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:

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

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:

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

• 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 $(M,d)$ is a disconnected metric space. But this is a contradiction.

• Therefore the assumption that there exists an $a\in M$ and an $r_{0}>0$ such that $\{x\in M:d(x,a)=r_{0}\}\emptyset$ was false.

• So for all $a\in M$ and for all $r>0$ the set $\{x\in M:d(x,a)=r\}$ is nonempty. $\blacksquare$

Continuous Functions on Connected Sets of Metric Spaces

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:

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

If $(M,d)$ is not disconnected then we say $(M,d)$ is connected.

Furthermore, we said that a subset $S\subseteq M$ is connected (or disconnected) if the metric subspace $(S,d)$ is connected (or disconnected).

We will now look at a nice theorem which tells us that if $f$ is continuous on a connected set then the image $f(S)$ is also connected in the codomain.

Theorem 1: Let $(S,d_{S})$ and $(T,d_{T})$ be metric spaces, $C\subseteq S$ and $f:C\to T$ be continuous. If $C$ is connected in $S$ then $f(C)$ is connected in $T$.

• Proof: Let $A$ be a connected set and suppose that $f(C)$ is not connected, i.e., disconnected. We will show that a contradiction arises.

• Suppose that $(f(C),d_{T})$ is disconnected. Then there exists nonempty open sets $A,B\subset f(C)$ such that $A\cap B=\emptyset$ and:

${\begin{aligned}\quad f(C)=A\cup B\end{aligned}}$

• Since $f$ is continuous we have that:

${\begin{aligned}\quad C=f^{-1}(A)\cup f^{-1}(B)\end{aligned}}$

• Note that $f^{-1}(A)$ and $f^{-1}(B)$ are nonempty, otherwise, $A$ or $B$ would be empty which cannot happen. Furthermore, both of these sets are open from the continuity of $f$. We claim that $f^{-1}(A)\cap f^{-1}(B)=\emptyset$. Suppose not. Then there exists an $x\in f^{-1}(A)\cap f^{-1}(B)$ and so $f(x)\in A\cap B$ which implies that $A\cap B\neq \emptyset$ which is a contradiction.

• Therefore $f^{-1}(A)\cap f^{-1}(B)=\emptyset$ and so $C$ is a disconnected set. But this is a contradiction.

• Hence the assumption that $f(C)$ was disconnected is false. Therefore, if $C$ is a connected set in $S$ and $f:C\to T$ is continuous then $f(C)$ is connected in $T$

Licensing

Content obtained and/or adapted from:

• Connected And Disconnected Metric Spaces, mathonline.wikidot.com under a CC BY-SA license
• Basic Theorems Regarding Connected and Disconnected Metric Spaces, mathonline.wikidot.com under a CC BY-SA license
• Continuous Functions on Connected Sets of Metric Spaces, mathonline.wikidot.com under a CC BY-SA license