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Definition 1: An algebra of sets on a non-empty set $X$ is a non-empty collection $\cal{A}$ of subsets of $X$ that is closed under taking complements and finite unions.

Definition 2: An elementary family of sets on a non-empty set $X$ is a collection $\cal{E}$ of subsets of $X$ such that (i) $\emptyset \in \cal{E}$, (ii) $\cal{E}$ is closed under finite intersection, (iii) if $E \in \cal{E}$, then $E^c$ is a finite disjoint union of members of $\cal{E}$.

Proposition: If $\cal{E}$ is an elementary of sets, then the collection $\cal{A}$ of finite disjoint unions of members of $\cal{E}$ is an algebra of sets.

Proof: It's easy to show that if $A,B \in \cal{E}$, then $A \setminus B \in \cal{A}$ (just write $A \setminus B = A \cap B^c$ and apply property (iii) to $B^c$). Using this observation, we write $A \cup B = (A \setminus B) \sqcup B$, and conclude that $A \cup B \in \cal{A}$, as well, for any $A,B \in \cal{E}$. The rest of the proof follows by induction.

Question: Is every elementary family of sets in fact an algebra of sets?

Look at the proof: we make an observation there that if $A,B \in \cal{E}$, then $A \cup B \in \cal{A}$. But $\emptyset \in \cal{E}$, so we may as well say that if $A \in \cal{E}$, then $A = A \cup \emptyset \in \cal{A}$.

Reference: Folland, Real Analysis, pp. 23-24.

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All your last paragraph shows is that $\mathcal{E} \subset \mathcal{A}$ in the proposition. So every elementary family of sets is contained in an algebra of sets. You did not show that $\mathcal{A} = \mathcal{E}$. – Trevor Wilson Sep 15 '12 at 19:51
Of course! I'm so sorry, I was being stupid! Thanks! – Rick Sep 17 '12 at 22:44
I dont see how we get that $A$\$B \in \epsilon$ ? – Albanian_EAGLE Oct 29 '13 at 1:31
up vote 1 down vote accepted

Here is a counterexample to your conjecture: $\{\emptyset, \{x\}, \{y\}\}$. When we say that $A \setminus B = A \cap B^c \in \mathcal{A}$, we are using the fact that we are taking disjoint unions over $\mathcal{E}$. This is because property (iii) only gives us that $B^c$ is a finite disjoint union.

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I see the counterexample, thanks. But I still don't understand what goes wrong. The observation is clear: the union of any two members of $\cal{E}$ is an element of $\cal{A}$. What am I missing? – Rick Sep 15 '12 at 15:00
Try to follow the proof and see what goes wrong. – Yuval Filmus Sep 15 '12 at 20:57
You're right. I was being sleep deprived or something... Thank you! – Rick Sep 17 '12 at 22:44

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