Suppose I have an infinite $\sigma$-algebra $\mathcal{A}$ generated by a fixed set of generators whose cardinality is stricly larger than $\aleph_0$. In other words, the set of generators of $\mathcal{A}$ is uncountable. Is $\mathcal{A}$ an atomless $\sigma$-algebra?
I always thought that it is indeed the case because only countable intersections are allowed in $\mathcal{A}$. Therefore, since I have a supply of uncountably many generators, there exist no minimal non-empty elements of $\mathcal{A}$. Is that correct? (I suddenly have a doubt.)
Also, the cardinality of $\mathcal{A}$ is at least $\mathfrak{c}=2^{\aleph_0}$. I am under the impression that the cardinality of $\mathcal{A}$ is $\mathfrak{c}$ if and only if $\mathcal{A}$ is atomic and strictly larger than $\mathfrak{c}$ if and only if $\mathcal{A}$ is atomless. Is that correct?
No. Take any uncountable set $X$ and look at the $\sigma$-algebra on $X$ generated by the singletons, that is $\{\{x\}\mid x\in X\}$ which has the same cardinality as $X$, which by assumption is uncountable.
Easily, this is exactly $\{A\subseteq X:\min\{|A|,|X\setminus A|\}\leq\aleph_0\}$. I leave you to convince yourself why this is an atomic $\sigma$-algebra.