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.