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If $\Omega\in \mathbb{R^n}$ is a connect open set,function $f:\Omega \to \mathbb{R}$ is continious and locally constant. Prove:$f$ is constant function.

I try to construct a "path" to connect $x$ and $y$, but in fact $E$ is not path connected，so this can't work.

Lagranngekmno4
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    what is $E$? Also, I'm not sure if you've seen this before, but if $\Omega \subset \mathbb{R}^n$ is connected and open, then you can conclude that $\Omega$ is path-connected. – Jonathan Pal Mar 14 '23 at 14:51
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    It follows from the definition of connectedness. That it is in $\mathbb R^n$ is not important. (See here) – Arctic Char Mar 14 '23 at 14:55

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