This is an exercise from Velleman's "How To Prove It":
Suppose $\mathcal{F}$ is a nonempty family of sets. Let $I = \bigcup \mathcal{F}$ and $J = \bigcap \mathcal{F}$. Suppose also that $J \neq \emptyset$, and notice that it follows that for every $X \in \mathcal{F}$, $X \neq \emptyset$, and also that $I \neq \emptyset$. Finally, suppose that $\{A_i | i \in I\}$ is an indexed family of sets.
d. Discover and prove a theorem relating $\bigcap_{i \in J}A_i$ and $\bigcup_{X \in \mathcal{F}}(\bigcap_{i \in X}A_i)$.
After doing a few examples on paper, I decided that $\bigcup_{X \in \mathcal{F}}(\bigcap_{i \in X}A_i) \subseteq \bigcap_{i \in J}A_i$. Here is a proof of this supposition:
Proof: Let $y \in \bigcup_{X \in \mathcal{F}}(\bigcap_{i \in X}A_i)$ be arbitrary. Then we can choose an $X \in \mathcal{F}$ such that $y \in \bigcap_{i \in X}A_i$. Now let $j \in J = \bigcap \mathcal{F}$ be arbitrary. Since $j \in \bigcap \mathcal{F}$ and $X \in \mathcal{F}$, we must have $j \in X$. Then since $j \in X$ and $y \in \bigcap_{i \in X}A_i$, $y \in A_j$. Since $j$ was arbitrary, $y \in \bigcap_{i \in J} A_i$. Since $y$ was arbitrary, $\bigcup_{X \in \mathcal{F}}(\bigcap_{i \in X}A_i) \subseteq \bigcap_{i \in J}A_i$. $\square$
I am struggling to understand how this is true intuitively. Right now, I am thinking of $\mathcal{F}$ as a family of sets containing sets of indices, e.g., {{1,2}, {2,3}, {2,4}} (notice $\bigcap \mathcal{F} \neq \emptyset$). Then $y \in \bigcup_{X \in \mathcal{F}}(\bigcap_{i \in X}A_i)$ means that there is a set of indices in $\mathcal{F}$ such that $y$ is contained in $A_i$ for every index $i$ in that set. $y \in \bigcap_{i \in J}A_i$ means that for every index $i$ that is contained in all sets $X \in \mathcal{F}$, we must have $y \in A_i$. The formal proof seems to work out, but I am not seeing the relationship between these two sets clearly.
Best Answer
If $X\in\mathscr{F}$, then $J\subseteq X$. This immediately implies that $\bigcap_{i\in X}A_i\subseteq\bigcap_{i\in J}A_i$: when you intersect the $A_i$ with $i\in X$, you’re intersecting all of the $A_i$ with $i\in J$ and possible some others as well, so if anything you get a smaller intersection. This is true for every $X\in\mathscr{F}$, so each of the intersections $\bigcap_{i\in X}A_i$ is contained in the big intersection $\bigcap_{i\in J}A_i$, and therefore their union is as well: $\bigcup_{X\in\mathscr{F}}\bigcap{i\in X}A_i\subseteq\bigcap_{i\in J}A_i$.
The key to intuitive understanding is recognizing that when you intersect a larger collection of sets, you’re imposing more restriction on what can be in the intersection, so you get a smaller intersection. Each $X$ is larger than $J$ (or, to be more accurate, at least as large as $J$).