Proving that $\bigcap \mathcal P\subseteq\left(\bigcap\mathcal M\right)\cup\left(\bigcup\mathcal N\right)$

Question

I'm trying to prove the statement:

Let $\mathcal M,\mathcal N,\mathcal P$ be families, none of which is the empty set. Assume that if $A\in\mathcal M$ and $B\in\mathcal N$, then $A\cup B\in\mathcal P$. Prove that $\bigcap \mathcal P\subseteq\left(\bigcap\mathcal M\right)\cup\left(\bigcup\mathcal N\right)$

Here is what I have written so far for my proof:

Let us suppose $x\in \bigcap \mathcal P$. Then, $x\in C$ for all $C\in \mathcal P$.

Let us now suppose $x\notin \bigcap\mathcal M$. Then, $x\notin A$ for some $A\in \mathcal M$.

Let us now suppose there exists $B\in \mathcal N$ such that $x\notin B$. Now we see that $A\cup B\in \mathcal P$, so $x\in A\cup B$ and since $x\notin B$ and $x\in A$, but $x\notin A$, we see there is a contradiction. Hence, there does not exist a $B\in \mathcal N$ such that $x\notin B$. Thus, $x\in B$ for all $B\in \mathcal N$. Hence, $x\in \bigcup\mathcal N$.

Thus, $x\notin \bigcap\mathcal M$ implies that $x\in \bigcup\mathcal N$. Therefore, $x\in \bigcap\mathcal M \cup \bigcup\mathcal N$.

Hence, $\bigcap \mathcal P\subseteq\left(\bigcap\mathcal M\right)\cup\left(\bigcup\mathcal N\right)$

I'm not sure if my proof is correct or if it even was going in the right direction. I get confused with families of sets and I struggle with proofs involving them. Any feedback or help is appreciated.

Answer 1

Hence, there does not exist a $B\in \mathcal N$ such that $x\notin B$. Thus, $x\in B$ for all $B\in \mathcal N$.

It's correct, but it means that $x\in \bigcap\mathcal N$.

Now use that $\mathcal N$ is not empty to properly conclude $x\in\bigcup \mathcal N$.

Otherwise, your proof is correct.

Answer 2

Consider the family $\mathcal F:=\{F:\mathcal X\to\mathcal Y\;\big|\;\forall(A,B)\in\mathcal X\;F(A,B)=A\lor F(A,B)=B\}$ where $\mathcal X=\mathcal M\times\mathcal N,\mathcal Y=\mathcal M\cup\mathcal N$. Let $\pi_1:(A,B)\mapsto A$ denote the usual projection map. $$F=\pi_1\implies\bigcap_{(A,B)\in\mathcal X}F(A,B)=\big(\bigcap_{(A,B)\in\mathcal X}A\big)=\bigcap\mathcal M$$ $$F\neq\pi_1\implies\exists(A_0,B_0)\in\mathcal X\;\bigcap_{(A,B)\in\mathcal X}F(A,B)\subseteq F(A_0,B_0)=B_0\subseteq\bigcap\mathcal N$$ $$\therefore\;\;\bigcap\mathcal P\subseteq\bigcap_{(A,B)\in\mathcal X}A\cup B=\bigcup_{F\in\mathcal F}\bigcap_{(A,B)\in\mathcal X}F(A,B)=\big(\bigcap_{(A,B)\in\mathcal X}A\big)\cup\big(\bigcup_{F\neq\pi_1}\bigcap_{(A,B)\in\mathcal X}F(A,B)\big)$$

$$\subseteq\bigcap\mathcal M\;\cup\bigcup\mathcal N$$