If I have a subset $A \subseteq \mathbb{R}$, a function $f: A \to \mathbb{R}$, and a point $a \in A$, then (according to the definitions I came across) for $f$ to be differentiable at $a$, then there must exist some open interval $I$ so that $a \in I$ and $I \subseteq A$.

I tried understanding why this requirement exists, by first discarding it and seeing if it naturally arises from the other bits of the definition.

NOTATION: For any $x \in X\subseteq \mathbb{R}$, I define $D[x;X] \subseteq \mathbb{R}$ as: $$D[x;X] = \{ \ h \in \mathbb{R} \setminus \{ 0 \} \ | \ x + h \in X \ \}$$ and for any $\phi: X \to S$ such that $S \subseteq \mathbb{R}$, I define $\mathrm{Q}^{\phi}_{x} : D[x;X] \to \mathbb{R}$ as: $$\mathrm{Q}^{\phi}_{x}(h) = \frac{\phi(x+h) - \phi(x)}{h}$$ DEFINITION: First I will fix the point $a \in A$, and then define the derivative of $f$ at $a$ as: $$f'(a) = \lim_{ h \to 0 } \mathrm{Q}^{f}_{a}(h)$$ REFLECTION: If $f'(a)$ exists, then $\lim_{ h \to 0 } \mathrm{Q}^{f}_{a}(h)$ exists. For that to exist, $0\in\mathbb{R}$ has to be an accumulation point of $D[a;A]$ (every punctured neighbourhood of $0\in\mathbb{R}$ contains at least one point from $D[a;A]$). From that, it follows that $a\in A$ must be an accumulation point of $A$. But now I am stuck, I am missing some sort of step or requirement or something, that would allow me to say: $$a_{0} \ \text{is an accumulation point of} \ A \ \ \wedge \ \ \text{something else} \implies \text{there must exist some open interval $I$ so that $a_{0} \in I$ and $I \subseteq A$.}$$ Now, I could always restrict the definition of limits to require that the domain of a function includes within it a punctured neighbourhood of the limit point. But that kind of feels like cheating? What is it that I am missing to make this step?


I thought maybe some common properties of derivatives cannot be proven with this definition. For example: $$f'(a) = \lim_{ h \to 0 } \mathrm{Q}^{f}_{a}(h) \implies \lim_{ x \to a } f(x) = f(a)$$ But sure enough, if I consider the function $I: D[a;A] \to \mathbb{R}; \ \ I(h) = h$, I can deduce that $\lim_{ h \to 0 } I(h) = 0$, so I can use the limit composition to say that: $$ \begin{align} &\lim_{ h \to 0 } I(h) = 0 \ \ \text{and} \ \ \lim_{ h \to 0 } \mathrm{Q}^{f}_{a}(h) = f'(a) &\implies& \begin{aligned} \lim_{ h \to 0 } \left[ I(h) \cdot \mathrm{Q}^{f}_{a}(h) \right] &= \lim_{ h \to 0 } \left[ h \cdot \frac{f(a+h)-f(a)}{h} \right] \\ &= \lim_{ h \to 0 } \left[ f(a+h)-f(a) \right] \\ &= 0 \cdot f'(a) \\ &= 0 \\ \end{aligned} \\ &&\implies& \lim_{ x \to a } f(x) = a \\ \end{align} $$ In a similar vein, Hans Lundmark suggested that you need extra assumptions to prove the chain rule, but that doesn't seem to be true either (I might be wrong please check my proof🙏).

For any subsets $A,B \subseteq \mathbb{R}$, and for any functions $f: A \to \mathbb{R}$ and $g: B \to \mathbb{R}$ such that $f[A] \subseteq B$, the composite function $g \circ f: A \to \mathbb{R}$ is well defined, which I will denote as $u$. I will now:

  1. Fix the point $a \in A$ and assume that $f'(a) \in \mathbb{R}$ exists.
  2. Define $b \in B$ as $b = f(a)$ and assume that $g'(b) \in \mathbb{R}$ exists.

