both <math>U\cap | + | ** Both <math>U\cap Y\ne\emptyset</math> and <math>V\cap Y\ne\emptyset</math> |
| − | This is | + | 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== | ==Results== | ||
Revision as of 00:42, 22 June 2015
Contents
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:
- [math]U\cap V=\emptyset[/math] and
- [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
