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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: enter image description here (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.

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  • $\begingroup$ Doesn't sound like something that can possibly be true. Since $X$ appears only on one side of the "iff", if would imply that any given $R$ is both an algebra (by choosing an $X$ that is in $R$) and is not an algebra (by choosing an $X$ that is not in $R$). That is absurd. $\endgroup$ Commented Sep 2, 2018 at 4:57
  • $\begingroup$ I'm not sure what you mean. This question was taken from Folland's Text. $\endgroup$
    – kemb
    Commented Sep 2, 2018 at 5:02
  • $\begingroup$ Suppose what you state holds. Then if you give me any ring, I could use it to prove that the ring is not an algebra, simply by applying your property with $X$ being something that is not in the ring. And then I could show again that your ring is an algebra, by applying it to a different $X$ that is an element of your ring. $\endgroup$ Commented Sep 2, 2018 at 5:05
  • $\begingroup$ So are you saying that there is something wrong with the way the question is stated? I've written the question as it is stated in the textbook. $\endgroup$
    – kemb
    Commented Sep 2, 2018 at 5:08
  • $\begingroup$ Yes, there must be something wrong with the way the question is stated. What it claims cannot possibly be true in general. $\endgroup$ Commented Sep 2, 2018 at 5:09

1 Answer 1

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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.'

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