I am having some problems joining sigma algebras. So I have: $\textit{K} =\left \{ A\cap B: A\in \sigma (D),B\in \sigma (E) \right \}$ and I need to show $\sigma(K)= \sigma (D,E)$
What I've done so far
I intend to do this by showing $\sigma(K)\subseteq \sigma (D,E)$ and $\sigma(K)\supseteq\sigma (D,E)$
The first part is easy enough, my problem is $\sigma(K)\supseteq\sigma (D,E)$ as I'm not sure where to begin.
I've started by saying by definition $\sigma(D,E)= \sigma(\sigma(D)\cup\sigma(E))$ and $\sigma(\sigma(D)\cup\sigma(E))=\sigma(\sigma(D)\cap\sigma(E))$ by De Morgan and $\sigma(D)\cap\sigma(E)=K$, but I think that is wrong as $\sigma(D)\cap\sigma(E)$ may be a sigma algebra and thus its smaller than $ \sigma (A,B)$.
I've also considered something like $\sigma(K^{c})$ to somehow argue that $\sigma(\sigma(D)\cup\sigma(E))$ is a subset of that but intuitively it feels like what I did with the De Morgan above.
Thanks
EDIT: sorry for the confusing notation everyone, everything has been changed.