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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!

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    $ \epsilon := m' - m $ – 2017-01-22
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    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? – 2017-01-22
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    I'm sorry now I can see it isn't true - consider the set $ S := \{ 1, 2 \} $ and $\epsilon = 1/2$ . – 2017-01-22
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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. – 2017-01-22
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    @user4894 mind elaborating on that point if you dont mind? – 2017-01-22
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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. – 2017-01-22
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    Thank you! here is some more: http://math.stackexchange.com/questions/2109339/infinitely-small-vs-arbitrarily-small – 2017-01-22

1 Answers 1

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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 $m0$, $m-\epsilon

This proves definition 2 in terms of definition 1.

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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. – 2017-01-22
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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 $m'0$. Take $0<\epsilon $\exists s\in S$ s.t. $m' – 2017-01-22
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    makes sense! (hypothetical +1) – 2017-01-22