Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Yes, this is correct. – 1015 Feb 21 '13 at 14:59
up vote 4 down vote accepted


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

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 $\epsilon$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.