Connectedness
6.1 Connected and Disconnected Spaces
Section titled “6.1 Connected and Disconnected Spaces”Definition. A topological space is disconnected if there exist nonempty disjoint open sets with . Such a pair is called a separation of .
is connected if it is not disconnected.
Equivalently, is connected if and only if the only clopen subsets of are and .
Example 6.1. is connected. is disconnected (with the subspace topology from ).
Example 6.2. with the subspace topology from is totally disconnected: the only connected subsets are singletons.
6.2 Connected Subsets of
Section titled “6.2 Connected Subsets of R\mathbb{R}R”Theorem 6.1. A subset of (with the standard topology) is connected if and only if it is an interval.
(Here an interval is any set with the property: if and , then .)
6.3 Path-Connectedness
Section titled “6.3 Path-Connectedness”Definition. A space is path-connected if for any two points , there exists a continuous function with and .
Proposition 6.1. Every path-connected space is connected. The converse is false.
Example 6.3 (Topologist”s sine curve). Let
is connected but not path-connected.
Example 6.4. is path-connected for all . Any convex subset of is path-connected.
6.4 Components
Section titled “6.4 Components”Definition. A connected component of is a maximal connected subset of . The connected components of form a partition of .
Proposition 6.2. Connected components are closed (in a Hausdorff space, they are always closed).
Definition. A path component of is a maximal path-connected subset. Path components also partition , and each path component is contained in a connected component.
6.5 Local Connectedness
Section titled “6.5 Local Connectedness”Definition. is locally connected if for every and every open neighbourhood of , there exists a connected open neighbourhood of with .
Proposition 6.3. Every open subset of is locally connected.
Example 6.5. The topologist’s sine curve is connected but not locally connected.
6.6 Properties of Connected Spaces
Section titled “6.6 Properties of Connected Spaces”Proposition 6.4. The continuous image of a connected space is connected. Therefore connectedness is a topological invariant.
Proof. If is continuous and is connected, suppose with open and disjoint in . Then and are open, disjoint, and cover , contradicting connectedness.
Proposition 6.5 (Intermediate Value Theorem). If is continuous and is connected, then attains every value between any two of its values. This generalises the classical IVT from to any connected space.
Proposition 6.6 (Products). The product of connected spaces is connected (with the product topology). Finite products follow from Proposition 6.4 by noting is homeomorphic to the image of the product space under a continuous map; arbitrary products require more care but also hold.
Proposition 6.7 (Closure). If is connected, then any set with is connected. In particular, the closure of a connected set is connected.
6.7 Total Disconnectedness and the Cantor Set
Section titled “6.7 Total Disconnectedness and the Cantor Set”Definition. A space is totally disconnected if its only connected subsets are singletons. Examples include (with the subspace topology from ) and the Cantor set .
Proposition 6.8. The Cantor set is totally disconnected, compact, uncountable, and perfect (every point is a limit point). It is homeomorphic to with the product topology.
6.8 Worked Example: Proving Disconnectedness
Section titled “6.8 Worked Example: Proving Disconnectedness”Problem. Show that (the set of invertible real matrices with the subspace topology from ) is disconnected.
Solution
The determinant map is continuous (it is a polynomial in the matrix entries). The image is , which is disconnected (separated by ). Since the continuous image of a connected space must be connected, cannot be connected.
In fact, has exactly two connected components: matrices with positive determinant and matrices with negative determinant.
6.9 Worked Example: Path-Connectedness of
Section titled “6.9 Worked Example: Path-Connectedness of Rn∖{0}\mathbb{R}^n \setminus \{0\}Rn∖{0}”Problem. For which is path-connected?
Solution
For , any two points can be joined by a path avoiding the origin. For example, if and are not antipodal, use the straight line segment; if they are antipodal (), take a path through a third point. Thus is path-connected for .
For , is disconnected (and hence not path-connected).