10.7 Connected spaces

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Intuitively, we can think of a topological space $X$ as being ‘disconnected’ if it can be decomposed into two disjoint subsets $X=A\cup B$ without these sets having any boundary points in common. Since the boundary of a set is at the same time the boundary of the complement of that set, the only way such a decomposition can occur is if there exists a set $A$ other than the empty set or the whole space $X$ that has no boundary points at all. Since $b(A)=\overline{A}-A^o$ , the only way $b(A)=\varnothing$ can occur is if $\overline{A}=A^o.$ As $\overline{A}\subseteq A\subseteq A^o$ the set $A$ must equal both its closure and interior; in particular, it would need to be both open and closed at the same time. This motivates the following definition: a topological space $X$ is said to be connected if the only subsets that are both open and closed are the empty set and the space $X$ itself. A space is said to be disconnected if it is not connected. In other words, $X$ is disconnected if $X=A\cup B$ where $A$ and $B$ are disjoint sets that are both open and closed. A subset $A\subset X$ is said to be connected if it is connected in the relative topology.

Example 10.17

The indiscrete topology on any set $X$ is connected, since the only open sets are $\varnothing$ or $X$. The discrete topology on any set $X$ having more than one point is disconnected since every non-empty subset is both open and closed.

Example 10.18

The real numbers $\mathbb{R}$ are connected in the standard topology. To show this, let $A\subset \mathbb{R}$ be both open and closed. If $x\in A$ set $y$ to be the least upper bound of those real numbers such that $[x,y)\subset A$ . If $y<\infty$ then $y$ is an accumulation point of $A,$ and therefore $y\in A$ since $A$ is closed. However, since $A$ is an open set there exists an interval $(y-a,y+a)\subset A$ . Thus $[x,y+a)\subset A$ , contradicting the stipulation that $y$ is the least upper bound. Hence $y=\infty$ . Similarly $(-\infty ,x]\subset A$ and the only possibility is that $A=\varnothing$ or $A=\mathbb{R}$

Theorem 10.15 · Closure of a connected set

The closure of a connected set is connected.

Proof

Let $A\subset X$ be a connected set. Suppose $U$ is a subset of the closure $\overline{A}$ of $A,$ , which is both open and closed in $\overline{A},$ , and let $V=\overline{A}-U$ be the complement of $U$ in $\overline{A}.$ . Since $A$ is connected and the sets $U\cap A$ and $V\cap A$ are both open and closed in $A$, one of them must be the empty set, while the other is the whole set $A$. If, for example, $V\cap A=\varnothing$ then $U\cap A=A$ , so that $A\subset U\subseteq \overline{A}$ . Since $U$ is closed in $\overline{A}$ we must have that $U=\overline{A}$ and $V=\overline{A}-U=\varnothing$ . If $U\cap A=\varnothing$ then $U=\varnothing$ by an identical argument. Hence $\overline{A}$ is connected.

The following theorem is used in many arguments to do with connectedness oftopologica spaces or their subspaces. Intuitively, it says that connectedness is retained if any number of connected sets are ‘attached’ to a given connected set.

Theorem 10.16 · Unions of connected sets meeting a common connected set

Let $A_0$ be any connected subset of a topological space $X$ and $\{A_i\mid i\in I\}$ any family of connected subsets of $X$ such that $A_0\cap A_i\ne \varnothing$ for each member ofthefamily Then the set $A=A_0\cup \left({\bigcup }_{i\in I}{A}_i\right)$ is a connected subset of $X$.

Proof

Suppose $A=U\cup V$ where $U$ and $V$ are disjoint open sets in the relative topology on $A$. For all $i\in I$ the sets $U\cap A_i$ and $V\cap A_i$ are disjoint open sets of $A_i$ whose union is $A_i$ Since $A_i$ is connected, either $U\cap A_i=\varnothing$ or $V\cap A_i=\varnothing$ . This also holds for $A_0:$ either $U\cap A_0=\varnothing$ or $V\cap A_0=\varnothing .$ , say the latter. Then $U\cap A_0=A_0$ , so that $A_0\subseteq U$ . Since $A_0\cap A_i\ne \varnothing$ we have $U\cap A_i\ne \varnothing$ for all $i\in I$ . Hence $V\cap A_i=\varnothing$ and $U\cap A_i=A_i$ ; that is, $A_i\subseteq U$ for all $i\in I$ . Hence $U=A$ and $V=\varnothing$ , showing that $A$ is a connected subset of $X$.

A theorem similar to Theorem 10.10 is available for connectedness: the image of a connected space under a continuous map is connected. This also shows that connectedness is a topological property, invariant under homeomorphisms.

Theorem 10.17 · Continuous images of connected sets

If $f:X\to Y$ is a continuous map from a connected topological space $X$ into a topological space $Y$, its image set $f(X)$ is a connected subset of $Y$.

Proof

Let $B$ be any non-empty subset of $f(X)$ that is both open and closed in the relative topology. This means there exists an open set $U\subseteq Y$ and a closed set $C\subseteq Y$ such that $B=U\cap f(X)=C\cap f(X)$ . Since $f$ is a continuous map, the inverse image set $f^{-1}(B)=f^{-1}(U)=f^{-1}(C)$ is both open and closed in $X$. As $X$ is connected it follows that $B=f(X);$ ; hence $f(X)$ is connected.

A useful application of these theorems is to show the topological product of two connected spaces is connected.

