# Continuous image of connected set is connected: Proof

In De La Fuente's Mathematical Methods and Models for Economists, the following is said:

Let $f:X\to Y$ be a continuous mapping between two metric spaces. If C is a connected subset of $X$, then $f(C)$ is connected.

The proof goes as in Rudin's Principles, and I cannot understand exactly what Rudin also does not explain:

Suppose $f(C)$ is not connected. Then $f(C)=P\cup Q$, where $P$ and $Q$ are nonempty, separated subsets of $Y$, that is,

$clP\cap Q = \emptyset$ and $P\cap clQ = \emptyset$

Let $$A = C\cap f^{-1}(P) \\ B = C\cap f^{-1}(Q)$$ and notice that then $$C = A\cup B$$ where neither $A$ nor $B$ is empty, and $$f(A)=P \\ f(B) = Q$$

The proof goes on, but this is where De La Fuente loses me. I can clearly see that $f(A)\subseteq P$, but not that $P\subseteq f(A)$. Any thoughts?

Thanks!

• I have added (proof-explanation) tag - see the tag-info - since your question seems to be about this specific proof (rather than asking for any proof of the fact mentioned in the title). – Martin Sleziak Sep 4 '17 at 22:15

## 2 Answers

Let $p\in P$. Then as $P\subset f(C)$, we know that $f(c)=p$ for some $c\in C$. Can you go from there?

• Yes! Thanks a lot! – Miguel Santana Sep 5 '17 at 8:42

The proof has to be properly aligned with the OP's definition of connectedness:

Definition: A subspace $C$ of a topological space $X$ is connected if and only if for any non-trivial partition $A \cup B$ of $C$,

$\qquad \overline{A} \cap B \ne \emptyset\; \text{ OR }\; A \cap \overline{B} \ne \emptyset$.

Proposition 1: A function $f: X \to Y$ is continuous $\text{ iff for every } A \subset X \text{, } f(\overline{A})\subseteq \overline{f(A)}$.

The above proposition and some general properties about sets and functions should be enough to construct the desired proof.