Given any $r\in\mathbb{R}$, $\exists n\in \mathbb{N}$ such that $r<n$. In this exercise, it is suggested to begin with "Suppose not ... and apply the Least Upper Bound Principle to the set $\mathbb{N}$". But I don't quite understand what needs to be done there.
Here's what I've tried:
"Suppose that, given any $r\in S\subset \mathbb{R}$, $\not\exists n\in T\subset\mathbb{N}$ s.t. $n>r$ (where $S$ and $T$ are bounded subsets). By LUPB, $T$ has a least upper bound, say, $\sup(T) = L$. Then $n\leq L$, $\forall n\in T$. Suppose that $\sup(S)\leq \sup(T)$, then $r \leq n$, a contradiction.
I feel something is not right with my proof. Would appreciate some suggestions.
For a proof by contradiction, you should assume the negation of what you're trying to prove, with the goal of arriving at a contradiction. In this case, the negation is "there exists $r\in\mathbb{R}$ such that $n\leq r$ for all $n\in\mathbb{N}$."
The existence of such an $r$ implies that the set $\mathbb{N}$ is bounded above, hence it has a least upper bound $\alpha$. That is, there exists a real number $\alpha$ such that $n\leq \alpha$ for all $n\in\mathbb{N}$, and such that for any $\beta<\alpha$ there exists a natural number $n$ such that $\beta<n$.
Now how can $\beta$ be chosen to lead to a contradiction?