1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$ May 18 '16 at 16:09
  • $\begingroup$ @M.G sorry I edited my post $\endgroup$
    – Wolfy
    May 18 '16 at 16:09
  • $\begingroup$ Are all $A_i$ disjoint, or just $A_1$ from $A_2$? $\endgroup$
    – Carser
    May 18 '16 at 16:11
  • $\begingroup$ @Jed You changed the second equation in your edit! Yours is true, but the original equation was also true (and less trivial). $\endgroup$
    – user169852
    May 18 '16 at 16:30
  • $\begingroup$ @Bungo Apologies, it looked very much like a typo. Are you sure it wasn't? He had made several other typos. $\endgroup$
    – Carser
    May 18 '16 at 16:32
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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