Is every $\sigma$-algebra generated by a partition?
In the answer, in the first paragraph, it is written that if a finite set is used to generate a $\sigma$-algebra, every point is in a unique atom, so that they can be used as a partition.
Maybe I'm just stupid, but why is this stated in such a nonchalant way? Is this obvious? If so, what am I missing?
I don't know what you're missing, but yes it's more or less obvious. The $\sigma$-algebra generated by a finite set is certainly finite. Hence for every $x$ the intersection of all the measurable sets containing $x$ is measurable.
And there's your partition. (Strictly speaking I don't think it's correct to call those sets "atoms", because an atom has positive measure by definition and here there's no measure in the picture. If we do have a measure on our algebra the sets in the partition that happen to have positive measure are atoms, and if you take all the null sets and tack them on to one of those atoms it remains an atom, so we also have a partition into atoms.)
