1
$\begingroup$

I would like to get feedback on my proof. If it is not correct, I would appreciate it if you could only provide hints and not the full answer.

Proof that $A \cap B \in \mathcal{F}$.

$A,B \in \mathcal{F} \iff A^c, B^c \in \mathcal{F} \iff (A^c \cup B^c) \in \mathcal{F} \iff (A \cap B)^c \in \mathcal{F} \iff A \cap B \in \mathcal{F}.$

In the above proof, I am using De Morgan's Law to get the penultimate iff and the definition of a sigma-field for the final iff.

Proof that $A \setminus B \in \mathcal{F}$.

By definiton of a sigma-field, $\emptyset \in \mathcal{F}$ and $A \cap B \in \mathcal{F}$. Therefore, $A \cap B = \emptyset \iff A \setminus B.$

Proof that $A \triangle B$.

Stuck on this bit.

$\endgroup$
12
  • 1
    $\begingroup$ The second is not right. Use $A\setminus B=A\cap B^c$. For $A\triangle B$, I would use the fact that it is $(A\setminus B)\cup (B\setminus A)$. $\endgroup$ Feb 15, 2014 at 20:45
  • $\begingroup$ Is this correct for the second proof: $x \in (A \setminus B) \iff (x \in A \land x \notin B) \iff (x \notin A \lor x \in B) \iff x \in (A^c \cup B) \in \mathcal{F}$. $\endgroup$
    – user124485
    Feb 16, 2014 at 11:00
  • $\begingroup$ You left out a complement. The second iff is wrong. You want something like $(A^c \cup B)^c$. $\endgroup$ Feb 16, 2014 at 15:43
  • $\begingroup$ Revised proof: $x \in (A \setminus B) \iff (x \in A \land x \notin B) \iff (x \notin A \lor x \in B)^c \iff x \in (A^c \cup B)^c \in \mathcal{F}$. $\endgroup$
    – user124485
    Feb 16, 2014 at 23:20
  • $\begingroup$ Yes, that works fine. $\endgroup$ Feb 16, 2014 at 23:22

1 Answer 1

0
$\begingroup$

HINT: recall that $A\triangle B=(A\cup B)\setminus(A\cap B)=(A\setminus B)\cup(B\setminus A)$.

$\endgroup$

You must log in to answer this question.