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The major problem of 5.1 Definition in Baby Rudin

  1. Mar 10, 2012 #1
    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.

    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.

    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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    Last edited: Mar 10, 2012
  2. jcsd
  3. Mar 10, 2012 #2
    He admits that f may not be differentiable where the above quotient limit does not exist. If f is on [a,a] ( a = b ) , then the requirement ( x in [a,b] a < t < b , t != x ) cannot be satisfied and so the quotient ( limit ) does not exist.
    Last edited: Mar 10, 2012
  4. Mar 10, 2012 #3
    The quotient exists as an empty function. Doesn't it?
  5. Mar 10, 2012 #4
    whether or not you consider the quotient to exist as an empty function, the limit will not exist. ( And I guess the empty function will not count after imposing a "non-trivial" function restriction )
  6. Mar 10, 2012 #5
    I think that because it is undefined, the existence of it cannot be talked about.

    And more problematic thing is that if we think about the statement, which is used in defining the limit:

    [itex] \forall \epsilon > 0 \exists \delta > 0 \forall t \in \mbox{dom}(\phi) (0<|t-a|<\delta \to |\phi (t) - L| < \epsilon)[/itex],

    this statement is (vacuously) true for all [itex]L \in \mathbb{R}[/itex] which suggests that the derivative exists and is any real number.

    But anyway, in a formalistic view, if we constrain ourselves in the system of Rudin's text, the limit is undefined, thus we can't conceive whether it exists or not. Isn't it?
  7. Mar 10, 2012 #6
    The statement is not vacuously true for all L, since the implication does not follow if there is no t that satisfies | phi ( t ) - L | < epsilon.

    An undefined limit is a limit that does not exist
  8. Mar 13, 2012 #7
    You should have noted that [itex]\mbox{dom}(\phi)=\emptyset[/itex]. Thus the implication is (vacuously) true, as the statement becomes true. If you don't agree with this, please refer to Velleman, How to Prove It, p. 69.

    And also not everyone says nor thinks an object undefined does not exist.
    Last edited: Mar 13, 2012
  9. Mar 13, 2012 #8


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    All he's saying here is that you can have examples like f(x)=|x| (x in (-1,1), say) that are continuous everywhere but not differentiable at some isolated points in the domain, e.g. 0 for this particular f(x). The only reason he says "isolated points" here is because it's "easy" (to quote him) to give examples like the one I gave, but it's much more difficult to find, e.g., a function that's continuous on (-1,1) but not differentiable at, say, any point in [0,1/2].

    Anyway, I agree with you that his definition is problematic if the interval [a,a] is degenerate. But at the same time, does this really matter? It's kind of missing the point to talk about differentiability on {a}.
    Last edited: Mar 13, 2012
  10. Mar 14, 2012 #9
    Yes, I somewhat feel like it's a wasting-time thing. Especially it will be wasting time if there is no gain from the study of the derivative of this function defined on a singleton. (And I think it's indeed the case.) But anyway I wanted to check if my understanding on this definition and its application is rigorous. So, thanks for this point.

    And as you said, if Rudin had that kind of function in his mind, then well he is right thoguh his usage of 'isolated point' is totally wrong because the point 0 is not at all an isolated point in (-1,1). But I still kind of cannot help but tend to think that he used this expression correctly, i.e., he didn't have the function you referred to in his mind. So, I thought my understanding on his definition of derivative is wrong, but then his definition turns out kinda not right.. so.. what.. But anyway, I probably stop now thinking about this.
    Last edited: Mar 14, 2012
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