I know that every algebra is as semi-algebra, and the book (A course in Real Analysis, McDonald and Weiss) tells me that the opposite is not true: not every semi-algebra is an algebra. Why not?
A semi-algebra contains all finite intersections of its members. This satisfies the condition that an algebra must contain all finite intersection (or unions).
So I think the problem is with the complement-condition:
- If the complement to a member in the semi-algebra is $\emptyset$,
then this satisfies some of the condition to be an algebra (that
every complement must be in the algebra). - So I guess that something goes wrong with the complement being a
finite union of pairwise disjoint sets? But I don´t know why?
Do anyone have an example of semi-algebras that are not algebras?
Best Answer
Let $\Omega=\{a,b,c\}$ and $\mathcal{C}=\{\Omega,\emptyset,\{a\},\{b\},\{c\}\}$ then $\mathcal{C}$ is a semi-algebra but NOT an algebra.