Can there be a non constant continuous function from the open unit disc $D=\{z\in\Bbb{C}~:~|z|<1\}$to $\Bbb{R}$ which takes only irrational values.

Here I couldn't apply intermediate value theorem as could be done if it were a real function to disprove the statement.

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    $\begingroup$ "Here I couldn't apply intermediate value theorem" Why not? Look at the function on the segments $[u,v]=\{tu+(1-t)v\mid0\leqslant t\leqslant1\}$ for any $(u,v)$ in $D$. $\endgroup$ – Did May 4 '18 at 5:42
  • $\begingroup$ Oh thanks... This way I can apply ivp to any real valued function having a convex domain. $\endgroup$ – Abishanka Saha May 4 '18 at 5:53
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    $\begingroup$ Yes, or only path-connected. $\endgroup$ – Did May 4 '18 at 5:55

You can prove something a bit stronger, indeed this works if we replace the unit disc for any open path-connected set, or equivalently (in $\mathbb{C}$) a connected open set.

Let $f: U \subseteq \mathbb{C} \to \mathbb{R}$ be a continuous function with $U$ open and path connected. Let's fix $z,w \in U$ and a path $\gamma : [0,1] \to U$ from $z$ to $w$. Now, $f\gamma:[0,1] \to \mathbb{R}$ is continuous. Therefore, $f\gamma([0,1]) = f(\gamma[0,1]) \subseteq im(f)$ is connected, so $im(f)$ contains an interval and in particular, takes values in $\mathbb{Q}$.


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