Minimal field containing a class of subsets: prove equivalence of sets

Question

Let $\mathcal{C}$ be a class of subsets of $\Omega$ and $\mathcal{F}(\mathcal{C})$ be the minimal field containing $\mathcal{C}$. Define sets $$D:=\{A:A=\cap_{j=1}^nB_j,n\ge1\text{ where, }B_j\in\mathcal{C}\text{ or }B_j^c\in\mathcal{C}~\forall j\},\\G_1:=\{\cup_{i=1}^nA_i:n\ge1\text{ where, }A_i\in D~\forall i\}.$$ Prove that $G_1=\mathcal{F}(\mathcal{C})$.

This requires me to prove three parts.

1. $\mathcal{C}\subseteq G_1$
2. $G_1$ is a field
3. There cannot be any smaller field.

I have proved the first two parts but am stuck with the third. To prove it, I need to show that any other field containing $\mathcal{C}$ is a superset of $G_1$. Any help?

EDIT. field containing $\mathcal{C}$ means containing all elements of $\mathcal{C}$

Answer 1

Let $\mathcal F$ be a field with $\mathcal C\subseteq\mathcal F$.

Then for every $B^c\in\mathcal C$ we have $B^c\in\mathcal F$ and consequently $B\in\mathcal F$.

Then from [$B_j\in\mathcal C$ or $B_j^c\in\mathcal C$ for $j=1,\dots,n$] we can conclude that $B_j\in\mathcal F$ for $j=1,\dots,n$ and consequently that $\bigcap_{j=1}^nB_j\in\mathcal F$.

Proved is now that $\mathcal D\subseteq\mathcal F$ and it is immediate that every union of sets in $\mathcal D$ will again be an element of $\mathcal F$.

Proved is now that $\mathcal G_1\subseteq\mathcal F$ and we are ready.