# Erroneous Proof that the Derivative is Continuous

I have written the following which argues that for any arbitrary sequence of points such that $$x_n \to c, x_n\neq c$$ we must also have $$f'(x_n)\to f'(c)$$ which would imply that $$f'(x)\to f'(c)$$ as $$x\to c$$ , i.e., the derivative is continuous. However, I can't see where the error in my argument is.

Let $$f:[a,b] \to \mathbb{R}$$ be a differentiable function and $$c\in [a,b]$$. Fix $$\epsilon > 0$$. By assumption, for $$c\in [a,b]$$, $$f'(c)$$ exists so put $$\delta_1 > 0$$ so small that whenever $$0<\lvert x -c \rvert < \delta_1$$ we must have $$\left \lvert \frac{f(x)-f(c)}{x-c}-f'(c) \right \rvert < \frac{\epsilon}{2}$$ $$[a,b]$$ is closed so there exists a sequence such that $$x_n \to c$$ and $$x_n \neq c$$. By assumption, $$f'(x_n)$$ exists so there exists a $$\delta_2 >0$$ such that whenever $$0<\lvert x-x_n \rvert < \delta_2$$ we must have $$\left \lvert f'(x_n)-\frac{f(x)-f(x_n)}{x-x_n} \right \rvert < \frac{\epsilon}{2}$$ Now pick an integer $$N$$ so large that for $$n\geq N$$ we must have $$0<\lvert x_n - c\rvert < \mathrm{min}(\delta_1,\delta_2)$$. It follows that $$\left \lvert \frac{f(x_n)-f(c)}{x_n-c}-f'(c)\right \rvert < \frac{\epsilon}{2} \quad \textbf{and} \quad \left \lvert f'(x_n)-\frac{f(x_n)-f(c)}{x_n-c}\right \rvert < \frac{\epsilon}{2}$$ Applying the triangle inequality, we see that $$\lvert f'(x_n)-f'(c) \rvert < \epsilon$$ and we are done.

• Note that $\delta_2$ depends on $x_n$ (that is, depends on $n$). That is, the $\delta_2$ can change if $n$ changes. Commented Nov 4, 2021 at 21:00
• So let us write $\delta_2(n)$, since it depends on $n$. Then what you need is for $0\lt|x_n-c|\lt \min(\delta_1,\delta_2(n))$ to hold. That is a different condition for each $n$; thus, to make it a single condition satisfied by all of those $x_n$, you would want to define $\delta_2=\inf\{\delta_2(n)\mid n\geq N\}$ for some $N$. But that $\delta_2$ could be equal to $0$, which sinks your argument. Commented Nov 4, 2021 at 21:06
• @ArturoMagidin Very well spotted and very well put Commented Nov 4, 2021 at 21:10
• @ArturoMagidin I see, thank you! A very good answer indeed Commented Nov 4, 2021 at 21:13

Let me write it up so the question is not unanswered...

Note that your $$\delta_2$$ depends on $$x_n$$. So we should really write $$\delta_2(n)$$. That is for a fixed $$n$$, $$\delta_2(n)$$ is a positive real such that if $$0\lt |x-x_n|\lt \delta_2(n)$$, then $$\left|\frac{f(x)-f(x_n)}{x-x_n} - f'(x_n)\right| \lt \epsilon/2$$.

Then $$\delta=\min(\delta_1,\delta_2(n))$$ guarantees that $$|f'(c)-f'(x_n)|\lt \epsilon$$ for that fixed $$n$$, not for an arbitrary $$n$$.

So in order to ensure that this inequality holds for all $$n\geq N$$, you would need to find a $$\delta$$ that satisfies $$\delta\leq \inf(\{\delta_1\}\cup\{\delta_2(n)\mid n\geq N\})$$. Unfortunately, that infimum could be equal to $$0$$, which means that you have no guarantee that you can pick a $$\delta\gt 0$$ that ensures both inequalities you have at the end for all $$n\geq N$$.

• I think writing $\delta_2(n)$ is both a mathematical mistake and a pedagogical one. It makes $\delta_2$ look like a function, and it is not.
– Plop
Commented Nov 4, 2021 at 21:49
• @Plop: It's a choice function, certainly. I might agree with "pedagogical", but I don't see why it is a "mathematical mistake". There are other possible choice functions that work as well, of course. Commented Nov 4, 2021 at 21:58
• Well, transforming $\forall n, \exists \delta,\ P(n,\delta)$ into $\exists \delta,\ \forall n,\ P(n,\delta(n))$ isn’t at all an innocent logical manipulation.
– Plop
Commented Nov 4, 2021 at 22:02
• @Plop: "not innocent logical manipulation" $\neq$ "mathematical mistake". If for every $n\in\mathbb{N}$ there exists $\delta$ such that $P(n,\delta)$, then there exists a function $\delta_2$ with domain $\mathbb{N}$ such that for each $n\in \mathbb{N}$, $\delta_2(n)$ has property $P(n,\delta_2(n))$. At least, assuming the Axiom of Choice. It is not a quantifier exchange, because they are different types of objects. Commented Nov 4, 2021 at 22:03
• Ok, I may have been too rigid. What about « asserting implicitly that one will later use a Skolem form of an assertion is mathematically not careful enough »?
– Plop
Commented Nov 4, 2021 at 22:11

The problem is that your proof is not written well enough, so the mistake may be hard to find.

Something better would have been the following.

1. Blah blah.

...

1. For every $$n$$, there exists $$\delta_2$$ such that blah blah blah.

2. Let $$\delta_2$$ such that for every $$n$$, blah blah blah (such a $$\delta_2$$ exists from 23)).

...

1. So $$f’$$ is continuous.

In this proof, it is now quite obvious to spot the « mistake »: while $$\forall n, \ \exists \delta_2$$ is proved, you use the unproved assertion $$\exists \delta_2, \ \forall n$$.