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Given that $s$ is upper bound for set $A$ (of reals). And given $\forall \epsilon >0, \exists a \in A\ s.t.\ s - \epsilon < a$

Now I assume that $b$ be any other upper bound of $A$ and I need to prove that $s \leq b$.

Assume on the contrary that $s > b$. Lets take $\epsilon = \frac{b - s}{2}$. So, $\exists a \in A\ s.t.\ s - \epsilon < a .$ Simplifying we get $ \frac{s+b}{2} < a $ Now it suffices to prove that $ b < \frac{s+b}{2} $. Now we have $s > b$ and so $\frac{s}{2}+\frac{b}{2} > \frac{b}{2}+\frac{b}{2}$ and so $\frac{s+b}{2} > b $. so we have $ b < \frac{s+b}{2} < a $. This contradicts that $b$ is upper bound for $A$. Thus assumption $s >b$ is false.

Please check if its correct. Thanks a lot.

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    $\begingroup$ There is a typo. You want to take $\epsilon = \frac{s - b}{2}$ not $\frac{b - s}{2}$. Other than that you are fine. $\endgroup$
    – balddraz
    Aug 9, 2019 at 4:17
  • $\begingroup$ If $s>b$ then you choice of epsilon is negative! $\endgroup$
    – Sorfosh
    Aug 9, 2019 at 4:18
  • $\begingroup$ oh yes in my notebook I have taken mod.i forgot to mention it here. Thanks $\endgroup$
    – ReadThyOwnBook
    Aug 9, 2019 at 4:19
  • $\begingroup$ Note that based on your condition of "$\forall \epsilon >0, \exists a \in A\ s.t.\ s - \epsilon < a$" that the $s$ is actually the least upper bound, i.e., supremum, which you've basically proven (apart from the typo mentioned in the $2$ comments above). $\endgroup$
    – John Omielan
    Aug 9, 2019 at 4:20

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