# Proof of "the continuous image of a connected set is connected"

None of the existing questions is exactly answering my question so I'm posting a new question, but feel free to refer me to some already answered question!

In Rudin Theorem 4.22, we know that

If $f$ is a continuous mapping of a metric space $X$ into a metric space $Y$, and $E$ is a connected subset of $X$, then $f(E)$ is connected.

In the proof, we started with consider $f(E) = A \cup B$, where $A$ and $B$ are nonempty separated subsets. Then put $G = E \cap f^{-1}(A)$ and $H = E \cap f^{-1}(B)$. Then Rudin is claiming that $E = G \cup H$. I'm a little suspicious about this. What if $f$ is non-surjective, then $f^{-1}(A) \cup f^{-1}(B)$ is only a proper subset of $E$? Is there a property of $f$ being continuous that forces $f$ to be 1-1?

• If $x\in E$, then $f(x)\in f(E)=A\cup B$, so there must be some $a\in A$ or some $b\in B$ such that $f(x)=a$ or $f(x)=b$. But then $x\in f^{-1}(A)$ or $x\in f^{-1}(B)$ showing that $x\in G$ or $x\in H$. Dec 13, 2015 at 17:09
• A mapping is not connected, but its image might be
– zhw.
Dec 13, 2015 at 21:02
• Note that no matter which function $f : A \to B$ between any two sets, then we have $$f^{-1}(B)=A$$ This is one of the reasons why pre-images are easier to work with then direct images. Jul 28, 2017 at 20:16
• Which book are you referring to? Oct 16, 2018 at 18:21
• $G$ and $H$ are open sets? Jan 19, 2019 at 0:41

$f^{-1}(A) \cup f^{-1}(B)$ may be larger than $E$, but it must contain $E$: if $x \in E$, $f(x) \in f(E) = A \cup B$. Then either $f(x) \in A$ or $f(x) \in B$. In the first case, $x \in f^{-1}(A)$, and in the second $x \in f^{-1}(B)$. (The fact that $f^{-1}(A) \cup f^{-1}(B)$ may be larger than $E$ is the reason for intersecting with $E$ when defining $G$ and $H$.)

For all $$x\in X$$ we have \begin{align} x\in G\cup H &\iff [(x\in E \cap f^{-1}A)\lor (x\in E\cap f^{-1}B] \\ & \iff [x\in E \land (f(x)\in A\lor f(x)\in B)] \\ & \iff [x\in E\land f(x)\in f(E)] \\ & \iff x\in E. \end{align}

If $$e \in E$$, then $$f(e) \in f(E) = A \cup B$$.
So, $$f(e) \in A$$ or $$f(e) \in B$$.
So, $$e \in f^{-1}(A)$$ or $$e \in f^{-1}(B)$$.
So, $$e \in f^{-1}(A) \cup f^{-1}(B)$$.
So, $$E \subset f^{-1}(A) \cup f^{-1}(B)$$.

So, $$E = E \cap (f^{-1}(A) \cup f^{-1}(B)) = (E \cap f^{-1}(A)) \cup (E \cap f^{-1}(B)) = G \cup H$$.

$$A \cap B = \emptyset$$.
So, $$f^{-1}(A) \cap f^{-1}(B) = \emptyset$$.
So, $$\emptyset = E \cap \emptyset = E \cap (f^{-1}(A) \cap f^{-1}(B)) = (E \cap f^{-1}(A)) \cap (E \cap f^{-1}(B)) = G \cap H$$

Suppose $$E$$ is connected subset of $$X$$. We need to prove that $$f(E)$$ is connected. Assume the contrary. $$f(E)=C\cup D$$ forms a separation of $$f(E)$$. Since $$f$$ is surjective therefore both $$f^{-1}(C)$$ and $$f^{-1}(D)$$ are non-empty subsets of $$E$$. due to continuity $$f^{-1}(C)$$ and $$f^{-1}(D)$$ are open in $$X$$. $$f^{-1}(C) \cap f^{-1}(D)=f^{-1}(C\cap D)=f^{-1}(\phi)=\phi$$. Contradicting the fact that $$E$$ is connected.

• @Zargles I voted to reject your edit as it was not immediately clear that was what the answerer meant. You should instead comment below the post. Nov 5, 2020 at 9:12