# Topology/Connectedness
## 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 $(X,{\mathcal {T}})$ is said to be disconnected iff a pair of disjoint, non-empty open subsets $X_{1},X_{2}$ exists, such that $X=X_{1}\cup X_{2}$.
A space $X$ that is not disconnected is said to be a connected space.

### Examples

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

## Theorems about connectedness

If $X$ and $Y$ are homeomorphic spaces and if $X$ is connected, then $Y$ is also connected.
Proof: Let $X$ be connected, and let $f$ be a homeomorphism. Assume that $Y$ is disconnected. Then there exists two nonempty disjoint open sets $Y_{1}$ and $Y_{2}$ whose union is $Y$. As $f$ is continuous, $f^{-1}(Y_{1})$ and $f^{-1}(Y_{2})$ are open. As $f$ is surjective, they are nonempty and they are disjoint since $Y_{1}$ and $Y_{2}$ are disjoint. Moreover, $f^{-1}(Y_{1})\cup f^{-1}(Y_{2})=f^{-1}(Y)=X$, contradicting the fact that $X$ is connected. Thus, $Y=f(X)$ 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 $A$ and $B$ be two non-disjoint, connected sets. Let $X$ and $Y$ be non-empty open sets such that $X\cup Y=A\cup B$. Let $a_{0}\in A$. Without loss of generality, assume $a_{0}\in X$.
As $A$ is connected, $a\in X\forall a\in A$ ...(1).
As $Y$ is non-empty, $\exists b\in B$ such that $b\in Y$.
Hence, similarly, $b\in Y\forall b\in B$ ...(2) Now, consider $c\in A\cap B$. From (1) and (2), $c\in X\cap Y$, and hence $X\cap Y\neq \emptyset$. As $X,Y\in {\mathcal {T}}$ are arbitrary, $A\cup B$ 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

* Show that a topological space $X$ is disconnected if and only if it has clopen sets other than $\emptyset$ and $X$ (Hint: Why is $X_{1}$ clopen?)
* Prove that if $f:X\to Y$ is continuous and surjective (not necessarily homeomorphic), and if $X$ is connected, then $Y$ is connected.
* Prove the Intermediate Value Theorem: if $f:[a,b]\to \mathbb {R}$ is continuous, then for any $y$ between $f(a)$ and $f(b)$, there exists a $c\in [a,b]$ such that $f(c)=y$.
* Prove that $\mathbb {R}$ is not homeomorphic to $\mathbb {R} ^{2}$ (hint: removing a single point from $\mathbb {R}$ 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)