Intersection is Decreasing

Theorem

Let $U$ be a set.

Let $\mathcal F$ and $\mathcal G$ be sets of subsets of $U$.

Then $\mathcal F \subseteq \mathcal G \implies \bigcap \mathcal G \subseteq \bigcap \mathcal F$, where by convention $\bigcap \varnothing = U$.

That is, $\bigcap$ is a decreasing mapping from $(\mathcal P(\mathcal P(U)), \subseteq)$ to $(\mathcal P(U), \subseteq)$, where $\mathcal P(U)$ is the power set of $U$.

Proof

Let $\mathcal F \subseteq \mathcal G$.

Let $x \in \bigcap \mathcal G$.

Then for each $S \in \mathcal F$, $S \in \mathcal G$.

By the definition of intersection, $x \in S$.

Since this holds for all $S \in \mathcal F$, $x \in \bigcap \mathcal F$.

Since this holds for all $ x \in \bigcap \mathcal G$:

$\bigcap \mathcal G \subseteq \bigcap \mathcal F$

$\blacksquare$