Prove that if $\mathcal F \subseteq \mathcal{G}_1 \cap \mathcal{G}_2$, then $\cap \mathcal{G}_1 \cup \cap \mathcal{G}_2 \subseteq \cap \mathcal{F}$ I was hoping for a second pair of eyes on my proof attempt. I'm having a hard time wrapping my head around this one:
Question:

Prove that if $\mathcal F \subseteq \mathcal G_1 \cap \mathcal G_2$, then $\cap \mathcal G_1 \cup \cap \mathcal G_2 \subseteq \cap \mathcal F$.
Given: $\mathcal F , \mathcal G_1, \mathcal G_2$ are non-empty families of sets.

My Attempt:
Suppose $\mathcal F \subseteq \mathcal G_1 \cap \mathcal G_2$. Then let $A$ be an arbitrary set where $A \in \mathcal F$.
Then, by definition of a subset, $A \in \mathcal G_1 \cap \mathcal G_2$.
Now, let $x$ be arbitrary and $x \in \cap \mathcal G_1 \cup \cap \mathcal G_2$.
Then by definition of the union of a family of sets, $\cap \mathcal G_1 = \left \{ x | \forall A \in \mathcal G_1 (x \in A) \right \}$,  $\cap \mathcal G_2 = \left \{x | \forall A \in \mathcal G_2 (x \in A) \right \}$.
It follows $x \in A$ and since $A$ is an arbitrary element of $\mathcal F$, we can conclude $x \in \cap \mathcal F$.
We have $x \in \cap \mathcal G_1 \cup \cap \mathcal G_2$ and $x \in \cap\mathcal F$, therefore $\cap \mathcal G_1 \cup \cap \mathcal G_2 \subseteq \cap \mathcal F$.
 A: You're right to chase elements through the the definitions, but your proof contains a flaw, which is that you seem to have used the symbol $A$ on the one hand to refer to a fixed, arbitrary element of $\mathcal{F}$, as well as a variable which you're quantifying over. You also seem to have mixed up what you need to do to check one set is a subset of another.
Here's a suggestion for a proof.
Assume that $\mathcal{F} \subseteq \mathcal{G}_1 \cap \mathcal{G}_2$, and let $x \in (\bigcap \mathcal{G}_1) \cup (\bigcap \mathcal{G}_2)$ be arbitrary. Your goal is now to prove that $x \in \bigcap \mathcal{F}$, then you'll be done.
Unpacking the definitions:


*

*You know that every set in $\mathcal{F}$ is in both $\mathcal{G}_1$ and in $\mathcal{G}_2$.

*You know that either $x \in \bigcap \mathcal{G}_1$ or $x \in \bigcap \mathcal{G}_2$; thus either $x$ is an element of all the sets in $\mathcal{G}_1$, or $x$ is an element of all the sets in $\mathcal{G}_2$.

*Your goal is to prove $x \in \bigcap \mathcal{F}$; that is, you need to prove that $x$ is an element of all the sets in $\mathcal{F}$.


So now let $A \in \mathcal{F}$ be arbitrary. You need to prove that $x \in A$, from the assumptions outlined in the first two bullet points above. Once you've done this, you'll have proved that $\forall A \in \mathcal{F},\ x \in A$, so that $x \in \bigcap \mathcal{F}$, which is exactly what you needed to do.
I leave the rest to you.
