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I was reading a bit about open sets and wondered whether one could say that a subset of $\mathbb R $ is open if and only if it has neither a maximum nor a minimum. I tried searching for an answer online but didn't find anything on the matter. I can imagine that no one uses it since it doesn't translate that well to the $ \mathbb R^n$ as the $ \epsilon$-Neighbourhoods.

But nevertheless I'd like to know if my statement is wrong and if so where, just to get a better grasp of the topic.

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    If $U$ is open, then it has neither a maximum nor a minimum. But the converse need not be true: take $U := ]-1,-\frac{1}{2}]\cup[\frac{1}{2},1[$. – 2017-02-20
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    Ok yeah, as soon as you know the answer it seems so obvious, thanks for clearing it up for me ! – 2017-02-20

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Your statement is not quite correct. If the set is assumed to be connected (or equivalently, an interval) then your statement is true. But not all subsets of $\mathbb{R}$ are intervals.