Is a complete Boolean $\sigma$-algebra ccc?

Question

A Boolean $\sigma$-algebra that satisfies the countable chain condition (ccc) is complete. Is the converse true, i.e., is a complete Boolean $\sigma$-algebra ccc? If not (which is probably the case), what would be an example the complete BA which is not ccc?

Answer 1

By BA having ccc I understand that any family of disjoint elements in it is countable. By Completeness I understand that every downwards/upward directed subset has a $\inf$/$\sup$ respectively.

To get an complete BA without ccc consider power set algebra $\mathcal{P}(X)$ of any uncountable set $X$. If you don't want to be too abstract we can take $X = \mathbb{R}.$ Clearly $\mathcal{P}(X)$ has no ccc, just take the set of singletons $\Big\{\{x\} | x \in X \Big\}$, which is uncountable. On the other hand $A \subset \mathcal{P}(X)$, then the limits can be computed as $\inf A = \bigcap A$ and $\sup A = \bigcup A$. So, clearly $\mathcal{P}(X)$ is complete.

Intuitively, It seems that ccc is a property of BA not to be too broad, so its more about 'size' while completeness is more about 'topolgy'. And 'topolgy' can't control the 'size' as any BA admits a completion by a regular open algebra of its Stone space. I. E. You can make 'topology' more regular by adding more elements.