# 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

## 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:

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

## Notes

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.

# OLD PAGE

## Definition

A topological space $(X,\mathcal{J})$ is connected if there is no separation of $X$[1] A separation of [ilmath]X[/ilmath] is:

• A pair of non-empty open sets in [ilmath]X[/ilmath], which we'll denote as $U,\ V$ where:
1. $U\cap V=\emptyset$ and
2. $U\cup V=X$

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 $(X,\mathcal{J})$ is connected if and only if the sets $X,\emptyset$ are the only two sets that are both open and closed.

Connected$\implies$only sets both open and closed are $X,\emptyset$

Suppose $X$ is connected and there exists a set $A$ that is not empty and not all of $X$ which is both open and closed. Then as :this is closed, $X-A$ is open. Thus $A,X-A$ is a separation, contradicting that $X$ is connected.

Only sets both open and closed are $X,\emptyset\implies$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:

• $\exists U,V\in\mathcal{J}$ such that:
• $Y\subseteq U\cup V$ and
• $U\cap V\subseteq C(Y)$ and
• Both $U\cap Y\ne\emptyset$ and $V\cap Y\ne\emptyset$

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 $\exists U,V\in\mathcal{J}$ such that: $A\subseteq U\cup V$, $U\cap V\subseteq C(A)$, $U\cap A\ne\emptyset$ and $V\cap A\ne\emptyset$

TODO: Mendelson p115

Theorem: The image of a connected set is connected under a continuous map

TODO: Mendelson p116

## References

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