1

I have a problem. I have to prove that

$(\forall\,\epsilon > 0)(\exists\,\delta > 0)[0 < |x − a| < \delta \implies |f(x) − L| < \epsilon$ (formal definition of a limit)

$(\exists\,\delta > 0) (\forall\,\epsilon > 0)[0 < |x − a| < \delta \implies |f(x) − L| < \epsilon$

are the same. Basically the only difference is the order of the quantifiers.

My first instinct was to use DeMorgans Law and negate the quantifier but that is not quite the same proposition. I am not sure how to prove this equivalence. Can anybody help?

Tim Raczkowski
  • 20,046
user208628
  • 39
  • If the quantifier orders are switched and the rest unchanged, the meaning of the statement is different. – coffeemath Mar 03 '15 at 00:45
  • 1
    I agree with @coffeemath. Good luck proving those statements are equivalent. The mathematical world will tear itself apart if you do. –  Mar 03 '15 at 00:53
  • Then how am I supposed to prove that these two are the same??? – user208628 Mar 03 '15 at 00:55
  • 1
    Where is this problem coming from? – Tim Raczkowski Mar 03 '15 at 00:57
  • 1
    https://imgur.com/t9lRQvW – user208628 Mar 03 '15 at 01:01
  • i must have just misinterpreted it :( – user208628 Mar 03 '15 at 01:01
  • @user208628 Yes, the exercise is to prove that "$(\forall \epsilon > 0)(\exists \delta > 0)$..." is equivalent to "every $\epsilon$-neighborhood of $L$..." – BaronVT Mar 03 '15 at 01:02
  • but i dont really understand that either. how am i supposed to prove that? – user208628 Mar 03 '15 at 01:05
  • 1
    @user208628 Looking at that link, it looks like the authors give you the true quantified statement of a limit at the top of the page, and then they give you a different quantified statement and ask you to consider what it means. They make no claim the two statements are generally equivalent. Maybe that will help clear things up. –  Mar 03 '15 at 01:06
  • Having looked, I agree with therapist here. The second statement, with the quantifiers switched, was just put in there for the reader to notice how it was different from the standard one, so as to emphasize that the order of quantifiers was essential in defining what a limit is. – coffeemath Mar 03 '15 at 01:10
  • I still don't get how to prove the neighborhood statement is equivalent. If you have time, could you help me? Or should I make a new question? – user208628 Mar 03 '15 at 01:19
  • Does this answer your question? What's wrong with this "backwards" definition of limit? –  Oct 09 '23 at 01:03

2 Answers2

4

$\forall \epsilon \;\;\exists \delta$ means for any given $\epsilon$ we have $\delta$ which depends on $\epsilon$.

$\exists \delta\;\;\forall \epsilon $ means there is a $\delta$ for all $\epsilon$. In this case $\delta$ is independent of $\epsilon$

So your statements are NOT same.

Extremal
  • 5,785
  • $\exists \delta$ doesn't really say anything about a unique $\delta$ though; otherwise, we'd have $\exists !\delta$. –  Mar 03 '15 at 00:54
  • Ya the word 'unique' is not a correct word here. I will edit. – Extremal Mar 03 '15 at 00:58
  • Well, saying there is "one" $\delta$ still doesn't quite sound right (in my opinion; it sounds restrictive); I think it'd be best to simply interpret it literally: $\exists\delta;\forall\epsilon$ means "there exists $\delta$ for all $\epsilon$." –  Mar 03 '15 at 01:01
0

Another consequence of $(\exists\delta>0)(\forall\epsilon>0)(|x-a|<\delta\implies |f(x)-L|<\epsilon)$. Since $|f(x)-L|<\epsilon$ for all $\epsilon>0$, we must have $f(x)=L$ whenever $|x-a|<\delta$.

Tim Raczkowski
  • 20,046