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Are the following statements equivalent?

Statement 1

For any $\epsilon>0$, there exists $\delta>0$ such that for all real number $x$, if $|x-x_0|<\delta$, then $|f(x)-f(x_0)|<\epsilon$

Statement 2

There exists $\delta>0$ such that for any $\epsilon>0$, for all real number $x$, if $|x-x_0|<\delta$, then $|f(x)-f(x₀)|<\epsilon$

If so, can the second statement also be used as a definition for limits?

Anne Bauval
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    The second statement is equivalent to "There exists $\delta > 0$ such that $f(x)=f(x_0)$ for any $x \in (x_0-\delta,x_0+\delta)$" (why?), so the answer is NO, those statements are not equivalent. In general swapping the order of quantifiers "for any" and "there exists" does not give equivalent statemets. – Presage Sep 22 '23 at 20:54
  • No. In the standard definition, $\delta$ may depend on $\epsilon$ (you give me an $\epsilon$, and I can find a $\delta$ that works for that $\epsilon$). Your definition would require the same $\delta$ to work for every $\epsilon$, which will not generally be possible. – paw88789 Sep 22 '23 at 20:55
  • Similarly, you cannot restate it as "for any $\delta>0$ there exists $\epsilon>0$ such that for all $x$, $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\epsilon$". It is maybe a bit harder to see why this statement doesn't work compared to the one Dominik discusses – FShrike Sep 22 '23 at 21:02

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