Let $M$ be an infinite $\sigma$ algebra on X. Show that if $A \in M$, then the collection $M_A$={$B \cap A: B \in M$} is a $\sigma$ algebra on A.
My thoughts/attempt (Really struggling)
To show $M_A$ is a $\sigma$-algebra, I need to show that $M_A$ is closed under complements and countable unions.
I'm really confused on how to show either but here is my attempt anyways.
First I show finite unions:
Let $B_1 \cap A$ $\in M_A$ and $B_2 \cap A$ $\in M_A$ (thus $A$, $B_1$ and $B_2 \in M$).
$(B_1 \cap A)$ $\cup$ $(B_2 \cap A)$=($B_1 \cup B_2)$ $\cap$ $A$ $\in M_A$ $(B_1 \cup B_2 \in M$ as M is a $\sigma$ algebra). Thus we closed under finite union. Can do the same process to show for countable union.
Second: I show closure under complements:
Let $B_1 \cap A$ $\in M_A$. Need to show that $(B_1 \cap A)$$^{c}$ $\in$ $M_A$.
I can write $(B_1 \cap A)$$^{c}$ = ($B_1 \cap A^C$) $\cup$ ($B_1^C \cap A$).
$B_1 \cap A^C \in M_A$ as $A^c \in M$ as $M$ is a $\sigma$ algebra. $B_1^c \cap A \in M_A$ as $B_1^c \in M$ as $M$ is a $\sigma$ algebra.
Thus, $(B_1 \cap A)$$^{c}$ $\in M_A$ as it is the union of two things in $M_A$ and we previously showed that $M_A$ is closed under finite union.
Is my proof correct? Or am I very off in my approach? Thank you.
The main problem is that $B_1 \cap A^C \not\subset A$ !, so $B_1 \cap A^C \notin M_A$. The complement should be found in $A$, so the complement in $A$ for $B_1 \cap A$ is $B_1^C \cap A$.