Which means that: $$f'(a) = \lim_{ h \to 0 } \mathrm{Q}^{f}_{a}(h) \ \ \text{and} \ \ g'(b) = \lim_{ h \to 0 } \mathrm{Q}^{g}_{b}(h)$$ I define the function $k: D[a;A] \to \mathbb{R}; \ \ k(h) = f(a + h) - f(a)$; I know that $\lim_{ h \to 0 } k(h) = 0$ by the proof above. I then define the function $\mathrm{Q}^B: D[a;A] \to \mathbb{R}$ as: $$ \mathrm{Q}^B(h) = \begin{cases} \mathrm{Q}^{g}_{b}(k(h)) & k(h) \ne 0\\ g'(b) & k(h) = 0 \end{cases} $$ We can then deduce the following: $$ \begin{align} &\begin{aligned} \underline{\textbf{We know that:}} &&& \lim_{ h \to 0 } k(h) = 0 &&\iff \forall \varepsilon > 0, \exists \delta > 0, \forall h \in D[a;A]: 0<|h|<\delta \implies |k(h)| < \varepsilon \\ &&& \lim_{ h \to 0 } \mathrm{Q}^{g}_{b}(h) = g'(b) &&\iff \forall \varepsilon > 0, \exists \delta > 0, \forall h \in D[b;B]: 0<|h|<\delta \implies |\mathrm{Q}^{g}_{b}(h) - g'(b)| < \varepsilon \\ \end{aligned} \\ \\ &\begin{aligned} &\text{Take an arbitrary $\epsilon>0$, hence} && \text{$\exists \delta > 0, \forall h \in D[b;B]: 0<|h|<\delta \implies |\mathrm{Q}^{g}_{b}(h) - g'(b)| < \epsilon$} \\ &\text{Take the particular $d>0$ for which this is true, hence} && \text{$\forall h \in D[b;B]: 0<|h|<d \implies |\mathrm{Q}^{g}_{b}(h) - g'(b)| < \epsilon$} \\ &\text{Since $d>0$, we also have}&&\text{$\exists \delta > 0, \forall h \in D[a;A]: 0<|h|<\delta \implies |k(h)| < d$} \\ &\text{Take the particular $\varDelta>0$ for which this is true, hence}&&\text{$\forall h \in D[a;A]: 0<|h|<\varDelta \implies |k(h)| < d$} \\ &\text{Take an arbitrary $H \in D[a;A]$, hence}&&\text{$0<|H|<\varDelta \implies |k(H)| < d$} \end{aligned} \\ \\ &\begin{aligned} \underline{\text{Assume that $0<|H|<\varDelta$:}} &&&\text{Therefore we have $|k(H)|<d$} \\ &&&\text{By law of excluded middle, we have $k(H) = 0$ or $k(H) \ne 0$} \\ &&&\begin{aligned} \underline{\text{When $k(H) = 0$:}} &&&\text{$\mathrm{Q}^{B}(H) = g'(b)$, and therefore $0 = |\mathrm{Q}^{B}(H) - g'(b)| < \epsilon$} \end{aligned} \\ &&&\begin{aligned} \underline{\text{When $k(H) \ne 0$:}} &&&\text{$\mathrm{Q}^{B}(H) = \mathrm{Q}^{g}_{b}(k(H))$, and therefore $k(H) \in B[b;B]$} \\ &&&\text{Hence, we have $0<|k(H)|<d \implies |\mathrm{Q}^{g}_{b}(k(H)) - g'(b)| < \epsilon$} \\ &&&\text{Since $|k(H)| > 0$, we have $0<|k(H)|< d$} \\ &&&\text{By implication, we have $|\mathrm{Q}^{g}_{b}(k(H)) - g'(b)| = |\mathrm{Q}^{B}(H) - g'(b)| < \epsilon$} \end{aligned} \\ &&&\text{By cases, we conclude that $|\mathrm{Q}^{B}(H) - g'(b)| < \epsilon$} \end{aligned} \\ \\ &\begin{aligned} &\text{Therefore, we have} && 0<|H|<\varDelta \implies |\mathrm{Q}^{B}(H) - g'(b)| < \epsilon \\ &\text{$H$ is arbitrary, $\varDelta$ is particular, $\epsilon$ is arbitrary, so we have}&&\text{$\forall \varepsilon > 0, \exists \delta > 0, \forall h \in D[a;A]: 0<|h|<\delta \implies |\mathrm{Q}^{B}(h) - g'(b)| < \varepsilon$} \\ &&& \iff \lim_{ h \to 0 } \mathrm{Q}^{B}(h) = g'(b) \end{aligned} \\ \\ &\begin{aligned} \underline{\textbf{We conclude that:}} &&& \lim_{ h \to 0 } \mathrm{Q}^{B}(h) = g'(b) \end{aligned} \end{align} $$ From here it is rather trivial to show that $\mathrm{Q}^{u}_{a}(h) = \mathrm{Q}^{B}(h) \cdot \mathrm{Q}^{f}_{a}(h)$, just consider the cases $k(h) \not= 0, k(h) = 0$ in turn. Finally, by limit composition, we have $$ \begin{align} &\lim_{ h \to 0 } \mathrm{Q}^{B}(h) = g'(b) \ \ \text{and} \ \ \lim_{ h \to 0 } \mathrm{Q}^{f}_{a}(h) = f'(a) &\implies& \begin{aligned} \lim_{ h \to 0 } \left[ \mathrm{Q}^{B}(h) \cdot \mathrm{Q}^{f}_{a}(h) \right] &= \lim_{ h \to 0 } \mathrm{Q}^{u}_{a}(h) \\ &= u'(a) = (g \circ f)'(a) \\ &= g'(b)f'(a) = g'(f(a))f'(a) \end{aligned} \\ &&\implies& u'(a) = g'(f(a))f'(a) \\ \end{align} $$ Hence, the chain rule holds. To the best of my knowledge, I don't think I made use of any additional assumptions to deduce this, beyond the definitions I provided.


