# Topology/Connectedness

## Motivation[edit | edit source]

To best describe what is a connected space, we shall describe first what is a disconnected space. A disconnected space is a space that can be separated into two disjoint groups, or more formally:

A space is said to be **disconnected** iff a pair of disjoint, non-empty *open* subsets exists, such that .

A space that is not disconnected is said to be a connected space.

### Examples[edit | edit source]

- A closed interval is connected. To show this, suppose that it was disconnected. Then there are two nonempty disjoint open sets and whose union is . Let be the set equal to or and which does not contain . Let . Since X does not contain b, s must be within the interval [a,b] and thus must be within either X or . If is within , then there is an open set within . If is not within , then is within , which is also open, and there is an open set within . Either case implies that is not the supremum.
- The topological space is disconnected:

A picture to illustrate:

As you can see, the definition of a connected space is quite intuitive; when the space cannot be separated into (at least) two distinct subspaces.

## Definitions[edit | edit source]

Definition 1.1

A subset of a topological space is said to be **clopen** if it is both closed and open.

Definition 1.2

A topological space X is said to be totally disconnected if every subset of X having more than one point is disconnected under the subspace topology

## Theorems about connectedness[edit | edit source]

If and are homeomorphic spaces and if is connected, then is also connected.

**Proof**:

Let be connected, and let be a homeomorphism. Assume that is disconnected. Then there exists two nonempty disjoint open sets and whose union is . As is continuous, and are open. As is surjective, they are nonempty and they are disjoint since and are disjoint. Moreover, , contradicting the fact that is connected. Thus, is connected.

Note: this shows that connectedness is a topological property.

If two connected sets have a nonempty intersection, then their union is connected.

**Proof**:

Let and be two non-disjoint, connected sets. Let and be non-empty open sets such that . Let .

Without loss of generality, assume .

As is connected, ...(1).

As is non-empty, such that .

Hence, similarly, ...(2)

Now, consider . From (1) and (2), , and hence . As are arbitrary, is connected.

If two topological spaces are connected, then their product space is also connected.

**Proof:**

Let X_{1} and X_{2} be two connected spaces. Suppose that there are two nonempty open disjoint sets A and B whose union is X_{1}×X_{2}. If for every x∈X, {x}×X_{2} is either completely within A or within B, then π_{1}(A) and π_{1}(B) are also open, and are thus disjoint and nonempty, whose union is X_{1}, contradicting the fact that X_{1} is connected. Thus, there is an x∈X such that {x}×X_{2} contains elements of both A and B. Then π_{2}(A∩{(x,y)}) and π_{2}(B∩{(x,y)}), where y is any element of X_{2}, are nonempty disjoint sets whose union is X_{2}, and which are a union of open sets in {(x,y)} (by the definition of product topology), and are thus open. This implies that X_{2} is disconnected, a contradiction. Thus, X_{1}×X_{2} is connected.

## Exercises[edit | edit source]

- Show that a topological space is disconnected if and only if it has clopen sets other than and (Hint: Why is clopen?)
- Prove that if is continuous and surjective (not necessarily homeomorphic), and if is connected, then is connected.
- Prove the
**Intermediate Value Theorem**: if is continuous, then for any between and , there exists a such that . - Prove that is not homeomorphic to (hint: removing a single point from makes it disconnected).
- Prove that an uncountable set given the countable complement topology is connected (this space is what mathematicians call 'hyperconnected')
- a)Prove that the discrete topology on a set X is totally disconnected.

b) Does the converse of a) hold (Hint: Even if the subspace topology on a subset of X is the discrete topology, this need not imply that the set has the discrete topology)