How to negate $\forall\epsilon > 0,\; \exists n_0=n_0(\epsilon)\in \Bbb{N},\ni |a_n-a^{*}| <\epsilon,\;\;n\geq n_0$

Question

This became somewhat a concern to me today, when asked to negate the following definition of convergence below;

$$\lim_{n\to \infty}a_n=a^{*},\;\text{i.e.},$$

$$\forall\; \epsilon > 0,\; \exists\;n_0=n_0(\epsilon)\in \Bbb{N},\;\ni |a_n-a^{*}| <\epsilon,\;\;n\geq n_0$$

Please, can anyone show me how to negate the above definition?

Answer 1

Negation "percolates" from outside to inside by changing all quantifiers (even the bounded quantifiers used here): $$ \begin{matrix}\neg&\forall \epsilon>0,&&\exists n_0\in\Bbb N,&& \forall n\ge n_0,&\ni&& |a_n-a^*|<\epsilon\\ & \exists \epsilon>0,&\neg&\exists n_0\in\Bbb N,&& \forall n\ge n_0,&\ni&& |a_n-a^*|<\epsilon\\ &\exists \epsilon>0,&&\forall n_0\in\Bbb N,& \neg&\forall n\ge n_0,&\ni&& |a_n-a^*|<\epsilon\\ &\exists \epsilon>0,&&\forall n_0\in\Bbb N, &&\exists n\ge n_0,&\ni& \neg&|a_n-a^*|<\epsilon\\ &\exists \epsilon>0,&&\forall n_0\in\Bbb N, &&\exists n\ge n_0,&\ni&& |a_n-a^*|\ge \epsilon\end{matrix}$$

Answer 2

The negation is

$\exists \varepsilon >0~~\forall n_0\in\mathbb{N} ~~\exists n\ge n_0: |a_n-a^*|\ge\varepsilon$