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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?

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  • $\begingroup$ If the quantifier orders are switched and the rest unchanged, the meaning of the statement is different. $\endgroup$ – coffeemath Mar 3 '15 at 0:45
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    $\begingroup$ I agree with @coffeemath. Good luck proving those statements are equivalent. The mathematical world will tear itself apart if you do. $\endgroup$ – user220080 Mar 3 '15 at 0:53
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    $\begingroup$ Where is this problem coming from? $\endgroup$ – Tim Raczkowski Mar 3 '15 at 0:57
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    $\begingroup$ imgur.com/t9lRQvW $\endgroup$ – user208628 Mar 3 '15 at 1:01
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    $\begingroup$ @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. $\endgroup$ – user220080 Mar 3 '15 at 1:06
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$\forall \epsilon \;\;\exists \delta$ 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.

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  • $\begingroup$ $\exists \delta$ doesn't really say anything about a unique $\delta$ though; otherwise, we'd have $\exists !\delta$. $\endgroup$ – user220080 Mar 3 '15 at 0:54
  • $\begingroup$ Ya the word 'unique' is not a correct word here. I will edit. $\endgroup$ – Extremal Mar 3 '15 at 0:58
  • $\begingroup$ 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 $\epsilon$." $\endgroup$ – user220080 Mar 3 '15 at 1:01
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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 $|x-a|<\delta$.

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