Prove if R is a Ring, then R is an algebra iff x $\in$ R (Proof verification)

Question

Question from Folland:

Question 1b: Prove if $R$ is a ring (resp $\sigma-$ring), then $R$ is an algebra (resp $\sigma-$algebra) if and only if $X \in R$. Screenshot of question: (part b).

Proof: (First for rings): Let $R$ be a ring. Suppose $R$ is an algebra on $X$. An algebra of sets on X is a nonempty collection of subsets of $X$ that is closed under finite unions and complements. Thus we have for E $\in$ $R$, $E^c \in R$ and hence $X=E$ $\cup$ $E^c$ $\in$ R. Thus X $\in$ R.

Other direction: Suppose $R$ is a ring and suppose $X \in R$. We know that the ring, $R$, is closed under finite union, by definition of a ring. We need to show that a ring, $R$, is closed under complementation to show that it is an algebra. Let $E\in R$. Since a ring is closed under differences we know that $X$ \ $E$=$X \cap E^c=E^c$ $\in$ R. Thus R is an algebra.

(Second for $\sigma-$rings): Let $R$ be a $\sigma-$ring. Suppose $R$ is an $\sigma-$algebra on $X$. A $\sigma-$algebra of sets on X is an algebra that is closed under countable unions. Thus we have for E $\in$ $R$, $E^c \in R$ and hence $X=E$ $\cup$ $E^c$ $\in$ R. Thus X $\in$ R.

Other direction: Suppose $R$ is a $\sigma-$ring and suppose $X \in R$. We know that the $\sigma-$ring, $R$, is closed under countable union, by definition of a $\sigma-$ring. We need to show that a $ \sigma-$ring, $R$, is closed under complementation to show that it is an $\sigma-$algebra. Let $E\in R$. Since a $\sigma$-ring is closed under differences we know that $X$ \ $E$=$X \cap E^c=E^c$ $\in$ R. Thus $R$ is an $\sigma-$algebra.

I just wanted to verify if my proof was correct or if there was anything I need to fix/ improve. Thank you very much.

Answer 1

Your proof is correct and detailed.
Improvements: You've just shown that for all algebras $\mathcal{S}$ on $X$, that $X\in\mathcal{S}$. You could then take this alternative definition of an algebra. An algebra of subsets of $X$ is a collection $\mathcal{S}\subseteq P(X)$ with the following properties:
(i): $X\in\mathcal{S}$
(ii): $S\in\mathcal{S}\implies S^c\in\mathcal{S}$
(iii): $S,T\in\mathcal{S}\implies S\cup T\in\mathcal{S}$.

This would shorten your proof by a lot. You could even omit the cases for $\sigma$-algebras and $\sigma$-rings since the proofs are almost identical.
Nitpicks: For your proofs of ($\sigma$-)algebras to rings, you should not write `Let $R$ be a ($\sigma$-)ring.'