I am having trouble understanding this question. What does it mean to negate the limit. Does it mean that n does not have a limit? Also I have done the symbolic negation, is this correct?
$\begin{equation}\exists \epsilon > 0, \forall N \in \mathbb Z, \exists n\in \mathbb Z, n > N \wedge \neg L – \epsilon < a_n < L + \epsilon.\end{equation}$.
Here is the question:
Recall from Calculus the definition of the limit of a sequence $a_n$. We say that the limit of the sequence $a_n$ as $n$ goes to infinity equals $L$ and write:
$lim_{n\rightarrow \infty} a_n = L$
We can write this using quantifiers as follows:
\begin{equation}\label{limit}\forall \epsilon > 0, \exists N \in \mathbb Z, \forall n\in \mathbb Z, n > N \rightarrow L – \epsilon < a_n < L + \epsilon.\end{equation}
Explain in words what the negation of this definition means. Now write the negation of the limit from the given formula.
Best Answer
The real sequence $(a_n)_{n\in\mathbb{N}}$ converges (to the limit $L$) if and only if the following is true: $$\exists L\in\mathbb{R}\,\forall \epsilon>0\,\exists N\in\mathbb{N}\,\forall n\geq N\Rightarrow |a_n-L|<\epsilon$$
The negation of $L$ is the limit of the real sequence $(a_n)_{n\in\mathbb{N}}$ is $$\forall L\in\mathbb{R}\,\exists \epsilon>0\,\forall N\in\mathbb{N}\,\exists n\geq N \Rightarrow |a_n-L|\geq \epsilon $$
What I have done so far? I turned $\exists$ into $\forall$ and backwards. Note that $<$ is replaced by $\geq$.
If you treat the real numbers as a metric space, then use $d(a_n,L)=|a_n-L|$ as metric.