As pointed out by Hans Lundmark (thankyou), my assumption that $f[A] \subseteq B$ is a little strong. The most general case in which composition still makes sense, is if we define the following:

  1. $A^* = \{ \ x \in A \ | \ f(x) \in B \ \}$ the restricted domain of $f$ for which composition can occur
  2. $f_{r} = \{ \ \langle x, y \rangle \in f \ | \ x \in A^* \ \}$ the corresponding restricted $f$

I can now define the composite function $g \circ f: A^* \to \mathbb{R}; \ \ x \mapsto g(f_{r}(x))$, and as before:

  1. $u = g \circ f$.
  2. Fix the point $a \in A^*$ and assume that $f'(a) \in \mathbb{R}$ exists.
  3. Define $b \in B^*$ as $b = f(a)$ and assume that $g'(b) \in \mathbb{R}$ exists.

That seems to be the most general case for which the concept of function composition can make sense. And so, if I pick $A = \left\{ \ \frac{1}{2x} \ | \ x \in \mathbb{Z} \setminus \{ 0 \}, \frac{x}{3} \not\in \mathbb{Z} \ \right\} \cup \{ 0 \}$ and $B = \left\{ \ \frac{1}{3x} \ | \ x \in \mathbb{Z} \setminus \{ 0 \}, \frac{x}{2} \not\in \mathbb{Z} \ \right\} \cup \{ 0 \}$ and $f(x)=x$ and $g(x) = x$ and $a=0$, then we see that $f'(0)=1$ and $g'(f(0))=1$, but $u'(0)$ does not exist. Because $A^*= \{ 0 \}$, so $0$ is not an accumulation point of $A^*$. Here is a plot I drew up real quick:

enter image description here

The extra assumption needed to make this work, is to require that $a$ is an accumulation point of $A^*$. So then $f_{r}'(a) = f'(a)$, and since $f_{r}[A^*] = B^* \subseteq B$, we can just use the above proof and deduce that $u'(a)=g'(f(a))f'(a)$. And none of these extra assumptions is needed when you assume the open interval. Thank you Hans Lundmark that one :)

  • 1
    $\begingroup$ There is the notion of a function that can be differentiated to the left or to the right. I couldn't find it in the English wikipedia; Also, I put a link (fr.wikipedia.org/wiki/D%C3%A9rivabilit%C3%A9) to the French Wikipedia. $\endgroup$ Mar 5 at 15:43
  • 2
    $\begingroup$ If you don't do this, you'll need to add extra assumptions for the chain rule to hold. $\endgroup$ Mar 5 at 20:34
  • $\begingroup$ @HansLundmark I tried proving the chain rule and see where it falls apart (without extra assumptions) but it looks like it holds without any extra assumptions. Could you just skim over my proof real quick? Just to make sure I am not smuggling any assumptions along the way. $\endgroup$ Mar 7 at 19:13
  • 1
    $\begingroup$ Well, you've added the extra assumption that $f[A] \subseteq B$. The version of the chain rule that I had in mind is just the following, with no mention of that assumption: “If $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$, then $u = g \circ f$ is differentiable at $a$, with $u'(a) = g'(f(a)) \, f'(a)$.” $\endgroup$ Mar 7 at 21:59
  • $\begingroup$ @HansLundmark thankyou, I have updated my question again to reflect your input $\endgroup$ Mar 11 at 0:15

1 Answer 1


The idea of derivative, in general, is "linear approximation".

In fact, a function is differentiable at $a_0$ if there exists a linear map, which is the derivative, that approximates the function in the neighborhood of $a_0$. This is represented in the equation $$ f(a_0 + h) - f(a) = \frac{df}{dx}(a_0)h + ε(h); \quad \lim_{h\to 0} \frac{\varepsilon(h)}{h} = 0. $$ To do justice to this idea and to garuantee the function is defined for $a_0 + h$ for any $h$ small enough, we ask for the function be defined in an open set, which is a neighborhood of all of it's point.

  • $\begingroup$ when you say linear map, what is the linear map? is $f'$ the linear map? So then do we have $f': \mathbb{R} \to \mathbb{R}$, $\forall x,y \in \mathbb{R}: \ \ f'(x + y) = f'(x) + f'(y)$ and $f'(x \cdot y) = x \cdot f'(y)$? $\endgroup$ Mar 10 at 23:39
  • 1
    $\begingroup$ @thethinker The derivative $\frac{df}{dt}(a_0)$ is a number, then the map $h\mapsto \frac{df}{dt}(a_0)h$ is linear. $\endgroup$ Mar 10 at 23:42
  • 1
    $\begingroup$ @thethinker In higher dimensions, though, we define $f'(a_0)$ to be itself a linear map $\endgroup$ Mar 10 at 23:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .