Prove that any upper bounded set have a supremum

Asked Sep 18 '18 at 06:54

Active Sep 18 '18 at 08:46

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Using Dedekind's theorem and the property of real numbers, prove that if a set S is bounded above, then it has a supremum.

I just started real analysis and I thought this is so "trivial". But when I actually approach the problem, it feels difficult. Here is my proof (not sure if it's correct):

Consider a set $T=\{x \mid \exists y\in S\;$ such that $\;x\le y \}$, and $U=\{x \mid x\in \Bbb R\;x\notin T\}$ then using dedekind's theorem, there exists a real number $k$ such that every integer in $T$ is $\le k$ and every integer in $U$ is $\ge k$. Therefore the supremum is $k$.

Is this correct? Thank you in advance.

edited Sep 18 '18 at 08:46

limeeattack

asked Sep 18 '18 at 06:54

abc...

5,010

Your proof is OK, but some typos. Replace integer with "real number". And further you might want to show that $T, U$ have the properties for which Dedekind theorem can be applied. – Paramanand Singh Sep 18 '18 at 08:11
1

And once you are armed with Dedekind theorem or its equivalent most of the elementary real analysis proofs are trivial. – Paramanand Singh Sep 18 '18 at 08:12
@ParamanandSingh thank you :) – abc... Sep 18 '18 at 08:14
You may visit the link in previous comment to see Dedekind theorem equivalents. – Paramanand Singh Sep 18 '18 at 08:15

Prove that any upper bounded set have a supremum

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