Let $X$ be any Banach space and $M\subset X$ be bounded. We know the $\sup(M)\in\overline{M}$ in general. Since $M$ is bounded $\sup_{u\in M}\|u\|<\infty$.
Question: Can we somehow write that $$\sup_{u\in M}\|u\|=\sup_{u\in\overline{M}}\|u\|,$$ where $\overline{M}$ denotes the closure of the set $M$.
Yes, as by definition, every member of $\bar{M}$ is a limit to a sequence of elements of $M$, you can show that $\bar{M}$ is also bounded and that its maximum can not be greater that $M$'s one, otherwise, a sequence of elements of $M$ would converge to something greater that its supremum.
Note that in the case of $\bar{M}$, as we reach it, it is also a maximum.