# Set theory basics involving unions,intersections, disjointness

Let $B_n = \bigcup_{1}^{n}A_i$ where $A_i$ are a sequence of disjoint sets.

Suppose $A_1$ and $A_2$ are disjoint sets in some space $X$ and we have a set $E\subset X$. Suppose further that $B_2 = A_1\cup A_2$. Then would $$((E\cap B_2)\cap A_2)) = E \cap A_2?$$ and would $$((E\cap B_2)\cap A_2^c)) = E \cap A_1?$$

As I proceed in doing problems in measure theory I am confronted with stuff from set theory that stop me up. If anyone has any recommendations on getting a better understanding of set theory or more advanced set theory problems involving unions and intersections please let me know.

• Assuming that the $B_1$ should have been $B_2$ or vice versa: Drawing a few Venn diagrams ought to convince you that both of these are true. May 18 '16 at 16:09
• @M.G sorry I edited my post May 18 '16 at 16:09
• Are all $A_i$ disjoint, or just $A_1$ from $A_2$? May 18 '16 at 16:11
• @Jed You changed the second equation in your edit! Yours is true, but the original equation was also true (and less trivial).
– user169852
May 18 '16 at 16:30
• @Bungo Apologies, it looked very much like a typo. Are you sure it wasn't? He had made several other typos. May 18 '16 at 16:32

Given that $B_2 = A_1 \cup A_2$ and that $A_1$ and $A_2$ are disjoint, i.e., $A_1 \cap A_2 = \emptyset$, $$B_2 \cap A_2 = (A_1 \cup A_2)\cap A_2 = A_2$$ Therefore $$E \cap B_2 \cap A_2 = E \cap A_2$$ The same argument works for the other.
The first statement is is true even if $A_1$ and $A_2$ are not disjoint. Clearly we have $B_2 \cap A_2 \subset A_2$, and the opposite containment $A_2 \subset B_2 \cap A_2$ holds because $A_2 \subset A_1 \cup A_2 = B_2$. Since both containments hold, we have $A_2 \cap B_2 = A_2$.
On the other hand, the second statement requires the disjointness of $A_1$ and $A_2$. For example, if $A_1 = A_2$ then the left hand side is $\emptyset$ but the right hand side need not be.
To prove that the second statement is true if $A_1$ and $A_2$ are disjoint, note that in this case $$B_2 \cap A_2^c = (A_1 \cup A_2) \cap A_2^c = (A_1 \cap A_2^c) \cup (A_2 \cap A_2^c) = A_1 \cap A_2^c = A_1$$ where the last equality is true because $A_1$ and $A_2$ are disjoint.