Why $o(\log n)$ bits instead of $O(\log n)$ bits needed for succinct data structure?

Succinct data structure = succinct representation of data + bits for indexing. Link

Let $$S$$ be subset of of $$\{1,2,\ldots,n\}$$ then $$\lceil \log (2^n) \rceil + o(\log n )$$ bits are needed in succinct representation. The $$o(\log n)$$ bits needed for the indexing of the succinct data structure. I have a doubt, why $$o(\log n)$$ not $$O(\log n)$$ bits? Where $$O(\log n)$$ do not play any role when talking about conventional representation of data.

Note that $$\lceil \log(2^n) \rceil = n$$.
An algorithm that uses $$n$$ bits to represent the set would be fine (it would count as a succinct data structure). An algorithm that uses $$n + 2 \sqrt{n}$$ bits would also be fine.
An algorithm that uses $$2n$$ bits to represent the set would not be fine (it would not count as a succinct data structure). Similarly, an algorithm that uses $$1.01n$$ bits would not be fine.