Prove that $(a_n)$ is divergent In the euclidean metric space $\mathbb{R}$, the sequence $(a_n)$ where $a_n:=n$ is divergent.
Suppose $(a_n)$ is convergent. Then $\exists l \in \mathbb{R}$ for which $(a_n) \to l$. This means that given $\varepsilon >0$, we can find a natural number $\mathbb{N}$ such that $|a_n-l|<\varepsilon$ for every $n \ge N$.
Let $a_n=n$. Then $|n-l|<\varepsilon$.
Because $\varepsilon \in \mathbb{R}$ such that $\varepsilon>0$, then $|n-l|<\varepsilon \iff -\varepsilon<n-l<\varepsilon \iff l-\varepsilon<n<l+\varepsilon$.
However, this is impossible since $\forall N \in \mathbb{N}, \exists n \in \mathbb{N}$ for which $n>N$ and $a_n \ne a_N$. Thus $(a_n)$ is not eventually constant, and hence eventually $n>l+e$.
Would it be correct to say here instead that $\forall (l+\varepsilon) \in \mathbb{R}$, $\exists n \in \mathbb{N}$ such that $n>(l+\varepsilon)$; and therefore, $(a_n)$ must be divergent?


In the euclidean metric space $\,\mathbb{R}\,,\,$ the sequence $\,\big(a_n\big)_{\!n\in\mathbb N}\,$ where $\;a_n\!:=n\;$ is divergent.

Here is a direct proof :
For any $\,k>0\,,\,$ there exists $\;n_k=\lfloor k\rfloor+1\in\mathbb N\;$ such that
for all $\;n\in\mathbb N\;\land\;n>n_k\;$ it results that
$a_n=n>n_k=\lfloor k\rfloor+1>k\;.$
Consequently ,
for any $\,k>0\,,\,$ there exists $\;n_k\in\mathbb N\;$ such that , $\,$for all $\;n\in\mathbb N\;\land\;n>n_k\;,\;$ it results that $\;a_n>k\;.$
And, by definition, it means that the sequence $\,\big(a_n\big)_{\!n\in\mathbb N}\,$ is divergent .
