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This question is for me to better understand the beginning of a real analysis course.

We are provided with two definitions of supremum as follows:
Def 1 : Let $S$ be a set in $\mathbb{R}$ be bounded above, then $m$ is called the least upper bound (supremum) if $m \ge s\space, \forall s\in S $ and if $m'$ is some other upper bound, then $m < m'$
Def 2: Let $S$ be a set in $\mathbb{R}$ be bounded above, then $m$ is a supremum if for some arbitrary $\epsilon>0$ $\exists s \in S, m-\epsilon < s$
I understand how both statements are true, however, would it be possible to prove the Def 2 based on Def 1?

Any hint is appreciated!

qwr
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rannoudanames
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  • $ \epsilon := m' - m $ – ninjaaa Jan 22 '17 at 19:41
  • how would that work? When we take the $\epsilon$ we intend that it can be infinitly small, implying that $m- \epsilon$ leaves no space for any m' between m and all of s in S. How would your argument using $\epsilon := m'-m$ go? Would we assume $m'$ some arbitrary upper bound making $\epsilon$ also arbitrary? – rannoudanames Jan 22 '17 at 19:46
  • I'm sorry now I can see it isn't true - consider the set $ S := { 1, 2 } $ and $\epsilon = 1/2$ . – ninjaaa Jan 22 '17 at 19:59
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    @rannoudanames "When we take the ϵϵ we intend that it can be infinitly small," No no no. No such thing as "infinitely small" in the real numbers. Arbitrarily small is how we express it. Huge difference. – user4894 Jan 22 '17 at 20:02
  • @user4894 mind elaborating on that point if you dont mind? – rannoudanames Jan 22 '17 at 20:06
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    There are no infinitely small real numbers. The idea of replacing the vague idea of "infinitely small" with the precise idea of arbitrarily small is the key breakthrough in the modern formalization of the real numbers. Note that $\epsilon$ is always a positive real number. – user4894 Jan 22 '17 at 21:13
  • Thank you! here is some more: http://math.stackexchange.com/questions/2109339/infinitely-small-vs-arbitrarily-small – rannoudanames Jan 22 '17 at 21:14
  • Your def 2 was wrong, there only exists some $s \in S$ with $s > m - \epsilon$, not for all $s \in S$ – qwr Feb 15 '19 at 17:44
  • It is clearer to say "for all $\epsilon>0$" instead of "for some arbitrary $\epsilon>0$". – DanielWainfleet Feb 15 '19 at 20:53

1 Answers1

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Suppose $m$ is a supremum by definition 1. Suppose $\exists$ $\epsilon>0$ such that $m-\epsilon\geq s$ $\forall s\in S$, then $m'=m-\epsilon$ is another upper bound so it must be $m<m'=m-\epsilon$ which is impossible. So we must have $\forall \epsilon>0$, $m-\epsilon <s$ $\exists s\in S$.

This proves definition 2 in terms of definition 1.

mathma
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  • I see you are doing a proof by contradiction, and it makes sense to me. m cannot be inferior to some number smaller than m. – rannoudanames Jan 22 '17 at 20:07
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    Also let me add that definition 2 implies definition 1. If $m$ is a supremum by definition 2 then $m\geq s$ $\forall s\in S$. Now suppose $m'$ is another upper bound and $m'<m$ then $m-m'>0$. Take $0<\epsilon <m-m'$ so by def 2 $\exists s\in S$ s.t. $m'<m-\epsilon<s$ so $m'$ is not an upper bound, so we get again a contradiction. – mathma Jan 22 '17 at 20:24
  • makes sense! (hypothetical +1) – rannoudanames Jan 22 '17 at 20:29