Difference between revisions of "Connected (topology)"
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Revision as of 21:29, 30 September 2016
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
