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How do you write this in logical symbols: For each $\epsilon>0$ there exist a number $N$ such that $|f_n(x)-f(x)|<\epsilon$ for all $x\in S$ and all $n>N$.

Is this correct? $\forall \epsilon>0 \exists N,(x\in S\land n>N\implies |f_n(x)-f(x)|<\epsilon)$

And how can I negate this?

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Yes, this is correct. –  1015 Feb 21 '13 at 14:59
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1 Answer 1

up vote 4 down vote accepted

It’s

$$\forall\epsilon>0\,\exists N\in\Bbb N\,\forall x\in S\,\forall n>N\big(|f_n(x)-f(x)|<\epsilon\big)\;;$$

its negation, after the negation is pulled inside all the quantifiers, is

$$\exists\epsilon>0\,\forall N\in\Bbb N\,\exists x\in S\,\exists n>N\big(|f_n(x)-f(x)|\ge\epsilon\big)\;.$$

This negation can be performed mechanically, using the equivalence of $\neg\forall x\varphi(x)$ with $\exists x\neg\varphi(x)$ and the equivalence of $\neg\exists x\varphi(x)$ with $\forall x\neg\varphi(x)$.

Informally, the negation says that there is a ‘bad’ $\epsilon$, meaning one such that no matter how far out in the sequence $\langle f_n:n\in\Bbb N\rangle$ you go, you can find an $n$ at least that far out and a point $x\in S$ such that $f_n(x)$ and $f(x)$ differ by at least $\epsilon$.

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Is $\forall\epsilon>0\,\exists N\in\Bbb N\,\forall x\in S\,\forall n>N\big(|f_n(x)-f(x)|<\epsilon\big)\;$ equivalent with $\forall \epsilon>0 \exists N,(x\in S\land n>N\implies |f_n(x)-f(x)|<\epsilon)$ ? –  Kasper Feb 21 '13 at 15:10
    
@Kasper: My version contains a little more information, in that it explicitly specifies that $N\in\Bbb N$, but otherwise theyre equivalent. I would omit the comma from yours, however: it serves only to clutter up the expression, since it adds nothing to the information conveyed by the parentheses. –  Brian M. Scott Feb 21 '13 at 15:12
    
Okay, thanks for the explanation ! –  Kasper Feb 21 '13 at 15:31
    
@Kasper: You’re welcome! –  Brian M. Scott Feb 21 '13 at 15:32
    
Can't I pull out "$\forall N\in\Bbb N,\exists n>N$" from the negation and replace it (in the position where $\forall N\in\Bbb N$ was) with "$\forall n\in\Bbb N$"? They seem equivalent to me. –  Ryan Sep 28 '13 at 13:56
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