TITLE: Uncountable unions of nested sets in a sigma field
QUESTION [6 upvotes]: I know that in a $\sigma$-algebra, uncountable unions may not exist. However suppose I have a directed system $\{A_i,i\in I\}$ where for each $i\in I$, $A_i\in\mathcal A$ (the $\sigma$-algebra) with the property that for any $i,j\in I$, we have $A_i\subseteq A_j$ or $A_j\subseteq A_i$. Then I believe $\cup_{i\in I}A_i\in\mathcal A$ but I am having difficulty in writing a proof. Notice here that $I$ may be uncountable.

REPLY [5 votes]: Here is a very general counterexample. Suppose $\mathcal A$ is any proper subcollection of the power set $\mathcal P(E)$ and suppose that all finite subsets of $E$ belong to $\mathcal A.$ Choose an element $X\in\mathcal P(E)\setminus\mathcal A$ of minimum cardinality. By well-ordering the infinite set $X$ we can write it as the union of a chain of subsets of smaller cardinality. Thus $X$ is the union of a chain of elements of $\mathcal A,$ but $X\notin\mathcal A.$