Proof of complement of intersection defined using an arbitrary set. I hate asking questions like these (ones where I have no idea what it is talking about).

Let $ \xi$ be a  collection of sets and define 
  $$I= \bigcap \{ F|F \in \xi \}\quad\text{and}\quad U= \bigcup \{ F|F \in \xi \}.$$
  Prove that $I^{c}= \bigcup \{ F^{c}|F \in \xi \}$ and $U= \bigcap \{ F^{c}|F \in \xi \}$ are true.

This looks a lot like just proving it for the basic case (intersections of sets and their complement, and union of 2 sets and their complement) but the language has made me a bit lost.
Homework problem so please just give help not a complete answer thx.
 A: Let $x\in I^c$. Then there exists at least one $F\in\xi$, call it $F'$ such that $x$ is not in $F'$, otherwise $x$ would be in every $F$ of $\xi$, hence in their intersection, which is $I$, contradicting our assumption, which is $x\in I^{c}$. So $x\in(F')^c$. But one single set is contained in the union of that set with others sets, so that $(F')^c$ with $F'\in\xi$ is contained in the union of all $F^c$, $F$ running in $\xi$. This proves that $x\in\cup F^c$. But this holds for every $x$ in $I^c$, hence $I^c\subseteq\cup F^c$.
Conversely, take an element $y\in\cup F^c$. If an element lies in a union of sets, then it lies at least in one of them, so that there exists $F'\in\xi$ such that $y$ is in $(F')^c$, or equivalently $y$ is not in $F'$. But if $y$ is not in $F'$, $F'$ being a set of the family $\xi$, then $y$ cannot be in the intersection of all $F$ in $xi$, whose name is $I$, hence $y$ is in the complementary set of $I$, called $I^c$. This proves $y\in I^c$. But $y$ was arbitrarily chosen from $\cup F^c$, hence what we have proved is that the whole union $\cup F^c$ is contained in $I^c$.
Similarly for $U$ and $U^c$.
A: Hints:


*

*$a\in (X\cap Y)^c$ if and only if (  $a \notin X$ or $a\notin Y$ ).

*$a\in (X\cup Y)^c$ if and only if  ( $a\notin X$ and $a\notin Y$ ).
More Hints:
Let $a\in U^c$.  Let $X\in \xi$.  If $a\in X$, then $a\in X\cup \bigcup\{F:F\in \xi\setminus\{X\}\}=U$.  But $a\in U^c$, so $a\notin X$ and $a\in X^c$.  Since $X$ was arbitrary in $\xi$, $a\in X^c$, for all $X\in \xi$.  Putting these together, $a\in \bigcap\{X^c:X\in \xi\}$.  This proves $U^c\subseteq \bigcap\{F^c:F\in \xi\}$.  A similar proof, but reversed, proof is necessary for the reverse inclusion as pointed out by @user1 in the comments.
For the other part, with $I$, you can either do this sort of thing twice again, or USE the first part  together with the fact that $(X^c)^c=X$.  That is, define $\xi'=\{F^c:F\in \xi\}$.  Let $U'=\bigcup \{X:X\in \xi'\}$; by the part about $U$ we know $(U')^c=\bigcap\{X^c:X\in \xi'\}$.
