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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 $x \in \mathbb{R}$.

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  • $\begingroup$ How do you define "unbounded" and "bounded"? $\endgroup$ – Jack Mar 14 '17 at 0:27
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    $\begingroup$ Floor is usually defined using supremum of a set bounded above. Pretty well defined. $\endgroup$ – user251257 Mar 14 '17 at 0:28
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    $\begingroup$ Answers to this question would depend on what OP knows about real numbers. $\endgroup$ – Jack Mar 14 '17 at 0:33
  • $\begingroup$ @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 $n ∈ \Bbb{Z} \ge x$. $\endgroup$ – NounVerber Mar 14 '17 at 0:33
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The real numbers have the property that a subset which is upper bounded has a supremum.

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

Can you arrive to a contradiction?

Hint: $x+1\in\mathbb{Z}$.

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