Theorem 10.18 · Connectedness of product spaces

The topological product $X\times Y$ of two topological spaces is connected if and only ifboth $X$ and $Y$ are connected spaces.

Proof

By Theorem 10.3, the maps ${\iota }_x:Y\to X\times Y$ and ${\iota }_y^{\prime }:X\to X\times Y$ defined by ${\iota }_x(y)={\iota }_y^{\prime }(x)=(x,y)$ are both continuous. Suppose that both $X$ and $Y$ are connected topological spaces. Select a fixed point $y_0\in Y$ . By Theorem 10.17 the set of points $X\times y_0=\{(x,y_0)|x\in X\}={\iota }_{y_0}^{\prime }(X)$ is a connected subset of $X\times Y$ . Similarly, the sets $\{x\times Y={\iota }_x(Y)|x\in X\}$ are connected subsets of $X\times Y$ , each of which intersects $X\times y_0$ in the point $(x,y_0)$ . The union of these sets is clearly $X\times Y$ , which by Theorem 10.16 must be connected.

Conversely, suppose $X\times Y$ is connected. The spaces $X$ and $Y$ are both connected, by Theorem 10.17, since they are the images of the continuous projection maps ${\mathrm{pr}}_1:X\times Y\to X$ and ${\mathrm{pr}}_2:X\times Y\to Y$ , respectively.

Example 10.19

The spaces ${\mathbb{R}}^n$ are connected by Example 10.18 and Theorem 10.18. To show that the 2-sphere $S^2$ is connected consider the ‘punctured’ spheres $S^{\prime }=S^2-\{N=(0,0,1)\}$ and $S^{\prime \prime }=S^2-\{S=(0,0,-1)\}$ by removing the north and south poles, respectively. The set $S^{\prime }$ is connected since it is homeomorphic to the plane ${\mathbb{R}}^2$ under stereographic projection (Fig. 10.6),

$$x^{\prime }=\frac{x}{1-z},y^{\prime }=\frac{y}{1-z}\text{ where }z=\pm \sqrt{1-x^2-y^2},\text{\hspace{1em}(10.1)}$$

which has continuous inverse

$$x=\frac{2x^{\prime }}{r^{\prime 2}+1},y=\frac{2y^{\prime }}{r^{\prime 2}+1},z=\frac{r^{\prime 2}-1}{r^{\prime 2}+1}\left(r^{\prime 2}=x^{\prime 2}+y^{\prime 2}\right).\text{\hspace{1em}(10.2)}$$

Similarly $S^{\prime \prime }$ is connected since it is homeomorphic to ${\mathbb{R}}^2$ . As $S^{\prime }\cap S^{\prime \prime }\ne \varnothing$ and $S^2=S^{\prime }\cup S^{\prime \prime }$ it follows from Theorem 10.16 that $S^2$ is a connected subset of ${\mathbb{R}}^3$ . A similar argument can be used to show that the $n$-sphere $S^n$ is a connected topological space for all $n\ge 1$

Figure 10.6 Stereographic projection from the north pole of a sphere

A connected component $C$ of a topological space $X$ is a maximal connected set; that is, $C$ is a connected subset of $X$ such that if $C^{\prime }\supseteq C$ is any connected superset of $C$ then $C^{\prime }=C$ A connected component of a subset $A\subset X$ is a connected component with respect to the relative topology on $A$. By Theorem 10.15 it is immediate that any connected component is a closed set, since it implies that $C=\overline{C}$ . A topological space $X$ is connected if and only if the whole space $X$ is its only connected component. In the discrete topology the connected components consist of all singleton sets $\{x\}$.

Exercise

Show that any two distinct components $A$ and $B$ are separated, in the sense that $\overline{A}\cap B=A\cap \overline{B}=\varnothing$

Theorem 10.19 · Uniqueness of the connected component containing a connected set

Each connected subset $A$ of a topological space lies in a unique connected component.

Proof

Let $C$ be the union of all connected subsets of $X$ that contain the set $A$. Since these sets all intersect the connected subset $A$ it follows from Theorem 10.16 that $C$ is a connected set. It is clearly maximal, for ifthere exists a connected set $C_1$ such that $C\subseteq C_1$ then $C_1$ is in the family of sets of which $C$ is the union, so that $C\supset C_1$ . Hence $C=C_1$

To prove uniqueness, suppose $C^{\prime }$ were another connected component such that $C^{\prime }\supset A$ . By Theorem 10.16, $C\cup C^{\prime }$ is a connected set and by maximality of $C$ and $C^{\prime }$ we have $C\cup C^{\prime }=C=C^{\prime }$

Problems

Problem 10.18

Show that a topological space $X$ is connected if and only if every continuous map $f:X\to Y$ of $X$ into a discrete topological space $Y$ consisting ofat least two points is a constant map (see Problem 10.14).

Problem 10.19

From Theorem 10.16 show that the unit circle $S^1$ is connected, and that the punctured $n$-space ${\dot{\mathbb{R}}}^n={\mathbb{R}}^n-\{0\}$ is connected for all $n>1$ . Why is this not true for $n=1$?

Problem 10.20

Show that the real projective space defined in Example 10.15 is connected, Hausdorff and compact.

Problem 10.21

Show that the rational numbers $\mathbb{Q}$ are a disconnected subset of the real numbers. Are the irrational points a disconnected subset of $\mathbb{R}$? Show that the connected components of the rational numbers $\mathbb{Q}$ consist of singleton sets $\{x\}$.