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Connectedness

Definition. A topological space XX is disconnected if there exist nonempty disjoint open sets U,VU, V with X=UVX = U \cup V. Such a pair {U,V}\{U, V\} is called a separation of XX.

XX is connected if it is not disconnected.

Equivalently, XX is connected if and only if the only clopen subsets of XX are \emptyset and XX.

Example 6.1. [0,1][0, 1] is connected. [0,1)(2,3][0, 1) \cup (2, 3] is disconnected (with the subspace topology from R\mathbb{R}).

Example 6.2. Q\mathbb{Q} with the subspace topology from R\mathbb{R} is totally disconnected: the only connected subsets are singletons.

Theorem 6.1. A subset of R\mathbb{R} (with the standard topology) is connected if and only if it is an interval.

(Here an interval is any set IRI \subseteq \mathbb{R} with the property: if a,bIa, b \in I and a<c<ba < c < b, then cIc \in I.)

Definition. A space XX is path-connected if for any two points x,yXx, y \in X, there exists a continuous function γ:[0,1]X\gamma : [0, 1] \to X with γ(0)=x\gamma(0) = x and γ(1)=y\gamma(1) = y.

Proposition 6.1. Every path-connected space is connected. The converse is false.

Example 6.3 (Topologist”s sine curve). Let

S={(x,sin(1/x)):0<x1}{(0,y):1y1}R2.S = \{(x, \sin(1/x)) : 0 < x \leq 1\} \cup \{(0, y) : -1 \leq y \leq 1\} \subseteq \mathbb{R}^2.

SS is connected but not path-connected.

Example 6.4. Rn\mathbb{R}^n is path-connected for all n1n \geq 1. Any convex subset of Rn\mathbb{R}^n is path-connected.

Definition. A connected component of XX is a maximal connected subset of XX. The connected components of XX form a partition of XX.

Proposition 6.2. Connected components are closed (in a Hausdorff space, they are always closed).

Definition. A path component of XX is a maximal path-connected subset. Path components also partition XX, and each path component is contained in a connected component.

Definition. XX is locally connected if for every xXx \in X and every open neighbourhood UU of xx, there exists a connected open neighbourhood VV of xx with VUV \subseteq U.

Proposition 6.3. Every open subset of Rn\mathbb{R}^n is locally connected.

Example 6.5. The topologist’s sine curve is connected but not locally connected.

Proposition 6.4. The continuous image of a connected space is connected. Therefore connectedness is a topological invariant.

Proof. If f:XYf : X \to Y is continuous and XX is connected, suppose f(X)=UVf(X) = U \cup V with U,VU, V open and disjoint in f(X)f(X). Then f1(U)f^{-1}(U) and f1(V)f^{-1}(V) are open, disjoint, and cover XX, contradicting connectedness. \blacksquare

Proposition 6.5 (Intermediate Value Theorem). If f:XRf : X \to \mathbb{R} is continuous and XX is connected, then ff attains every value between any two of its values. This generalises the classical IVT from R\mathbb{R} 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 X×YX \times Y 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 AXA \subseteq X is connected, then any set BB with ABAA \subseteq B \subseteq \overline{A} 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 Q\mathbb{Q} (with the subspace topology from R\mathbb{R}) and the Cantor set CC.

Proposition 6.8. The Cantor set CC is totally disconnected, compact, uncountable, and perfect (every point is a limit point). It is homeomorphic to {0,1}N\{0, 1\}^{\mathbb{N}} with the product topology.

6.8 Worked Example: Proving Disconnectedness

Section titled “6.8 Worked Example: Proving Disconnectedness”

Problem. Show that GL(2,R)\operatorname{GL}(2, \mathbb{R}) (the set of 2×22 \times 2 invertible real matrices with the subspace topology from R4\mathbb{R}^4) is disconnected.

Solution

The determinant map det:GL(2,R)R{0}\det : \operatorname{GL}(2, \mathbb{R}) \to \mathbb{R}\setminus\{0\} is continuous (it is a polynomial in the matrix entries). The image is R{0}\mathbb{R}\setminus\{0\}, which is disconnected (separated by 00). Since the continuous image of a connected space must be connected, GL(2,R)\operatorname{GL}(2, \mathbb{R}) cannot be connected.

In fact, GL(2,R)\operatorname{GL}(2, \mathbb{R}) has exactly two connected components: matrices with positive determinant and matrices with negative determinant.

\blacksquare

6.9 Worked Example: Path-Connectedness of Rn{0}\mathbb{R}^n \setminus \{0\}

Section titled “6.9 Worked Example: Path-Connectedness of Rn∖{0}\mathbb{R}^n \setminus \{0\}Rn∖{0}”

Problem. For which nn is Rn{0}\mathbb{R}^n \setminus \{0\} path-connected?

Solution

For n2n \geq 2, any two points p,q0p, q \neq 0 can be joined by a path avoiding the origin. For example, if pp and qq are not antipodal, use the straight line segment; if they are antipodal (q=pq = -p), take a path through a third point. Thus Rn{0}\mathbb{R}^n \setminus \{0\} is path-connected for n2n \geq 2.

For n=1n = 1, R{0}=(,0)(0,)\mathbb{R} \setminus \{0\} = (-\infty, 0) \cup (0, \infty) is disconnected (and hence not path-connected).

\blacksquare