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I need help negating the following statement please, or if anyone could help putting it into words:

$$\forall \epsilon > 0 (\exists d>0(\forall x_0 (\forall x(|x - x_0| < d \implies |f(x) - f(x_0)| < \epsilon))))$$

Mankind
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Emilio B.
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  • I've edited your question to include MathJax - please verify that it's correct. Also note that this is one way to define uniform continuity. –  Sep 12 '13 at 23:50
  • Do you know how to negate this one? $\forall_\epsilon\left(\exists_\delta \left( \vert x-x_0\vert <\delta \implies \vert f(x)-f(x_0)\vert\right)\right)$ – gota Sep 12 '13 at 23:52
  • I think it would be: there exists a for epsilon such that(for all values of d(negation of all that's inside)) – Emilio B. Sep 12 '13 at 23:54

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The negation in words is: there exists a positive $\epsilon$ such that for all positive $\delta$ there exist $x_0$ and $x$ such that the distance between $x$ and $x_0$ is less than $\delta$, but the distance between $f(x)$ and $f(x_0)$ is at least $\epsilon$.

Symbolically, just push negation inside quantifiers by rewriting $\lnot\exists$ as $\forall\lnot$ and $\lnot\forall$ as $\exists\lnot$ and then use de Morgan's laws for the propositional connectives to rewrite, e.g., $\lnot(A \implies B)$ as $A \land \lnot B$.

Rob Arthan
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