Simple proof about induced $\sigma$-algebra Let $\cal F$ be a $\sigma$-algebra of subsets of $\Omega$ and $A$ a subset of $\Omega$. Show that $\mathcal C=\{A\cap B:B\in\mathcal F\}$ is a $\sigma$-algebra of subsets of $A$. Show that it is not a $\sigma$-algebra of subsets of $\Omega$.
I do not understand how to prove that $\mathcal C$ is closed under complementation.
Suggestions? 
 A: You'll want to show that $$A\setminus(A\cap B)=A\cap(\Omega\setminus B)$$ for any subset $B$ of $\Omega.$ Incidentally, you'll need to assume that $A$ is a proper subset of $\Omega$ to conclude that $\mathcal C$ is not a $\sigma$-algebra of subsets of $\Omega$ (why?).

Edit: It looks like the problem is arising from misleading notation. Given an arbitrary set $D,$ what is $D^c$? Well, it isn't the set of all objects that aren't in $D$--that turns out not to be a set at all! That notation only makes sense if we are working with subsets of one particular set. Let me provide an alternate phrasing that should make clearer what is necessary (and why the hint I gave above shows exactly what we want to show).

A set of sets $\mathcal S$ is said to be a $\sigma$-algebra of subsets of (a set) $X$ if the following conditions hold:

*

*Every element of $\mathcal S$ is a subset of $X$;


*$X\in\mathcal S;$


*for any countable subset $\mathcal E$ of $\mathcal S,$ we have $\bigcup\limits_{E\in\mathcal E}E\in\mathcal S;$


*for any $E\in\mathcal S,$ we have $X\setminus E\in\mathcal S.$

This last condition is the one you've been having trouble with. The notation I use makes it clearer exactly the set in which we're taking relative complements.
Since $\mathcal F$ is a $\sigma$-algebra of subsets of $\Omega$, then $\Omega\setminus B\in\mathcal F$ whenever $B\in\mathcal F.$ In order to show that $\mathcal C$ is a $\sigma$-algebra of subsets of $A,$ we must show that $A\setminus D\in\mathcal C$ whenever $D\in\mathcal C$. Does that clear up your confusion?
