Connected (topology)

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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


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

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:

See also

TODO: Flesh out, add more theorems, for example image of a connected set is connected, so forth


  1. 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.


  1. 1.0 1.1 1.2 1.3 Introduction to Topological Manifolds - John M. Lee
  2. 2.0 2.1 Introduction to Topology - Bert Mendelson



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:

  1. [ilmath]\emptyset[/ilmath] and
  2. [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


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:



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


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