Let ${\cal C} \in P(X)$ which $X$ is a non empty set and
$${\cal C}=\{A\in P(X); A \text{ or } A^c= X\smallsetminus A \}$$
Proof that $\cal{C}$ is a algebra but not a $\sigma$-algebra, and give an example.
I don't see what is the elements of set ${\cal C}$ (Are $A$ only or $A^c$ only or both?), and I don't know how to solve it.
The class $\mathscr{A}=\{F\subset \mathbb{N}: F \text{ is finite or } F^c \text{ is finite}\} $is an algebra of subsets of $\mathbb{N}$ but not a $\sigma$-algebra. The let $A_1, \ldots, A_n$ sets in $\mathscr{A}$, then or every $A$ is finite or at last one $A_i$ is infinite (and therefore A^i^c is finite) so or the union is finite or the complement of the union has finite complement.
Clearly $\mathscr{A}$ is closed to complement.
The union of even numbers do not lies in $\mathscr{A}$, so $\mathscr{A}$ is not a $\sigma$-álgebra.