Sub $\sigma$-algebra Is my proof of the following correct?

Let $\mathcal{A}$ be a $\sigma$-algebra on $\Omega$ and let $B\in\mathcal{A}$; then $\mathcal{B}=\{A\cap B:A\in\mathcal{A}\}$ is a $\sigma$-algebra on $B$

Pf. We have to show that $\mathcal B$ is closed under complements ($A\cap B\in \mathcal B\implies B-(A\cap B)\in \mathcal B$) and countable unions.


*

*$A^c\in \mathcal A\implies A^c\cap B\in \mathcal B$.  $A^c\cap B=B-A=B-(A\cap B)$ and we have closure under taking complements.

*$\bigcup A_i\in \mathcal A\implies \bigcup A_i \cap B\in \mathcal B$, but $\bigcup A_i\cap B=\bigcup (A_i\cap B)$ and we have closure under countable unions.

 A: I think that you have all the ingredients for the proof, but the way you have put them together makes it a little hard to follow.  In particular, you tend to write things a little backwards, so that, for instance, rather than starting with a countable set of elements of $\mathcal{B}$, you start with a countable set of elements of $\mathcal{A}$. 
Well, there is one little thing you have left out - that $\mathcal{B}$ has a maximal element:

$B$ is an element of $\mathcal{B}$ since $B=\Omega\cap B$. Furthermore, any other element $C\in\mathcal{B}$ satisfies $C\subseteq B$, since $C=A\cap B$ for some $A$. 

I like the fact that you defined the relative complement, although the $\mapsto$ arrow ("\mapsto") is probably more suitable than the $\Longrightarrow$ arrow in this case. But you haven't shown an argument that this actually is the relative complement (yes, I know it's "obvious", but the whole thing is obvious, so you have to be quite meticulous in these cases).
So, you have your proposed complement function $C\mapsto B\setminus C$, where $C$ has the form $A\cap B$, for some $A\in\mathcal{A}$. For convenience, lets denote this purported complement $\tilde{C}$, to distinguish it from the complement in $\mathcal{A}$ (which you denote $C^{c}$).
Then we must show that:


*

*$C\cap\tilde{C}=\emptyset$.

*$C\cup\tilde{C}=B$. 

*$\tilde{C}\in\mathcal{B}$. 
Fix an $A\in\mathcal{A}$ such that $C=B\cap A$. Then each of the above statements can be translated to statements about objects in $\mathcal{A}$. For example: 
\begin{eqnarray}
C\cap\tilde{C}&=&(A\cap B)\cap(B\setminus(A\cap B)) \\
& = & (A\cap B)\cap(B\cap(A\cap B)^{c}) \\
& = & (A\cap B)\cap(A\cap B)^{c} \\
& = & \emptyset
\end{eqnarray}
A similar approach is used for the countable unions, but here the situation is simpler, since you don't have to redefine unions the way you did complements (or the "top" element).  Let $\{C_{i}\vert i\in I\}\subseteq\mathcal{B}$ be a countable.  Then there is a countable set $\{A_{i}\vert i\in I\}\subseteq\mathcal{A}$ such that 
$$\forall i\in I\, A_{i}\cap B=C_{i}.$$
Since 
$$\bigcup_{i\in I}(A_{i}\cap B)=B\cap\left(\bigcup_{i\in I}A_{i}\right)
$$
we have again translated our question about $\mathcal{B}$ (namely, is $\bigcup_{i\in I}C_{i}$ an element of $\mathcal{B}$?) into a question about $\mathcal{A}$ (namely is $\bigcup_{i\in I}A_{i}$ an element of $\mathcal{A}?)$.  

You may have noticed that at no point did we use the fact that $B\in\mathcal{A}$.  This is because the result actually holds for any $B\subseteq \Omega$. Our proof is therefore of a more general result than the one stated! 
If you want to exploit the fact that $B\in\mathcal{A}$, things can be sped up quite a lot: specifically, since $\mathcal{A}$ is closed under finite intersections, we get 
$$
B\in\mathcal{A}\iff \mathcal{B}\subseteq\mathcal{A}.
$$
You can then show that $\mathcal{B}$ is closed under countable unions simply by observing that the countable union of elements of $\mathcal{A}$ that are subsets of $B$ is still an element of $\mathcal{A}$ that is a subset of $B$. There is no need to pick out all those $A_{i}s$. 
A: It is correct although you might want to include the fact that $B = \Omega \cap B \in \mathcal{B}$.
