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In the video lectures on real analysis by Professor Su he uses the following definition for the derivative:

A function $f: [a,b] \to \mathbb{R}$ is differentiable at $x \in [a,b]$ if the limit $\lim_{t \to x} \frac{f(t)-f(x)}{t-x}$ exists. In this case we say $f'(x)=\lim_{t \to x} \frac{f(t)-f(x)}{t-x}$ is the derivative of $f$ at $x$.

The definition of limit in $\mathbb{R}$ is as follows:

Let $f:E \to \mathbb{R}$ where $E \subset \mathbb{R}$ and let $p$ be a limit point of $E$, then we say $\lim_{x \to p} f(x)=q$ if $\exists q \in \mathbb{R}: \forall \epsilon>0$ $\exists \delta>0$ s.t. $\forall x \in E$, $0<\lvert x-p \rvert<\delta \implies \lvert f(x)-q \rvert<\epsilon$.

I was wondering why we restrict ourselves to functions with an interval as its domain. Usually the definitions in metric spaces are based on examples in $\mathbb{R}$ and can be generalized to arbitrary metric spaces, e.g. the definition of limit of a function or continuity of a function can be generalized by replacing the absolute value by a distance function. Of course, the derivative defined as a limit of a quotient cannot be generalized to arbitrary metric spaces since division might not be defined, but why would we restrict ourselves to functions defined on an interval?

In light of the definition of the limit we can make sense of this limit for any function defined on a subset $E \subset \mathbb{R}$. Of course in the case where the domain is $\mathbb{R}$ we do not need to specify the condition that $x$ must be a limit point for the derivative at $x$ to exist since every point of $\mathbb{R}$ is a limit point. However, if we generalize this to functions $f:E \to \mathbb{R}$ where $E \subset \mathbb{R}$, then we require that $x$ is a limit point for the limit to make sense.

This definition would also cover the special cases of intervals. For example, we can let $E=[a,b]$, then every point in this interval is a limit point as long as $a \neq b$ and we can consider the limit at any point without the need to separately introduce one-sided limits. This is because we can consider $[a,b]$ as a metric space in its own right and for the $\delta$ ball of any radius about $a$ or $b$ in $[a,b]$ is simply cut off at one side (because we only consider points in the metric space).

To sum up why do most textbooks/lectures on real analysis restrict to open or closed intervals for derivatives?

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  • $\begingroup$ Derivatives in discontinuous sets are not so useful. For instance, they would not allow to discuss the growth of the functions, nor use Rolle's theorem. $\endgroup$
    – user65203
    Commented Feb 4, 2020 at 13:16

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Actually, there are several options available here. Take, for instance, the definition of derivative in Spivak's Calculus:

The function $f$ is differentiable at $\mathbf a$ if$$\lim_{h\to0}\frac{f(a+h)-f(a)}h\text{ exists.}$$In this case the limit is denoted by $\mathbf{f'(a)}$ and is called the derivative of $\mathbf f$ at $\mathbf a$.

As you can see, no reference is made to the domain $D_f$ of $f$; it is simply implicit that it is such that $a$ is an accumulation point of $D_f$; otherwise, that limit doesn't have to be unique.

However, such an approach assumes that the students are comfortable with the concept of “accumulation point”. If that's a problem, then it is simpler and almost as general to assume that $D_f$ is an interval (with more than one point).

Besides, many standard Calculus theorems (such as Rolle's theorem, the extreme value theorem, or the mean value theorem) are theorems about functions $f$ for which $D_f$ is an interval which is closed and bounded.

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  • $\begingroup$ Thanks for your comment. I agree that calculus textbooks simplify most things, but I think in real analysis one should be more rigorous about the definitions. Good point about the theorems though. Another thing I had in mind was that requiring that $x$ is a limit point basically says that there is an open ball in the domain that contains infinitely many points and I guess one could then always restrict the function to that open ball if one is only interested in the point $x$. I am not entirely sure though if the definitions are equivalent then. $\endgroup$ Commented Feb 1, 2020 at 9:58
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    $\begingroup$ What I wrote was an answer, not a comment. And no, asserting that $x$ is a limit point does not say “that there is an open ball in the domain that contains infinitely many points”. If$$D_f=\{0\}\cup\left\{\frac1n\,\middle|\,n\in\mathbb N\right\},$$then $0$ is a limit point, but $D_f$ contains no open ball. $\endgroup$ Commented Feb 1, 2020 at 10:04
  • $\begingroup$ Ah sorry, of course limit point only says that every open ball around it contains some other point of the domain and not that the open ball is entirely in $\mathbb{R}$. $\endgroup$ Commented Feb 1, 2020 at 10:12

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