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From a textbook I got the following definition for a local maximum:

$x_0$ is a local maximum of the function $f$ if there exists an open interval $(a, b)$ such that $x_0 \in (a, b)$ and $f (x_0 ) \ge f (x)$ for all $x \in (a, b)$.

I am wondering why this needs to be an open interval? Wouldn't this definition work with a closed interval as well? What would be the problem with a closed interval?

Benjamin Wang
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mediation_boy
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    You might replace it with a closed interval where $x_0$ is not one of the endpoints, that would be equivalent. Point is, there has to be a two sided neighborhood (might be a very small one) where $x_0$ is a maximum. – Mark Dec 26 '23 at 14:48
  • alright thanks a lot! – mediation_boy Dec 26 '23 at 15:05
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    @HectorBlandin - derivatives are a nice tool for finding local extrema, but local extrema can exist even when the function is not differentiable anywhere. The derivative has nothing to do with the definition. The reason for using open intervals is that open intervals define what "local" means: We want an extrema over all points near to the given point. If the given point is the endpoint of the closed interval, then we are not looking at every point near to it to determine if it is the extreme. Points on one side of it are ignored. However, with open intervals, we are comparing on both sides. – Paul Sinclair Dec 28 '23 at 03:46

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