Theorem $5$. If $M$ is $\sup S$ and $\epsilon > 0$ , then there is at least one number $s$ in $S$ such that $M −\epsilon < s \le M$

I know theorems are always true... but if $\sup S$ is a "unique" number which is the largest number in the upperbound of $S$, how can $s \leq M$? Shouldn't it be $s < M$


It is possible for the supremum to be in the set.

For example let $S$ be $ \{2\}$.

Then the supremum is $2$ and we have to pick $s=M$.


Supremum is the $\color{blue}{\text{least}}$ upper bound.

  • $\begingroup$ Sorry I'm still confused.. how does this answer my question? I'm saying if there is a s such that $s \leq M$, that means there is a chance for s = M, then there would be two Sup S, which does not make sense since Sup 'x' is a unique number? $\endgroup$ – user13123 Dec 11 '17 at 2:11
  • $\begingroup$ If $s=M$, there would be two $\sup S$? which $2$? $M$ and $s$? but they are equal, there is only one such number isn't it? $\endgroup$ – Siong Thye Goh Dec 11 '17 at 2:16
  • $\begingroup$ Ah.... Upvoted! $\endgroup$ – user13123 Dec 11 '17 at 2:26

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