Continuous Functions on Connected Sets of 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 nonempty open sets $A$ and $B$ such that $A \cap B = \emptyset$ and:

(1)

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

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:

(2)

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

• Since $f$ is continuous we have that:

(3)

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

• 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 $T$