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.
Best Answer
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 $U^c$.