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 $x \in \mathbb{R}$.

  • $\begingroup$ How do you define "unbounded" and "bounded"? $\endgroup$ – Jack Mar 14 '17 at 0:27
  • 1
    $\begingroup$ Floor is usually defined using supremum of a set bounded above. Pretty well defined. $\endgroup$ – user251257 Mar 14 '17 at 0:28
  • 1
    $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.