Continuous Image of Connected Set is Connected

Theorem

The continuous image of a connected set is connected.

Proof

Suppose $C \subseteq R$ be a connected set. Let $f : C \to R$ which is continuous (on $C$ ).

(We want to show here that $f (C)$ is connected, using definition of connectedness).

Suppose $\exists A, B \subseteq R$ , where $A, B \neq \emptyset$ , such that $f (C) = A \cup B$ . We want to show that a limit point of $A$ is in $B$ , or vice versa.

Let $A^{'} = {x \in C : f (x) \in A}$ . Similarly $B^{'} = {x \in C : f (x) \in B}$ .

Sometimes we say that

f^{- 1} (A)

is the a shorthand to

A^{'}

, even though

f

may not be bijective (and thus non-invertible).

First off since $A \neq \emptyset$ then $\exists a \in A$ , so then $A^{'}$ is nonempty. Thus then $\exists x_{a}$ where $f (x_{a}) \in A$ and thus $x_{a} \in A^{'}$ . A similar argument shows that $\exists x_{b} \in B^{'}$ . Further notice that $C = A^{'} \cup B^{'}$ , because $f (C) = A \cup B$ .

Now because $C$ is connected, then (without loss of generality) then $\exists (x_{n}) \subseteq A^{'}$ such that $x_{n} \to x$ where $x \in B^{'}$ .

Now because $f$ is continuous at $x \in B^{'} \subseteq C$ then $f (x_{n}) \to f (x) \in B$ by definition. Now because each $f (x_{n}) \in A$ then clearly $B$ contains the limit points of the set $A$ . This shows then that $\overset{―}{A} \cap B \neq \emptyset$ . The same argument shows that $A \cap \overset{―}{B} \neq \emptyset$ .

☐

As an immediate corollary we get the Intermediate Value Theorem.