Let $\sigma(\cdot)$ denote the $\sigma$-algebra of its input. So $\sigma(\mathbb{R})$ is the $\sigma$-algebra of real numbers.
Question: I am wondering whether $\sigma(\sigma(\mathbb{R}))$ is equal to $\sigma(\mathbb{R})$, or it is different. (If the analysis is simpler, I am only wondering Borel $\sigma$-algebras.)
More explanation: The reason for my confusion is the following: $\sigma(\mathbb{R})$ is already closed under complements, countable unions, and countable intersections. So $\sigma(\sigma(\mathbb{R}))$ should be equal to $\sigma(\mathbb{R})$. However, the reason I do not fully agree with this statement is that $\sigma(\mathbb{R})$ is a set of intervals, hence $\sigma(\sigma(\mathbb{R}))$ should be a set of sets. I think $\sigma(\sigma(\mathbb{R}))$ is the powerset of $\sigma(\mathbb{R})$, but I have never heard of a powerset of an uncountable set.
Edit: Thanks to the amazing comments, this question is now resolved.
The power set of $X$ is a $\sigma$-algebra of $X$. After all, it's closed under all the required operations. And, you can have power sets of uncountable sets just fine.
This is true for both $\mathbb{R}$ and $\sigma(\mathbb{R})$.
But, it's probably not the only one on $\sigma(\mathbb{R})$.
Your intuition that it can't be the same thing as $\sigma(\mathbb{R})$ because it should be composed of sets of sets is correct.
I don't know what a smaller algebra over $\sigma(\mathbb{R})$ would be.