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.