# Prove that the set $\mathbb{Z}$ is unbounded above in $\mathbb{R}$?

My topology lesson casually made this assumption but I think it needs proof.

There is another thread with basically the same question here on this stackexchange but I'm not satisfied with the accepted answer. It makes use of the $⌊x⌋$ function but doesn't prove that it's defined for any $x \in \mathbb{R}$.

• How do you define "unbounded" and "bounded"? – Jack Mar 14 '17 at 0:27
• Floor is usually defined using supremum of a set bounded above. Pretty well defined. – user251257 Mar 14 '17 at 0:28
• Answers to this question would depend on what OP knows about real numbers. – Jack Mar 14 '17 at 0:33
• @Jack What I mean with $\Bbb{Z}$ being unbounded above in $\Bbb{R}$ is that there exists no $x ∈ \Bbb{R}$ such that there exists no $n ∈ \Bbb{Z} \ge x$. – NounVerber Mar 14 '17 at 0:33

Suppose $s=\sup\mathbb{Z}$. By definition of supremum, there exists $x\in\mathbb{Z}$ such that $x\ge s-\frac{1}{2}$.
Hint: $x+1\in\mathbb{Z}$.