What is the Analyse definition of local maximum point I am not talking about this : If a function f is defined for all $ \hspace{0.33em}{x}\mathrm{\in}{I} $ Not indeed the full definition set / I mean ${I}$ can be a subset of the definition set of the function f/
then if $ {f}{\mathrm{(}}{a}{\mathrm{)}}\mathrm{\geq}{f}{\mathrm{(}}{x}{\mathrm{)}} $ For all $ \hspace{0.33em}{x}\mathrm{\in}{I} $ Then f(a) is a local maximum point ..
But I read a book about this That said
Again :If a function f is defined for all $ \hspace{0.33em}{x}\mathrm{\in}{I} $ And we have the OPEN interval J Where $ \hspace{0.33em}{a}\mathrm{\in}{J} $ And we take the Intersection $ {I}\mathrm{\cap}{J} $ and we called it D So if $ {f}{\mathrm{(}}{a}{\mathrm{)}}\mathrm{\geq}{f}{\mathrm{(}}{x}{\mathrm{)}} $ for all $ \hspace{0.33em}{x}\mathrm{\in}{D} $ then f(a) is local maximum point.
I want to know what is the difference between the above definitions of a local maximum point? indeed why J is an Open interval ? I will be thankful for anyone who helps me
Let $f:D\rightarrow \mathbb{R}$ be a function with domain $D$.
When we say that $x_0$ is a local maximum (minimum) of $f$, intuitively, we want a neighborhood of $x_0$ satisfying that $f(x_0)$ gives the maximum value in that neighborhood. So here we need to think about what a neighborhood means. From the point of view in topology, $N$ is a neighborhood of $x_0$ if $\exists$ open set $G$ such that $x_0\in G\subset N$.
In the case $D\subset \mathbb{R}$, an open set can be written into countable union of open intervals, so without loss of generality, we can regard $J$ as an open interval. Also, it does not necessarily be an open interval (as long as it is a neighborhood of $x_0$ is okay. i.e. $\epsilon>0$ s.t. $(x_0-\epsilon,x_0+\epsilon)\subset J$. Indeed they are equivalent statements.). Using $J$ as an open interval in some sense keeps the consistency with topology (I mean the more general case that $D$ may be a subset of a topological space rather than $\mathbb{R}$).
To answer your first question, $I$ is just an arbitrary subset of $D$, so it may not be a neighborhood of $x_0$, say $I=\{1\}\cup(2,3)$, and $f(x)=|x|$. Then it satisfies the first definition but not the second. However, we know that $1$ is not a local minimum nor a local maximum.