Showing that a union of the subsets of two $\sigma$-algebras is a $\sigma$-algebra.

I got back an assignment for a first course in analysis and I have made a very basic error, and I'm having a lot of trouble pinpointing exactly what piece of information I'm missing.

You have two measure spaces $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\mu)$. Assume that $X$ and $Y$ are disjoint. Let $Z=X\cup Y$

Show that $\mathcal{C}=\{A\cup B:A\in \mathcal{A} \,and\,B\in \mathcal{B}\}$ is $\sigma$-algebra on $Z$.

Showing that $\emptyset \in \mathcal{C}$ and that $\mathcal{C}$ is closed under countable unions was easy, but it's the

If $C\in \mathcal{C}$, then $C^c \in \mathcal{C}$

which I've gotten wrong.

My thinking is like this: We have $\mathcal{A} \subset X$ and $\mathcal{B} \subset Y$, where $A^c \in \mathcal{A}$ and $B^c \in \mathcal{B}$. Since $X$ and $Y$ are disjoint, the intersection between two elements that are respectively in $X$ and $Y$ is just the empty set.

$C=(A\cup B)$, so then $C^c=(A \cup B)^c=A^c \cap B^c=\emptyset$

I guess I'm making a very basic error because the feedback I got was just "Wrong!", so I'd really appreciate if someone pointed it out for me.

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For $C \subseteq Z$ you have $$C^c = Z \backslash C$$ i.e. the complement of the set is taken with respect to the set $Z$. Hence $$A^c = Z \backslash A$$ since you consider $A$ as a subset of $Z$ (and not $X$!). Therefore in general $A^c \cap B^c \not= \emptyset$.
Example: $X=[0,\frac{1}{2})$, $Y= \left[\frac{1}{2},1 \right]$, $A=\left(\frac{1}{4},\frac{1}{2}\right)$, $B=\left[\frac{1}{2},\frac{3}{4}\right]$, then $$A^c \cap B^c = \left[0,\frac{1}{4}\right] \cup \bigg(\frac{3}{4},1 \bigg]$$ where the complement is taken with respect to to $Z=[0,1]$.
I see, but doesn't that make $C^c=(A\cup B)^c=(A^c \cap B^c)=(Z\setminus A\cap Z\setminus B)=Z\setminus (A\cup B)$? Doesn't this imply that $C^c$ is not in $\mathcal{C}$? –  john.abraham May 12 '13 at 14:56
@john.abraham No. Note that $$Z \backslash A = \underbrace{Y}_{\in \mathcal{B} \subseteq \mathcal{C}} \cup \underbrace{(X \backslash A)}_{\in \mathcal{A} \subseteq \mathcal{C}}$$ i.e. $Z \backslash A \in \mathcal{C}$ for $A \subseteq X$. And this implies $(Z \backslash A) \cap (Z \backslash B) \in \mathcal{C}$. –  saz May 12 '13 at 15:05
How do you then interpret $Z\setminus (A\cup B)$? –  john.abraham May 12 '13 at 15:44
@john.abraham There's nothing to interpret - it's simply the complement of $A \cup B$...? I don't see your point, unfortunately. –  saz May 12 '13 at 15:54
Well $(Z\setminus A) \cap (Z\setminus B) = Z\setminus (A\cup B)$ right? Which is the set $\{s\in Z|s\notin (A\cup B)\}$. Why does it follow that this set is an element of $\mathcal{C}$? (I'm really struggling to wrap my mind around this, it's embarassing, frankly, so thank you for your patience) –  john.abraham May 12 '13 at 16:48