# Topology/Connectedness

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## Motivation

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 ${\displaystyle (X,{\mathcal {T}})}$  is said to be disconnected iff a pair of disjoint, non-empty open subsets ${\displaystyle X_{1},X_{2}}$  exists, such that ${\displaystyle X=X_{1}\cup X_{2}}$ .

A space ${\displaystyle X}$  that is not disconnected is said to be a connected space.

### Examples

1. A closed interval ${\displaystyle [a,b]}$  is connected. To show this, suppose that it was disconnected. Then there are two nonempty disjoint open sets ${\displaystyle A}$  and ${\displaystyle B}$  whose union is ${\displaystyle [a,b]}$ . Let ${\displaystyle X}$  be the set equal to ${\displaystyle A}$  or ${\displaystyle B}$  and which does not contain ${\displaystyle b}$ . Let ${\displaystyle s=\sup X}$ . Since X does not contain b, s must be within the interval [a,b] and thus must be within either X or ${\displaystyle [a,b]\setminus X}$ . If ${\displaystyle s}$  is within ${\displaystyle X}$ , then there is an open set ${\displaystyle (s-\varepsilon ,s+\varepsilon )}$  within ${\displaystyle X}$ . If ${\displaystyle s}$  is not within ${\displaystyle X}$ , then ${\displaystyle s}$  is within ${\displaystyle [a,b]\setminus X}$ , which is also open, and there is an open set ${\displaystyle (s-\varepsilon ,s+\varepsilon )}$  within ${\displaystyle [a,b]\setminus X}$ . Either case implies that ${\displaystyle s}$  is not the supremum.
2. The topological space ${\displaystyle X=(0,1)\setminus \{{\frac {1}{2}}\}}$  is disconnected: ${\displaystyle A=(0,{\frac {1}{2}}),B=({\frac {1}{2}},1)}$
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

Definition 1.1

A subset ${\displaystyle U}$  of a topological space ${\displaystyle X}$  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

If ${\displaystyle X}$  and ${\displaystyle Y}$  are homeomorphic spaces and if ${\displaystyle X}$  is connected, then ${\displaystyle Y}$  is also connected.

Proof:
Let ${\displaystyle X}$  be connected, and let ${\displaystyle f}$  be a homeomorphism. Assume that ${\displaystyle Y}$  is disconnected. Then there exists two nonempty disjoint open sets ${\displaystyle Y_{1}}$  and ${\displaystyle Y_{2}}$  whose union is ${\displaystyle Y}$ . As ${\displaystyle f}$  is continuous, ${\displaystyle f^{-1}(Y_{1})}$  and ${\displaystyle f^{-1}(Y_{2})}$  are open. As ${\displaystyle f}$  is surjective, they are nonempty and they are disjoint since ${\displaystyle Y_{1}}$  and ${\displaystyle Y_{2}}$  are disjoint. Moreover, ${\displaystyle f^{-1}(Y_{1})\cup f^{-1}(Y_{2})=f^{-1}(Y)=X}$ , contradicting the fact that ${\displaystyle X}$  is connected. Thus, ${\displaystyle Y=f(X)}$  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 ${\displaystyle A}$  and ${\displaystyle B}$  be two non-disjoint, connected sets. Let ${\displaystyle X}$  and ${\displaystyle Y}$  be non-empty open sets such that ${\displaystyle X\cup Y=A\cup B}$ . Let ${\displaystyle a_{0}\in A}$ .
Without loss of generality, assume ${\displaystyle a_{0}\in X}$ .

As ${\displaystyle A}$  is connected, ${\displaystyle a\in X\forall a\in A}$  ...(1).

As ${\displaystyle Y}$  is non-empty, ${\displaystyle \exists b\in B}$  such that ${\displaystyle b\in Y}$ .

Hence, similarly, ${\displaystyle b\in Y\forall b\in B}$  ...(2)
Now, consider ${\displaystyle c\in A\cap B}$ . From (1) and (2), ${\displaystyle c\in X\cap Y}$ , and hence ${\displaystyle X\cap Y\neq \emptyset }$ . As ${\displaystyle X,Y\in {\mathcal {T}}}$  are arbitrary, ${\displaystyle A\cup B}$  is connected.

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

Proof:
Let X1 and X2 be two connected spaces. Suppose that there are two nonempty open disjoint sets A and B whose union is X1×X2. If for every x∈X, {x}×X2 is either completely within A or within B, then π1(A) and π1(B) are also open, and are thus disjoint and nonempty, whose union is X1, contradicting the fact that X1 is connected. Thus, there is an x∈X such that {x}×X2 contains elements of both A and B. Then π2(A∩{(x,y)}) and π2(B∩{(x,y)}), where y is any element of X2, are nonempty disjoint sets whose union is X2, and which are a union of open sets in {(x,y)} (by the definition of product topology), and are thus open. This implies that X2 is disconnected, a contradiction. Thus, X1×X2 is connected.

## Exercises

1. Show that a topological space ${\displaystyle X}$  is disconnected if and only if it has clopen sets other than ${\displaystyle \emptyset }$  and ${\displaystyle X}$  (Hint: Why is ${\displaystyle X_{1}}$  clopen?)
2. Prove that if ${\displaystyle f:X\to Y}$  is continuous and surjective (not necessarily homeomorphic), and if ${\displaystyle X}$  is connected, then ${\displaystyle Y}$  is connected.
3. Prove the Intermediate Value Theorem: if ${\displaystyle f:[a,b]\to \mathbb {R} }$  is continuous, then for any ${\displaystyle y}$  between ${\displaystyle f(a)}$  and ${\displaystyle f(b)}$ , there exists a ${\displaystyle c\in [a,b]}$  such that ${\displaystyle f(c)=y}$ .
4. Prove that ${\displaystyle \mathbb {R} }$  is not homeomorphic to ${\displaystyle \mathbb {R} ^{2}}$  (hint: removing a single point from ${\displaystyle \mathbb {R} }$  makes it disconnected).
5. Prove that an uncountable set given the countable complement topology is connected (this space is what mathematicians call 'hyperconnected')
6. 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)

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