Connectedness

Connected and Disconnected Metric Spaces

Definition: A metric space $(M,d)$ is said to be Disconnected if there exists nonempty open sets $A$ and $B$ such that $A\cap B=\emptyset$ and $M=A\cup B$ . If $M$ is not disconnected then we say that $M$ Connected. Furthermore, if $S\subseteq M$ then $S$ is said to be disconnected/connected if the metric subspace $(S,d)$ is disconnected/connected.

Intuitively, a set is disconnected if it can be separated into two pieces while a set is connected if it’s an entire piece.

For example, consider the metric space $(\mathbb {R} ,d)$ where $d$ is the Euclidean metric on $\mathbb {R}$ . Let $S=(a,b)\subset \mathbb {R}$ , i.e., $S$ is an open interval in $\mathbb {R}$ . We claim that $S$ is connected.

Suppose not. Then there exists nonempty open subsets $A$ and $B$ such that $A\cap B=\emptyset$ and $(a,b)=A\cup B$ . Furthermore, $A$ and $B$ must be open intervals themselves, say $A=(c,d)$ and $B=(e,f)$ . We must have that $A\cup B=(c,d)\cup (e,f)$ . So $c=a$ or $e=a$ and furthermore, $d=b$ or $f=b$ .

If $c=a$ then this implies that $f=b$ (since if $d=b$ then $A=(a,b)$ which implies that $B=\emptyset$ ). So if $A\cup B=(c,d)\cup (e,f)=(a,d)\cup (e,b){\text{ we must have that }}a<d,e<b$ . If $d=e$ then $A\cup B=(a,d)\cup (d,b)$ and so $d\not \in (a,b)$ so $A\cup B\neq (a,b)$ . If $d<e$ then $A\cup B=(a,d)\cup (e,b)$ and $(d,e)\not \in (a,b)$ so $A\cup B\neq (a,b)$ . If $d>e$ then $A\cap B=(e,d)\neq \emptyset$ . Either way we see that $(a,b)\neq A\cup B$ .

We can use the same logic for the other cases which will completely show that $(a,b)$ is connected.

Basic Theorems Regarding Connected and Disconnected Metric Spaces

A metric space $(M,d)$ is said to be disconnected if there exists $A,B\subseteq M$ , $A,B\neq \emptyset$ where $A\cap B=\emptyset$ and:

(1)

{\begin{aligned}\quad M=A\cup B\end{aligned}}

We say that $(M,d)$ is connected if it is not disconnected.

Furthermore, we say that $S\subseteq M$ is connected/disconnected if the metric subspace $(S,d)$ is connected/disconnected.

We will now look at some important theorems regarding connected and disconnected metric spaces.

Theorem 1: A metric space $(M,d)$ is disconnected if and only if there exists a proper nonempty subset $A\subset M$ such that $A$ is both open and closed.

$\Rightarrow$ Suppose that $(M,d)$ is disconnected. Then there exists open $A,B\subset M$ , $A,B\neq \emptyset$ , where $A\cap B=\emptyset$ and $M=A\cup B$ .

Since $A$ is open in $M$ we have that $A^{c}=B$ is closed in $M$ . But $B$ is also open. Similarly, since $B$ is open in $M$ , $B^{c}=A$ is closed in $M$ . So in fact $A$ and $B$ are both nonempty proper subsets of $M$ that are both open and closed.

$\Leftarrow$ Suppose that there exists a proper nonempty subset $A\subset M$ such that $A$ is both open and closed. Let $B=A^{c}$ . Then $B$ is also both open and closed. Furthermore, since $B\neq \emptyset$ and $A\cap B=\emptyset$ . Additionally, $M=A\cup B$ , so $M$ is disconnected. $\blacksquare$

Theorem 2: If $(M,d)$ is a connected unbounded metric space, then for every $a\in M$ and for all $r>0$ , $\{x\in M:d(x,a)=r\}$ is nonempty.

Proof: Let $(M,d)$ be a connected unbounded metric space and suppose that there exists an $a\in M$ and there exists an $r_{0}>0$ such that:

(2)

{\begin{aligned}\quad \{x\in M:d(x,a)=r_{0}\}=\emptyset \end{aligned}}

We will show that a contradiction arises. Let $A=\{x\in M:d(x,a)<r_{0}\}$ and let $B=\{x\in M:d(x,a)>r_{0}\}$ . Then $A$ is open since it is simply an open ball centered at $a$ . Furthermore, $B$ is open since $B^{c}$ is a closed ball centered $a$ . $A$ is nonempty since $a\in A$ and $B$ is nonempty since $(M,d)$ is unbounded (if it were empty then this would imply $(M,d)$ is bounded). Clearly $A\cap B=\emptyset$ and $M=A\cup B$ . So $Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (M, d)}$ is a disconnected metric space. But this is a contradiction.