Theorem: If $f$ is continuous on a connected set then the image $f(S)$ is also connected in the codomain.

Proof: Suppose that $f(C)$ is not connected, i.e., disconnected. We will show that a contradiction arises.

Suppose that $f(C)$ is disconnected. Then there exists nonempty open sets $A,B \subset f(C)$ such that $A \cap B= \emptyset$ and $f(C)=A \cup B$.

Then a necessary step is to say $C=f^{−1}(A) \cup f^{−1}(B)$. I don't know how to verify this condition.

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)∈A∩B which implies that A∩B≠∅ 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. Thus, we prove that f(C) is connected.


If $x\in C,$ then $f(x)\in f(C)=A\cup B$ so either $f(x)\in A$ or $f(x)\in B.$ Thus either $x\in f^{-1}(A)$ or $x\in f^{-1}(B).$

  • $\begingroup$ Thanks. Does this specfic step need continuous? $\endgroup$
    – Mariana
    Jun 13 at 2:53
  • 1
    $\begingroup$ @Mariana No, this is just set-theoretic properties of the inverse image: $f^{-1} \left ( \bigcup_{i \in I} A_i \right ) = \bigcup_{i \in I} f^{-1}(A_i)$. $\endgroup$
    – Ian
    Jun 13 at 2:57

So let $C$ be a connected set and view the continuous function $f: C\to f(C)$. (Note that this function is surjective.)

We have $f(C)$ (the image of C) is disconnected.

We then we can write $f(C)=A\cup B$ with the according properties. Now we look at the preimage.

This gives $C=f^{-1}(A\cup B)$. Why is this true? Then an important rule for the preimage is that $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$. If you do not know this, try to proof it. The proof is not difficult.

Remember that everything you have to show is just equality of sets.

Now $f^{-1}(A)\cap f^{-1}(B)=\emptyset$. Why?

Leading to a contradiction. Why?


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