Problem on $\sigma$-algebra from Rudin Does there exist an infinite $\sigma$-algebra which has only countably many members ?
Proof: Suppose that $\sigma$-algebra $\mathfrak{M}$ has countably many members, namely $\{A_i\}_{i=1}^{\infty}$. By definition of $\sigma$-algebra the set $F=\bigcap \limits_{i=1}^{\infty}A_i^c$ lies in $\mathfrak{M}$. Since it lies in $\mathfrak{M}$ then $F=A_j$ for some $j\in \mathbb{N}$, i.e. $\bigcap \limits_{i=1}^{\infty}A_i^c=A_j$ which is equivalent to $$A_j^c\cap \bigcap\limits_{i\geqslant 1 \atop{i\neq j}}A_i=A_j$$ but the last equality is false since $x\in A_j$  then $x\notin A_j^c$ then $x$ does not lie in LHS.
But one moment seems to me confusing it's when $A_j$ is empty.
Sorry if this  topic  appeared in MSE before.
 A: Suppose the sigma algebra is countably infinite. Then the set $X$ on which the sigma algebra acts on must be infinite. (Why?) Let $A$ be an infinite measurable set. Suppose there does not exist a proper infinite subset of $A$. This is a contradiction to the assumption that the sigma algebra is countable infinite. (Why? Consider intersections and complements of measurable sets). Since $X$ is infinite, we can make a strictly decreasing infinite sequence of measurable sets. Hence, we can partition $X$ into disjoint sequence of nonempty measurable sets. However, their unions must be in the sigma algebra too. Hence $|\mathfrak{M}| \geq 2^{\aleph_0}$.
When constructing a partition, instead of considering the intersection of a strictly decreasing sequence $\{A_n\}$, take $B_n=A_{n-1} \setminus A_{n}$ and consider the union of $B_n$'s.
A: Suppose $\mathcal{B}$ is an infinite algebra of sets (we don't need $\sigma$-algebra just yet). We will show that there are infinitely many pairwise disjoint non-empty elements of $\mathcal{B}$.
There are in fact two cases to consider, related to possible atoms:
A definition: a subset $A \in \mathcal{B}$ is called an atom if it is non-empty and there is no proper subset $B$ under $A$ in the algebra, i.e. for no $B \in \mathcal{B}$ do we have $\emptyset \subsetneq B \subsetneq A$.
Basic facts: two distinct atoms $A_1$ and $A_2$ are disjoint. Also, for any set in $C \in \mathcal{B}$, and any atom $A$, $C \cap A = \emptyset$ or $C \cap A = A$ (or it would be a proper set in-between).
The first case is where $\mathcal{B}$ is atomic, which means that every non-empty member of $\mathcal{B}$ has a subset that is an atom. 
In that case, there cannot be finitely many atoms. Because if there are only atoms $A_1,\ldots,A_n$, then the map that sends $A \in \mathcal{B}$ to $(A \cap A_1,\ldots, A \cap A_n)$ has $2^n$ many values (as the intersections are empty or $A_i$ for every $i$) and is a bijection (here we use that $\mathcal{B}$ is atomic), showing that $\mathcal{B}$ is finite, contradicting the assumption. So $\mathcal{B}$ has infintiely many atoms if it is atomic, and so an infinite pairwise dijsoint family (of atoms).  
In the other case, $\mathcal{B}$ is not atomic, so there is some $Y \neq \emptyset$ in $\mathcal{B}$ such that $Y$ contains no atoms. So construct $Y_1,Y_2,\ldots$, strictly decreasing, by recursion: $Y_1 = Y$ and $Y_n$ constructed, let $Y_{n+1}$ be a proper non-empty subset of $Y_{n}$ (as $Y_{n}$ is not an atom). Then $Y_{n} \setminus Y_{n+1}$, $n=1,2,\ldots$ are a family of pairwise disjoint members of $\mathcal{B}$.
Having such a countable pairwise disjoint family $A_n, n \in \mathbb{N}$ of non-empty subsets, for every subset $I$ of $\mathbb{N}$ we define $A(I) = \bigcup_{n \in I} A_n$ is in $\mathcal{B}$ (here we use that we have a $\sigma$-algebra!) and for different $I$, these unions are all different (by the disjointness condition). 
So $|\mathcal{B}| \ge 2^{\aleph_0}$. So any infinite $\sigma$-algebra is uncountable.
