How to prove this statement?
Suppose $E\subset \mathbb{R}$. If $\sup(E)=\infty$, then for all $n\in \mathbb{N}$, $n< \sup(E)$. Moreover, we can construct a sequence $(x_n)\subset E$ such that $x_n>n$, $\forall n \in \mathbb{N}$.
From that statement, I know that $E$ is unbounded above, i.e. $\forall M\in \mathbb{R}, \exists x \in E$, such that $x>M$.
Then, since $E$ is unbounded above on $\mathbb{R}$, then $(x_n)$ is unbounded above, i.e. $\forall M\in \mathbb{R}, |x_n|>M, \forall n\in \mathbb{N}$.
I tried to prove that statement by using definition of supremum below: $\sup(E)=s \Leftrightarrow \forall \epsilon>0, \exists x\in E$ such that $s-\epsilon<x$.
And I think to use Archimedian property, but I am still get stuck to prove that.
Any help will be apreciated.
E has no finite upper bound $\implies$ is unbounded above, i.e., $\forall n\in\mathbb{N},\exists x_n\in E$, such that $x_n>n$. Let $S=\{x_n\}$ be set of these $x_n$ which is a subset of these E. $\implies x_n<\text{Sup(E)}\implies n<\text{Sup(E) } \forall n$.