Can the set of all sigma-algebras over an arbitrary set be uncountable

Question

In my reading material, in the proof for the claim that

Let $\mathcal{L} \in \mathcal{P}(\Omega)$ be a collection of subsets of $\Omega$. Then the smallest $\sigma$-algebra on $\Omega$ which contains $\mathcal{L}$ exists and is unique

the existence step is performed by first considering the intersection over the elements of the set of all $\sigma$-algebras on $\Omega$, and then using the property of $\sigma$-algebras that the intersection over non-empty collection of $\sigma$-algebras is also a $\sigma$-algebra. However the proof that

the intersection over non-empty collection of $\sigma$-algebras is also a $\sigma$-algebra

is itself proved by using the definition and inducing property of $\sigma$-algebras, that if $\mathcal{F}$ is a $\sigma$-algebra and $E_1,E_2,E_3,\dots \in \mathcal{F}$, then i.) $\bigcup_{n \in \mathbb{N}}E_n \in \mathcal{F}$, ii.) $\bigcap_{n \in \mathbb{N}}E_n \in \mathcal{F}$.

In light of the fact that these two set operations are over a countably infinite index set, I would conclude that either necessary the set of all $\sigma$-algebras is countable, or there is a missing remark in my reading material. However, I can't say for sure whether I am missing some small technical detail. What is your thought on this?

Answer 1

Based on a comment "If it is so that this property can be extended to uncountable intersections..." I finally see the problem.

Fact 1. If $\mathcal F$ is a sigma algebra, $E_1,E_2,\dots\in\mathcal F$, and $E=\bigcap_{j=1}^\infty E_j$ then $E\in\mathcal F$.

Fact 2. If $\Omega$ is a set, $S$ is any nonempty collection of sigma algebras on $\Omega$ and $\mathcal F=\bigcap_{A\in S}A$ then $\mathcal F$ is a sigma-algebra.

No, Fact 1 cannot be extended to uncountable intersections. This is not contradicted by Fact 2, because Fact 2 is not an extension of Fact 1! The two facts are talking about different things.