Suppose F, G, and H are nonempty families of sets and for every A∈F and every B∈G, AUB ∈ H.
Prove ∩H ⊆ (∩F) U (∩G)

Question

Suppose F, G, and H are nonempty families of sets and for every A∈F and every B∈G, AUB ∈ H. 
Prove ∩H ⊆ (∩F) U (∩G)

reemii · Answer

$\cap F$ is a set of $F$.
$\cap G$ is a set of $G$.
$(\cap F) \cup (\cap G)$ is a set $K$ of $H$ by hypothesis. 
Therefore ...

anonymous · Answer

what do you mean by ∩F is a set of F?

reemii · Answer

It is an element of F (and it also is a set).

anonymous · Answer

No, that isn't true.
consider F = { {1,2} , {1,3} }
∩F = {1}, but {1} isn't an element of F

reemii · Answer

ok.

reemii · Answer

did you try anything?

reemii · Answer

@sourwing

reemii · Answer

Consider $E = \{A\cup B | A\in F, B\in G\}$. Obviously $E\subseteq H$, therefore $\cap H\subseteq\cap E$.


We would like to show that $\cap E$ is a subset of $(\cap F) \cup (\cap G)$. (doesn't matter if it's equal).

Suppose there exists $x\in \cap E$ but $x\not\in \cap F \cup \cap G$. Then x is neither in $\cap F$ nor in $\cap G$. Then there exists A and B such that $x\not\in A$ and $x\not\in B$.

But we also have $A\cup B \in E$ by definition.
Therefore $\cap E \subseteq A\cup B $ beause $A\cup B$ is one set involved in $\cap E$.

This is a contradiction, because $x\in \cap E\subseteq A\cup B \not\ni x$ is impossible.
--
Conclusion.  $\cap H\subseteq \cap E \subseteq \cap F \cup \cap G$.

reemii · Answer

idk if it's a "real question" or an exercise for us..
if the demo's flawed, don't hesitate to tell me.

anonymous · Answer

sounds good to me. thank you