Soft Question: Are sigma fields, fields?

Question

I'm sorry if this is a foolish question but: Is a $\sigma$-field (of sets) a field (in the sense of algebra) if we only consider finite intersections and finite unions?

Answer 1

No. Consider $\mathbb{R}$ with the Borel $\sigma$-algebra. The operation $\cap$ does not permit inverses. Indeed, if $E$ is the identity, then $E\cap\mathbb{R}=\mathbb{R}$ so $E=\mathbb{R}$. Now, if every element has an inverse with respect to $\cap$ we should find $A\subseteq\mathbb{R}$ such that $[0,1]\cap A=\mathbb{R}$. There is no such $A$.

EDIT: It is not even a ring because the other operation $\cup$ also does not permit inverses. Indeed, if $I$ is the identity then $I\cup\emptyset=\emptyset$ so $I=\emptyset$. But there is no $A\subseteq\mathbb{R}$ such that $[0,1]\cup A=\emptyset$.

Answer 2

Consider symmetric difference over $ \mathbb{R}$ with $\emptyset$ as the identity.

• $A \triangle \emptyset = A$
• $A \triangle A^c = \emptyset$

So it is a group over symmetric difference, but what is the other operation for a field?

If you use symmetric difference and intersection you get a ring. Can we get it to be a field?