Question: I want to find an example of a family of subsets $\mathcal{F}$ of a set $X$ such that $X$ belongs to the family, it is closed under complementation and finite disjoint unions, but $\mathcal{F}$ is not a field on $X$. However, I'm not sure on how to proceed.

I was thinking of considering a finite set $X$, however I think that it should be infinite for the counterexample to exist. I can find examples of families that are not sets, but they will not be closed under complementation. What would be a good example of this?

Thanks for the help.

Answer: 
**Hint:** Consider $X=\{1,2,\ldots,2n\}$ for some $n \in \mathbb{N}$ and $$\mathcal{F} := \{A \subseteq X; \sharp A \, \text{is even}\}.$$

($\sharp A$ denotes the cardinality of the set $A$.)
Let X=(0,1] and 
$\mathcal{B}=\{(\frac{i}{2^n},\frac{i+1}{2^n}]~:~i=0,1,...,(2^{n}-1)\}$ . Consider $\mathcal{F} =\{X,\phi\}\cup\{A : A~ \text{is obtained by finite unions of members of }\mathcal{B}\}\cup\{\mathbb{Q}\cap(0,1]~,~(0,1]\setminus\mathbb{Q}\}$. Here $\mathcal{F}$ is closed under complementation and disjoint finite unions but not a field, since $(\frac{0}{2^n},\frac{1}{2^n}]\in\mathcal{F}$ and $\mathbb{Q}\cap(0,1]\in\mathcal{F}$ But $(\frac{0}{2^n},\frac{1}{2^n}]\cup\{x~:~\frac{1}{2^n}<x\leq 1~\text{and}~x\in\mathbb{Q}\}\notin\mathcal{F}$. Here I consider some fixed $n\in\mathbb{N}$ and $n > 2$