Suppose b = Sup S and b is not in S. prove b is an accumulation point.

if b is the supremum of S but not in S then b is in R/S.

Since b is the supremum then for all s $\displaystyle \in$ S, s < b.

But, because b is not in S we can find a neighborhood across s(max) = c, such that s < c < b, where c $\displaystyle \in$ S, but still less than b.

is this ok?