Basic Theorems Regarding Connected and Disconnected Met. Spaces

# Basic Theorems Regarding Connected and Disconnected Metric Spaces

Recall from the Connected and Disconnected Metric Spaces page that 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{align} \quad M = A \cup B \end{align}

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{align} \quad \{ x \in M : d(x, a) = r_0 \} = \emptyset \end{align}

- 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. $\blacksquare$