The discussion focuses on proving that the image of a connected set under a continuous function remains connected. Specifically, for a continuous function $f: K \subseteq \mathbb{R}^n \to \mathbb{R}$ where $K$ is a connected subset, it is established that $f(K)$ is also connected. The proof utilizes contradiction by assuming $f(K)$ is disconnected and demonstrating that this leads to a contradiction with the connectedness of $K$. An alternative proof method involves showing that the only nonempty, open, and closed subset of $f(K)$ is $f(K)$ itself, leveraging the continuity of $f$.
PREREQUISITESMathematicians, students of topology, and anyone interested in the properties of continuous functions and connected sets will benefit from this discussion.
I would do this by contradiction. Suppose that $f(K)$ is not connected. Then it can be written as the disjoint union of two nonempty open subsets, $U$ and $V$ say. You need to show that $K$ is the disjoint union of the nonempty open sets $f^{-1}(U)$ and $f^{-1}(V)$. That contradicts the fact that $K$ is connected.Julio said:Let $f: K\subseteq \mathbb{R}^n\to \mathbb{R}$ continuous, $K$ an connexe subset. Show that $f(K)$ is connexe.Hello :). Any ideas for this problem? Thanks :)