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Question on ε in epsilon-delta definition of limits.

  1. Dec 29, 2015 #1
    I am using Spivak calculus. The reason why epsilon-delta definition works is for every
    ε>0, we can find some δ>0 for which definition of limit holds.
    Spivak asserts yhat if we can find a δ>0 for every ε>0, then we can find some δ1 if ε equals ε/2. How is this statement possible? Since ε>0, then (ε/2) must be greater than zero. So, naturally one would argue that, if we can find δ for an ε>0, we can also find δ1 for (ε/2)>0. But a question arises for me. Why can't we say that if ε=(ε/2) and is >0, then ε>0. or why cant we say the converse. Why cant the proof start in converse way.
    Suppose, I can find a δn for every ε= ε^(2) +ε, can we concude the converse that ε must be greater than 0.
     
    Last edited: Dec 29, 2015
  2. jcsd
  3. Dec 29, 2015 #2

    FactChecker

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    This is not true in general. Are you sure it says 'for a ε>0'? If it say 'every', then I can agree. There must be something more to it. Maybe they picked an arbitrary ε>0 and are proving that there is a related δ1. That would work and since the ε>0 was arbitrary, the proof would work for every ε>0.
     
  4. Dec 29, 2015 #3
    y
    yeah....corrected it..
     
  5. Dec 29, 2015 #4

    Svein

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    No. That statement is plain wrong, since ε never equals ε/2 (except when ε=0). The correct statement is "Given ε>0, we can find a δ such that <something> is less than ε/2 whenever <something else> is less than δ".
     
  6. Dec 29, 2015 #5

    Fredrik

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    Let P(ε,δ) be some statement that involves ε and δ. (P should be viewed as a "function" that takes pairs of numbers to statements about those numbers). It seems to me that he was trying to say that the statement
    For all ε>0, there's a δ>0 such that P(ε,δ).​
    implies the statement
    For all ε>0, there's a δ>0 such that P(ε/2,δ).​

    This can be proved in the following way: Suppose that the first statement holds. Let ε be a positive real number. We have ε/2>0. Let δ be a positive real number such that P(ε/2,δ). (Since ε/2 is a positive real number, the first statement ensures that such a δ exists). Since ε is an arbitrary positive real number, this means that the second statement holds.
     
  7. Dec 30, 2015 #6
    Svein...Yes, it is a mistake on my part to equate both of them. So, it looks ambiguous. Fredrik put it in a right way what I meant.
     
  8. Dec 30, 2015 #7
    Can we prove that P(ε,δ) is true provided P(ε/2,δ) is true?
     
  9. Dec 30, 2015 #8

    Fredrik

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    No, but we can prove that the second statement in my previous post implies the first. The proof is very similar to the other one.

    Suppose that the second statement holds. Let ε be a positive real number. We have 2ε>0. Let δ be a positive real number such that P((2ε)/2,δ). (Since 2ε>0, the second statement ensures that such a δ exists). Since (2ε)/2=ε, this implies that P(ε,δ). Since ε is an arbitrary positive real number, this means that the first statement holds.
     
  10. Dec 31, 2015 #9
    ε δ
    Yes, I get it....if the f(ε) is a function of ε. Then if, P(f(ε),δ) is true, the statement P(ε,δ) is true, if and only if the assertion, " f(ε) directly implies ε>0"

    So, f(ε) can be ε^3 or ε^5 which we assume to be greater than 0. This directly implies ε>0.
    But the assertion that f( ε)= ε^2 or ε^4 does not imply ε>0. because ε>0 or ε<0. Is this right?

    Now, I have a doubt in statement P(ε,δ). I assume that it is two variable function of ε and δ and it does not signify an ordered pair(according to definition of function)
     
  11. Dec 31, 2015 #10

    Svein

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    Well, it does not have to be a function of ε. All we require is a procedure to find a δ for whatever ε we specify. As such, starting with a requirement that something is "less than ε/8192" is just a way of saying "find a δ such that <something> is less than ε1 (where ε1≤ε/8192)".
    Sorry, this is meaningless. We specify ε and try to find a δ.
     
