# Is the empty set a $\sigma$-ring?

Folland defines a $\sigma$-ring to be a family of sets $\mathcal{R} \subset \mathcal{P}(X)$ that is closed under differences, and finite unions.

The empty set $\mathcal{R} = \varnothing$ satisfies this, but is it considered a $\sigma$-ring?

Here's another way to resolve this. Interpret

The family $\mathcal{R} \subset \mathcal{P}(X)$ is closed under finite unions.

to mean

For every finite subfamily $\mathcal{F} \subseteq \mathcal{R}$, we have $\bigcup \mathcal{F} \in \mathcal{R}$.

Now, no matter what $\mathcal{R}$ is, we do have $\varnothing \subseteq \mathcal{R}$ and $\varnothing$ is certainly a finite set. So, by the above assumption, we must have $\bigcup \varnothing \in \mathcal{R}$ with $\varnothing$ being viewed as a family of subsets of $X$. But now, if you think about it, you have $\bigcup \varnothing = \varnothing$. So, we get $\varnothing = \bigcup \varnothing \in \mathcal{R}$. This prevents us from taking $\mathcal{R} = \varnothing$ since $\varnothing \notin \varnothing$.

Added comment: The definitions of $\bigcup \mathcal{A}$ and $\bigcap \mathcal{A}$ for a family $\mathcal{A} \subseteq \mathcal{P}(X)$ are: $$\bigcup \mathcal{A} := \{ x \in X : \text{ there exists } A \in \mathcal{A} \text{ with } x \in A\}$$ $$\bigcap \mathcal{A} := \{x \in X : \text{ for all } A \in \mathcal{A}\text{, } x \in A\}.$$ If you understand how to apply these definitions in edge cases, you should agree that $\bigcap \varnothing = X$. This is reasonable since intersecting sets can be thought of as a process where different "filters" are being applied. If you apply no filters at all, then you have everything left.

• By the way, despite this answer, I would also stipulate that the collection be nonempty when defining a sigma-algebra. Sometimes a little redundancy is best. Jan 28, 2014 at 3:57