I'm trying to prove the next: Let $\mathbb{E}$ be a ring of subset of $X.$ Then the collection $$\mathcal{A}(\mathbb{E})=\{F\subset X:F\in\mathbb{E}\space\text{or}\space F^{c}\in\mathbb{E}\}$$ where $\mathcal{A}(\mathbb{E})$ is the minimum algebra generated by $\mathbb{E}$.
So, I'd like to prove that the collection $\{F\subset X:F\in\mathbb{E}\space\text{or}\space F^{c}\in\mathbb{E}\}$ is an algebra which contains the ring $\mathbb{E}.$
I'm stuck proving that if $(F_{n})_{n=1}^{k}\subset\{F\subset X:F\in\mathbb{E}\space\text{or}\space F^{c}\in\mathbb{E}\}$ then $\displaystyle\bigcup_{n=1}^{k}F_{n}$ belongs to such collection.
I'm considering three cases:
Case i): $\forall n\in\{1,\ldots,k\}\space F_{n}\in\mathbb{E}$ implies that $\displaystyle\bigcup_{n=1}^{k}F_{n}$ is element of such collection because $\mathbb{E}$ is ring.
Case ii) $\forall n\in\{1,\ldots,k\}\space F_{n}^{c}\in\mathbb{E}$ we have that $\displaystyle\bigcap_{n=1}^{k}F_{n}^{c}\in\{F\subset X:F\in\mathbb{E}\space\text{or}\space F^{c}\in\mathbb{E}\}$ because $\mathbb{E}$ is ring. Now, I've proved $\{F\subset X:F\in\mathbb{E}\space\text{or}\space F^{c}\in\mathbb{E}\}$ is closed under complementation. The conclution follows.
Case iii) $\exists m\in\{1,\ldots,k\}$ such that $F_{m}^{c}\in\mathbb{E}$ and $\forall n\neq m\space F_{n}\in\mathbb{E}.$
I'm stuck in case iii). I can't find a way to express such union in terms of $F_{n}$ and $F_{m}^{c}$ such that belongs to $\{F\subset X:F\in\mathbb{E}\space\text{or}\space F^{c}\in\mathbb{E}\}.$ I tried to get a contradiction utilizing that such union doesn't belong to such collection. So the only element that doesn't belong to $\{F\subset X:F\in\mathbb{E}\space\text{or}\space F^{c}\in\mathbb{E}\}$ is $F_{m}^{c},$ but I can't see how to find a contradiction using this.
Is there an easier way to prove this?
Any kind of help is thanked in advanced.