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Let $f\colon \mathbb{R} \mapsto \mathbb{R}$ be a function with the property that the image of every connected set is connected. Is $f$ necessarily continuous?

I've recently learned the definition of connected set and i'm still not totally confortable with it. I thought about the function $f(x)=\sin(1/x)$ for all real non zero $x$ and $f(0)=0$ (obviously discontinuous) for a counterexample but i'm not certain that the image of every connected set is connected...

Thank you in advance for your help!

Git Gud
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u1571372
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    The standard counterexample ;) What are the connected subsets of $\mathbb{R}$? – Daniel Fischer Oct 06 '13 at 18:54
  • Are the connected subsets of $\mathbb{R}$ the intervals? – u1571372 Oct 06 '13 at 19:00
  • Precisely those. Open, half-open, closed, degenerate (both endpoints the same), one or both ends may be infinite. Now, outside $0$, the function is continuous, so all intervals that don't contain $0$ have connected image. It remains to investigate the intervals containing $0$. – Daniel Fischer Oct 06 '13 at 19:07
  • There are more interesting examples as well, for instance, you can have a noncontinuous function so that the image of every open nonempty interval is the entire ${\mathbb R}$. – Moishe Kohan Oct 06 '13 at 21:36

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Recall the theorem that says

A subset $E$ of the real line $\mathbb{R}$ is connected if and only if it has the following property: If $x\in E, y\in E$ and $x<z<y$, then $z\in E$.

So your question reduces to does every function satisfying the following property: let $x,y$ be such that $f(x)<f(y)$ then there exists a $c\in (x,y)$ such that $f(x)<f(c)<f(y)$ (Darboux property) is continuous? This has already been answered Continuity $\Rightarrow$ Intermediate Value Property. Why is the opposite not true? (noticed by Andred Caicedo)

Community
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hrkrshnn
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