3
$\begingroup$

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.

  • 0
    I dont see how we get that $A$\$B \in \epsilon$ ?2013-10-29

1 Answers 1

3

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.

  • 0
    You're right. I was being sleep deprived or something... Thank you!2012-09-17