  12. Dec 31, 2015 #11
    But ε^3>0, implies ε*ε*ε>0
    ε>0.Can't we say that way?
     
  13. Dec 31, 2015 #12

    Fredrik

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    We haven't been talking about a function f yet. But if we're talking specifically about the definition of what it means to say that ##\lim_{x\to a}f(x)=L## when ##a## is a limit point of the domain of f, then P(ε,δ) is the implication
    $$0<|x-a|<\delta~\Rightarrow~|f(x)-L|<\varepsilon$$ and P(ε/2,δ) is the implication
    $$0<|x-a|<\delta~\Rightarrow~|f(x)-L|<\frac \varepsilon 2$$

    f(ε) would be a number. A number can't imply an inequality, or any other kind of statement. Only a statement can imply a statement.

    I used that if P(ε,δ) for all positive real numbers ε, and x is a positive real number, then P(x,δ). If this isn't obvious, you need to spend some time thinking about the role of the "for all" in that statement. It may help to consider a less abstract example: The statement "for all real numbers x, we have x2≥0" implies the statement "32≥0" (because 3 is a real number).

    I don't follow you here, but the choice of the function f doesn't imply anything about the number ε.

    A fuction "of two variables" always takes an ordered pair as input. f(2,3) is a simplified notation for f((2,3)). The biggest difference between my P and this f is that while this f takes an ordered pair of real numbers to a real number, my P takes an ordered pair of variables (i.e. symbols that represent real numbers) to a statement. You can think of P as the incomplete statement
    $$0<|x-a|< (\text{insert second variable here})~\Rightarrow~|f(x)-L|<(\text{insert first variable here})$$
    I thought the P notation would make things easier, because it would allow us to focus on the important part of the definition (i.e. the "for all"), but if it's causing you any confusion at all, it's probably best if you just replace every occurence of P(ε,δ) and P(ε/2,δ) in my two proofs with the two implications at the start of this post.
     
    Last edited: Dec 31, 2015
  14. Dec 31, 2015 #13
    Sorry,....I shouldn't have written as invoked function f....I will put it more clearly.
    1. w is a gunction of ε....So we can write the function as w(ε).....I mean w(ε) is a real number which corresponds to an ε...Since w is a function, for a single value of ε, we only have a single value of w(ε).....
    For example, w(ε) can be either ε/2, 2ε, ε^2,ε^5 and so on...
    2.....let w(ε) be postive number for a range of value of ε....Then P(w(ε),δ) is true means that, we can find a δ for a given positive w(ε).
    But we should be able to find a δ for a given ε... This can only happen if and only if we can prove that if w(ε) is postive, then ε is positive( ε is a positive number just like w(ε))...
    .Example, if w(ε)=ε/2, then the statement w(ε) is postive, simply implies ε/2 is positive..... So, naturally ε is positive...
    For a positive ε, we are able to find a δ....So, the statement P(ε,δ) is true provided P(ε/2,δ)...
     
  15. Dec 31, 2015 #14

    Fredrik

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    There's no "there exists" that targets δ in the statement P(w(ε),δ). The notation P(x,y) is used for statements such that every "for all" and every "there exists" inside it targets some variable other than x or y.

    This is an odd thing to say. The next thing after a "for all" (or equivalently, "for a given") should be a variable. I guess you must have meant "for a given positive ε and a given function w such that w(ε) is positive".

    If I understand you correctly, you have understood the general idea. But I have to recommend that you practice writing proofs. This will help you communicate your thoughts more clearly.
     
    Last edited: Jan 1, 2016
  16. Jan 1, 2016 #15
    Yes, didn't think about for "all ε" and "some δ" needed for the statement P (ε,δ)....
     
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