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In Theorem 5.2, Rudin proves that if a function on $[a,b]$ is differentiable at $x$, it is continuous at $x$. I think the go-to example is the absolute value function, but he then writes:

The converse of this theorem is not true. It is easy to construct continuous functions which fail to be differentiable at isolated points.

I assume that Rudin is using "isolated points" as in Definition 2.18: "If $p \in E$ and $p$ is not a limit point of $E$, then $p$ is an isolated point of $E$."

He defines derivatives in terms of the limit of a difference quotient $f'(x) = \lim_{t \to x} \frac{f(t) - f(x)}{t-x}$, but earlier he defines limits only as the domain points of $f$ approach a limit point. So it doesn't seem that one can even define differentiability at an isolated point.

Am I misinterpreting what Rudin said here?

Cardinality
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    The absolute value function is a great example of something but not of the thing you've described. A nice example of that is the product of $f(x) = x^2$ and $g(x) = $ 1 if $x$ is rational, $0$ otherwise. It's differentiable at $x = 0$, and therefore continuous there, but it's not differentiable or continuous anywhere else. – John Hughes Jan 17 '25 at 13:35
  • @JohnHughes Thank you very much for the answer. I'm not sure that I fully understand. It seems that $f$ and $g$ are defined on all of $\mathbb{R}$, so wouldn't no point of the domain be an isolated point? – Cardinality Jan 17 '25 at 14:37
  • @John Hughes: I think the problem here is that Rudin's wording is a bit ambiguous. The OP is taking the meaning to be that the points in question are isolated in the domain of the function (such as is discussed in Proving that a function is not differentiable at an isolated point), whereas you (and as Rudin intended, but it probably should have been worded more carefully) are taking the meaning to be that the points in question are isolated in the reals. – Dave L. Renfro Jan 18 '25 at 14:59
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    I was talking only about the first sentence and a half, which seemed to me to show a misunderstanding about something basic, NOT about the rest of the question. But for the rest of the question, I agree with Dave L. Renfro's interpretation of what Rudin is trying to say. – John Hughes Jan 18 '25 at 15:49

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