Showing Connexity of $f(K)$ for Continuous $f$ on Connexe $K$

[Julio1]

Let $f: K\subseteq \mathbb{R}^n\to \mathbb{R}$ continuous, $K$ an connexe subset. Show that $f(K)$ is connexe.

Spoiler

Hello :). Any ideas for this problem? Thanks :)

 

[Opalg]

Let $f: K\subseteq \mathbb{R}^n\to \mathbb{R}$ continuous, $K$ an connexe subset. Show that $f(K)$ is connexe.

Spoiler

Hello :). Any ideas for this problem? Thanks :)

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.

 

[Euge]

Another way to show that $f(K)$ is connected is by proving that the only nonempty, open and closed subset of $f(K)$ is $f(K)$ itself. So let $A$ be a nonempty subset of $f(K)$ that is open and closed. Using continuity of $f$ and the fact that $A \subset f(K)$, show that $f^{-1}(A)$ is a nonempty, open and closed subset of $K$. Since $K$ is connected, $f^{-1}(A) = K$. Thus $f(K) = f(f^{-1}(A)) \subset A$, showing that $f(K) = A$.