Question about definition of Semi algebra I am wondering if someone could help me with basic properties of semi algebra.
We say that $S$ is a semi algebra of subsets of X if 


*

*$\emptyset \in S$

*If $P_1$, $P_2 \in S$, then $P_1 \cap P_2 \in S$  

*If $P \in S$, then $X \backslash P$ can be written as a finite union of 
sets from $S$.


But I am finding that sometimes it is defined using the following 3'
instead of 3.
3'. If $P \in S$, then $X \backslash P$ can be written as a disjoint finite union of 
sets from $S$.
My question is are these definitions equivalent? If so can someone please show me how we can obtain 3' from the first three conditions?
Thank you.
 A: Here is a counterexample showing that 1,2, and 3 do not prove 3'.
Let $X$ be the nodes of an infinite complete binary tree.  Then for $x\in X$, let $L(x)$ denote all nodes in the left subtree from $x$, and similarly let $R(x)$ denote all nodes in the right subtree from $x$.  Then let
$S = \{\{x\}| x\in X\} \cup \{\{x\}\cup L(x)| x\in X\} \cup \{\{x\}\cup R(x)| x\in X\} \cup \{\{\} \}$
In other words, S is comprised of all singletons, all singletons with their left subtrees, and all singletons with their right subtrees.  One can check that this is a semi-algebra in the sense of 1,2,and 3.  But we will never be able to write $X$ (the complement of the empty set) as a finite disjoint union of elements of $S$.
A: My guess is that you cannot easily show this. Most good books I have seen, that use the concept of semi-algebra, take care to use your 1,2, and 3' - not 1,2, and 3 as its definition.
Answer 1 to your question is (as you have spotted yourself, I think) basically wrong: Alex has missed the fact that in his construction of the $B_i$ from the $A_i$, he is relying on complements being members of ${\cal S}$, which he is not entitled to do.
I don't know whether 3' can be deduced by 1,2, and 3. It's an interesting question. I suppose a disproof would be to exhibit a class of subsets of some set that satisfies 1, 2, and 3 but contains a member whose complement is not a disjoint union of members.
I would be interested if someone here could answer your conundrum one way or another.
A: Instead of (1) I use (1') $X\in S$.
It follows that under assumptions
(1') and (2), (3) and (3') are equivalent.
Do you need a proof, or it is obvious?
