Hi

Let $\displaystyle \mathcal{F}\;\mbox{and}\;\mathcal{G}$ be non empty families of sets.

I am trying to prove that

$\displaystyle \left[(\cup \mathcal{F})\cap(\cup \mathcal{G})=\emptyset \right] \Rightarrow \left[\mathcal{F} \cap \mathcal{G}=\{\emptyset \}\right]$

I needed to prove something else. I had to come up with some counterexample which gave me the idea

for the above theorem. I am trying to see if I can prove it. Following is my try.

Since this is an implication , Givens are

$\displaystyle (\cup \mathcal{F})\cap(\cup \mathcal{G})=\emptyset$

and the Goal is

$\displaystyle \mathcal{F} \cap \mathcal{G}=\{\emptyset\}$

So the givens are , in logical language ,

$\displaystyle \forall y[\exists A \in \mathcal{F}(y \in A)\Rightarrow \forall B \in \mathcal{G} (y \notin B)]$

Now, I put the goal in logical language. I am not very sure of this, please check it

$\displaystyle \forall z[z \in (\mathcal{F}\cap \mathcal{G})\Rightarrow \forall t(t \in z \Rightarrow t \notin z)]$

since the goal is of the form $\displaystyle \forall x P(x)$ , I let z be arbitrary and suppose that

$\displaystyle z \in (\mathcal{F}\cap \mathcal{G})$

So the new list of the givens is

Givens:

$\displaystyle \forall y[\exists A \in \mathcal{F}(y \in A)\Rightarrow \forall B \in \mathcal{G} (y \notin B)]$

$\displaystyle z \in (\mathcal{F}\cap \mathcal{G})$

and the goal now becomes

$\displaystyle \forall t(t \in z \Rightarrow t \notin z) $

again , since the goal is of the form $\displaystyle \forall x P(x)$ , I let t be arbitrary and suppose that

$\displaystyle t \in z $

Finally the givens are

$\displaystyle \forall y[\exists A \in \mathcal{F}(y \in A)\Rightarrow \forall B \in \mathcal{G} (y \notin B)]$

$\displaystyle z \in \mathcal{F}$

$\displaystyle z \in \mathcal{G}$

$\displaystyle t \in z $

and the goal is simply

$\displaystyle t \notin z $

So after breaking this logically , we can now see from the givens that we let y be z and

since $\displaystyle t \in z $ , we can conclude that

$\displaystyle t \notin z $

since z and t are arbitrary , the result holds for all values of z and t. But I am not

yet sure if this proves that

$\displaystyle \mathcal{F} \cap \mathcal{G}=\{\emptyset\}$

please comment......

I need this for some related problem in Velleman's book "How to prove it". Author wants the

readers to explore all the logical structure of the proof. Hence all the details.

thanks