We have an intuitive idea of what it means for a shape (or in this case, topological space) to be connected, owever our intuition may fail us when it comes to esoteric spaces.
For example, take the ‘topologist sine curve’.
TO-DO: Picture and explanation
TO-DO: Compare rations not being connected
Connected Spaces
We will view connectedness as a property belonging to a topological space, rather than as a property of sets in a space. However much like compactness the notion can be easily extended to sets using the subspace topology.
Let X be a topological space. A separation of X is a pair of disjoint, nonempty, open subsets whose union is X.
Symbollically, this means a separation of a space is a pair U,V of open subsets such that
\varnothing \not = U,V \text{ and } U\,\dot\cup\,V = X
Where \dot\cup represents a disjoint union. This leads to our definition of connectedness -
Let X be a topological space. If there exist no separation of X, then X is connected.
If a separation does exist, X is called disconnected.
And that’s it!
Properties
The most common property of connectedness is that
Let (X,\tau) be a topological space. Then (X,\tau) is connected if and only if \varnothing and X are the only sets that are both open and closed.
The proof is suprisngly simple.
(\Rightarrow) Let (X,\tau) be a connected space. By way of contradiction, let A \subset X be a nonempty proper subset that is both open and closed. Then there exist a separation
U = A \quad V = A^C.
But since (X,\tau) is connected by assumption, we have a contradiction. Thus there exist no such set A.
(\Leftarrow) Let (X,\tau) be a space such that the only sets that are both open and closed are \varnothing and X. Towards a contradiction assume that there exist some separation of (X,\tau) by sets U,V. Then since U \, \dot\cup \, V = X, we have that V = U^C and that both U and U^C are open.
Then that means U is both open and closed, contradicting our assumption that (X,\tau) contains no other closed and open set. Thus there exists no such separation, meaning (X,\tau) is connected.
Let \mathcal{T}_1 and \mathcal{T_2} be topologies on a set X, where \mathcal{T_1} is finer than \mathcal{T_2}. Then if (X,\mathcal{T}_1) is connected, so is (X,\mathcal{T}_2).