- #1

julypraise

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I will prove the following statement is true to show the flaw of 5.1 Definition in Baby Rudin. If in any case I'm wrong, please correct me. Thanks.

Statment:

Suppose [itex]f[/itex] is a function defined on [itex][a,a][/itex] with [itex]a \in \mathbb{R}[/itex]. Then it is impossible to apply 5.1 Definition in Baby Rudin for this function.

Proof.

As stated in 5.1 Definition, we form the [itex]\phi[/itex] function as the domain of this function becomes [itex]\{t:t \neq a, a<t<a\}[/itex] which is empty. Then this function is an empty function (i.e., the empty set) because of its domain. Now we consider this domain as a subset of a metric space [itex]\mathbb{R}[/itex] with the usual metric [itex]| \, |[/itex]. Then since no point in the metric space [itex]\mathbb{R}[/itex] is a limit point of the domain of [itex]\phi[/itex], the limit [itex]\lim_{t \to a} \phi (t)[/itex] is undefeind according to 4.1 Definition. Thus we conclude we cannot apply 5.1 Definition. This completes the proof.

Actually, by some intro calculus textbook definition, it is even possible to prove that a function is differentiable at a isolated point but the derivative value at this point is any number in [itex]\mathbb{R}[/itex] (Bartle, Introduction to Real Analysis, 6.1.1 Definition). I still can't find any textbook that proposes a definition of derivative purely by which I can prove that any function defined on a singleton does not have a derivative at that point.

Rudin, in his Baby Rudin, talks about the function defined at isolated points. Being specific, what he says is that "it is easy to construct continuous functions which fail to be differentiable at isolated points (p. 104)" but by his definition this is not provable. At least he could have just stated that a function at a isolated point is not differentiable. But even this thing he didn't do.

Anyway, my proof might be wrong. Please prove me wrong or propose some nice strong definition of derivative. Thanks.

Statment:

Suppose [itex]f[/itex] is a function defined on [itex][a,a][/itex] with [itex]a \in \mathbb{R}[/itex]. Then it is impossible to apply 5.1 Definition in Baby Rudin for this function.

Proof.

As stated in 5.1 Definition, we form the [itex]\phi[/itex] function as the domain of this function becomes [itex]\{t:t \neq a, a<t<a\}[/itex] which is empty. Then this function is an empty function (i.e., the empty set) because of its domain. Now we consider this domain as a subset of a metric space [itex]\mathbb{R}[/itex] with the usual metric [itex]| \, |[/itex]. Then since no point in the metric space [itex]\mathbb{R}[/itex] is a limit point of the domain of [itex]\phi[/itex], the limit [itex]\lim_{t \to a} \phi (t)[/itex] is undefeind according to 4.1 Definition. Thus we conclude we cannot apply 5.1 Definition. This completes the proof.

Actually, by some intro calculus textbook definition, it is even possible to prove that a function is differentiable at a isolated point but the derivative value at this point is any number in [itex]\mathbb{R}[/itex] (Bartle, Introduction to Real Analysis, 6.1.1 Definition). I still can't find any textbook that proposes a definition of derivative purely by which I can prove that any function defined on a singleton does not have a derivative at that point.

Rudin, in his Baby Rudin, talks about the function defined at isolated points. Being specific, what he says is that "it is easy to construct continuous functions which fail to be differentiable at isolated points (p. 104)" but by his definition this is not provable. At least he could have just stated that a function at a isolated point is not differentiable. But even this thing he didn't do.

Anyway, my proof might be wrong. Please prove me wrong or propose some nice strong definition of derivative. Thanks.

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