# Ask a step in the proof of continuous function of connected set

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 $$x\in f^{-1}(B).$$

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