Connected (topology)
The message provided is:
- 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
Contents
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:
- A topological space is connected if and only if the only sets that are both open and closed in the space are the entire space itself and the emptyset
- Some authors give this as the definition of a connected space, eg[2]
- A topological space is disconnected if and only if there exists a non-constant continuous function from the space to the discrete space on two elements
- A topological space is disconnected if and only if it is homeomorphic to a disjoint union of two or more non-empty topological spaces
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
See also
TODO: Flesh out, add more theorems, for example image of a connected set is connected, so forth
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]
- [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]
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
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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:
- [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