Let $A\subset\Omega$ and $\mathcal{B}_{A} = \{B\cap A:B\in\mathcal{B}\}$. Show that $\mathcal{B}_{A}$ is a $\sigma$-algebra on $A$. Let $\Omega$ be a nonempty set and $\mathcal{B}$ be a $\sigma$-algebra on $\Omega$. Let $A\subset\Omega$ and $\mathcal{B}_{A} = \{B\cap A:B\in\mathcal{B}\}$. Show that $\mathcal{B}_{A}$ is a $\sigma$-algebra on $A$.
MY ATTEMPT
To begin with, notice that $A\in\mathcal{B}_{A}$. Since $\Omega\in\mathcal{B}$, we conclude that $A = \Omega\cap A\in\mathcal{B}_{A}$.
Let us suppose that $S_{1},S_{2},\ldots, \in\mathcal{B}_{A}$. Then one has that $S_{i} = B_{i}\cap A$ for some $B_{i}\in\mathcal{B}$, where $i\geq 1$. Consequently,
\begin{align*}
S = \bigcup_{i=1}^{\infty}S_{i} = \bigcup_{i=1}^{\infty}(B_{j}\cap A) = \left(\bigcup_{i=1}^{\infty}B_{j}\right)\cap A = B\cap A \Rightarrow S\in\mathcal{B}_{A}
\end{align*}
Could someone help me to prove that $\mathcal{B}_{A}$ is closed under complementation?
Any help is appreciated.

If $S =A\cap B\in \mathcal B_A$ where $B \in \mathcal B$  then complement of $S$ in $A$ is $A\cap S^{c}$ and $S^{c} \in \mathcal B$ so $S^{c} \in \mathcal B_A$.
