What is the cardinality of the set of all $\sigma$-algebra containing $\mathcal{B}$

Question

Let $X$ be a set, and $\mathcal{B}$ be a subset in $\mathcal{P}(X)$ the power set of $X$

Let $\{\mathcal{F}_i\}$ be the set of all $\sigma$-algebra containing $\mathcal{B}$ such that $\mathcal{B} \subseteq \mathcal{F}_i, \forall i$

What would be the cardinality of $\{\mathcal{F}_i\}$? Finite, countably infinite or uncountable?

We know it is at least finite because the intersection of all $\{\mathcal{F}_i\}$ is a $\sigma$-algebra containing $\mathcal{B}$...

Answer 1

It depends on what $\mathcal{B}$ and $X$ are.

• If $\mathcal{B}=\mathcal{P}(X)$, then clearly the only $\sigma$-algebra of subsets of $X$ containing $\mathcal{B}$ is just $\mathcal{B}$. So in this case, the answer is "$1$."

• If $X$ is finite, then $\mathcal{P}(\mathcal{P}(X))$ is also finite, so there are only finitely many $\sigma$-algebras at all; then the answer is "finitely many," regardless of what $\mathcal{B}$ is. (Computing the specific possible values looks a bit messy.)

• If $X$ is uncountable and $\mathcal{B}$ has cardinality $<\vert\mathcal{P}(X)\vert$ - that is, $\mathcal{B}$ is "small" (note that this is the standard case of $X=\mathbb{R}$, $\mathcal{B}=\{$opens$\}$) - then we get lots of $\sigma$-algebras. Specifically, the $\sigma$-algebra $[\mathcal{B}]$ generated by $\mathcal{B}$ has size $\omega_1\times \vert\mathcal{B}\vert$, which must be less than $\vert\mathcal{P}(X)\vert$. It's now not hard to show that there are uncountably many $\sigma$-algebras containing $\mathcal{B}$.

It is not obvious to me whether there are some $X$ and $\mathcal{B}$ such that the answer is "countably infinitely many."