Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.