Connected and Disconnected Spaces

A SPACE THAT CANNOT BE CUT IN TWO

Two open sets cut a space in two when between them they cover it, they share no point of it, and each contains at least one point of it. A space that can be cut this way is disconnected; a space that cannot is connected.

U V the space is dark, the open sets dashed M ✓ together they cover the space ✓ they share no point of the space ✓ each holds part of the space disconnected U V U ∩ V ∩ M the space is one piece M ✓ together they cover the space ✗ they share no point of the space ✓ each holds part of the space connected
Left: two open sets cut the space. Right: no pair of open sets cuts it; the dashed sets may spill past it.

The definition

Definition · connected space

Let \((M, \mathcal{O})\) be a topological space. A separation of \(M\) is a pair of open sets \(U, V \in \mathcal{O}\) with

\[ U \cap V = \emptyset, \qquad U \neq \emptyset, \qquad V \neq \emptyset, \qquad U \cup V = M. \]

\(M\) is disconnected if a separation of it exists, and connected if none does.

Each of the four demands is load bearing. Without \(U \cup V = M\) any two disjoint open sets would do, so \(\mathbb{R}\) would count as disconnected on the strength of \((0,1)\) and \((2,3)\). Without the two non-emptiness demands, \(M = M \cup \emptyset\) is a separation of everything. Without disjointness, \(M = M \cup M\) is a separation of every non-empty space.

Where a space is met sitting inside a larger one, the open sets are usually taken from the larger space. The demands are then that they cover the space, that each meets it, and that they share no point of it, which leaves them free to overlap outside it. A subset carries the subset topology, so this is the same test.

Open and closed at once is the same demand

Suppose \(U\) and \(V\) separate \(M\). Then \(V\) is exactly the complement of \(U\), so \(U\) is an open set whose complement is open, which is to say \(U\) is closed as well, and it is neither empty nor all of \(M\). Run it the other way: if \(A\) is both open and closed and is neither, then \(A\) and \(M \setminus A\) are both open, disjoint, non-empty, and cover \(M\), so they separate it.

So the two forms are one statement. A space is connected exactly when the only subsets that are both open and closed are \(\emptyset\) and \(M\).

Three points, worked

Let \(M = \{1, 2, 3\}\) and give it \(\mathcal{O} = \{\emptyset, \{1\}, \{2,3\}, M\}\), which is a topology. Take \(U = \{1\}\) and \(V = \{2,3\}\). Both are open, they share no point, neither is empty, and together they are \(M\). That is a separation, so this space is disconnected.

Now keep the same three points and give them \(\mathcal{O}' = \{\emptyset, \{1\}, \{1,2\}, M\}\), again a topology. Its non-empty open sets are \(\{1\}\), \(\{1,2\}\) and \(M\), and every one of them contains the point \(1\), so no two of them are disjoint. There is nothing left to try, and the space is connected.

Nothing moved between those two paragraphs except which subsets were declared open. Connectedness is a fact about the topology, not about how far apart the points are drawn.