# Properties of Supremum

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 $s=M$.

Remark:

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

• 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? – user13123 Dec 11 '17 at 2:11
• 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? – Siong Thye Goh Dec 11 '17 at 2:16
• Ah.... Upvoted! – user13123 Dec 11 '17 at 2:26