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I have that convergence requires that the following be true

$$ \forall\epsilon>0\exists N: n\ge N:|a_n-a|<\epsilon $$

for some sequence $\{a_n\}_{n\in\mathbb{N}}$ that converges to $a$ for $n\to\infty$.

I want to use the reverse of this requirement. I do not know how to express it but I am guessing

$$ \exists\epsilon>0\forall N:n\ge N:|a_n-a|\ge\epsilon $$

it is expressed like the above. Can you help me out?

bubububub
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  • Seems fine to me if what you want to express is ${a_n}$ does not converge to $a$. You might want to add some quantifiers for $n\geq N$ though – Bill O'Haran May 06 '18 at 16:03

1 Answers1

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A somewhat cleaner definition of the limit of a sequence is: $$ \forall \epsilon>0, \exists N\in \mathbb{N}, \forall n \geq N, |a_n-a|<\epsilon $$ Then the opposite is: $$ \exists \epsilon>0, \forall N\in \mathbb{N}, \exists n \geq N, |a_n-a|\geq\epsilon $$

Bill O'Haran
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