This is Velleman's exercise 3.3.13. Suppose $\mathcal F $ and $\mathcal G$ are families of sets and $\mathcal F \subseteq \mathcal G$. Prove that $\bigcap\mathcal G \subseteq\bigcap\mathcal F$.
My approach so far:
Let $x$ be an arbitrary element of $\bigcap\mathcal G$. Then $x$ is an element of every $A\in\mathcal G$. To show that $x$ needs to be an element of $\bigcap\mathcal F$ it suffices to show that it has to be an element of every $A\in\mathcal F$. Suppose $A_0$ is an arbitrary element of $\mathcal F$. Due to $\mathcal F \subseteq \mathcal G$ any $A\in \mathcal F$ is an element of $\mathcal G$. So $A_0 \in \mathcal G$. Since $x$ is an arbitrary element of $\bigcap\mathcal G$, it follows that $x \in A_0$. This shows that every $x \in\bigcap\mathcal G$ will be an element of $\bigcap\mathcal F$.
Any comments, suggestions and improvements relating to my attempt are appreciated. I am a noob so feel free to bash it as good as you can. Thanks in advance.
Looks correct to me. Your proof is very detailed and commendably rigorous—I think that's a good thing, given that the result might, at first sight, seem counterintuitive.
To highlight the intuitive content of the result, let me present a less formal but perhaps more illuminating argument. Think of $\bigcap \mathcal G$ as the set of all points that satisfy all the restrictions embodied by the family of sets $\mathcal G$ (viz., such points must be contained in every $G\in\mathcal G$). Similarly, $\bigcap \mathcal F$ is the set of all points that satisfy all the restrictions embodied in $\mathcal F$. Since $\mathcal F\subseteq\mathcal G$, $\mathcal F$ embodies fewer restrictions. Therefore, the condition $\bigcap \mathcal F$ is, in fact, looser than $\bigcap \mathcal G$, so more points pass it. In other words, if a point clears all the requirements embodied by $\bigcap \mathcal G$, it must a fortiori clear the less stringent requirements embodied by $\bigcap \mathcal F$. This implies that $\bigcap \mathcal G\subseteq \bigcap \mathcal F$.