1. ## Family of sets

The definition of $\displaystyle x \in \cap \mathcal{F}$ is $\displaystyle \forall A (A\in \mathcal{F}\Rightarrow x \in A)$.

Similarly, the definition of $\displaystyle x \in \cap \mathcal{G}$ is $\displaystyle \forall B (B\in \mathcal{G}\Rightarrow x \in B)$.

Could someone please show me how to define $\displaystyle x\in \cap(\mathcal{F} \cup \mathcal{G})$?

2. Originally Posted by novice
The definition of $\displaystyle \cap \mathcal{F}$ is $\displaystyle \forall A (A\in \mathcal{F}\Rightarrow x \in A)$.
Are you quite sure that what you posted above is actually what your text material has?
It is common to use $\displaystyle \bigcap {\mathcal{F} = \left\{ {x:\left( {\forall A} \right)\left[ {A \in \mathcal{F} \to x \in A} \right]} \right\}}$.
That is from Suppes. There are other variations of that notation.
I have never seen what you posted.

3. Originally Posted by Plato
Are you quite sure that what you posted above is actually what your text material has?
It is common to use $\displaystyle \bigcap {\mathcal{F} = \left\{ {x:\left( {\forall A} \right)\left[ {A \in \mathcal{F} \to x \in A} \right]} \right\}}$.
That is from Suppes. There are other variations of that notation.
I have never seen what you posted.
Yes, sir, you are right. I didn't realize that

$\displaystyle \bigcap {\mathcal{F} = \left\{ {x:\left( {\forall A} \right)\left[ {A \in \mathcal{F} \to x \in A} \right]} \right\}}$.

and

$\displaystyle x \in \bigcap {\mathcal{F}$ means $\displaystyle \forall A (A\in \mathcal{F}\Rightarrow x \in A)$

There is another equivalent expression used in my book. $\displaystyle x\in \bigcap {\mathcal{F}$ means $\displaystyle \forall A \in \mathcal{F}(x \in A)$

For $\displaystyle \cap (\mathcal{F} \cup \mathcal{G})$, it seems that $\displaystyle \mathcal{F}\cup \mathcal{G}$

would look like $\displaystyle \mathcal{F}\cup \mathcal{G}=\{A_1, A_2,\cdots ,A_n\}\cup \{B_1, b_2, \cdots, B_n\}=\{A_1, B_1, A_2, B_2,\cdots , A_n, B_n\}$.

Then$\displaystyle \cap (\mathcal{F}\cup \mathcal{G})= \{A_1\cap B_1\cap A_2\cap B_2\cdots \cap A_n \cap B_n\}$

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

How would you express it in logic?

4. Originally Posted by novice
$\displaystyle \cap (\mathcal{F}\cup \mathcal{G})=(\cap \mathcal{F})\cap (\cap \mathcal{G})$
How would you express it in logic?
$\displaystyle \cap (\mathcal{F}\cup \mathcal{G})=(\cap \mathcal{F})\cap (\cap \mathcal{G})$
That is correct. The basic logic behind it is:
$\displaystyle \begin{gathered} \left( {P \vee Q} \right) \to R \hfill \\ \left( {\neg P \wedge \neg Q} \right) \vee R \hfill \\ \left( {\neg P \vee R} \right) \vee \left( {\neg Q \vee R} \right) \hfill \\ \left( {P \to R} \right) \wedge \left( {Q \to R} \right) \hfill \\ \end{gathered}$

5. Originally Posted by Plato
$\displaystyle \cap (\mathcal{F}\cup \mathcal{G})=(\cap \mathcal{F})\cap (\cap \mathcal{G})$
That is correct. The basic logic behind it is:
$\displaystyle \begin{gathered} \left( {P \vee Q} \right) \to R \hfill \\ \left( {\neg P \wedge \neg Q} \right) \vee R \hfill \\ \left( {\neg P \vee R} \right) \vee \left( {\neg Q \vee R} \right) \hfill \\ \left( {P \to R} \right) \wedge \left( {Q \to R} \right) \hfill \\ \end{gathered}$
This example you gave is so fascinating.

I found it by experimenting your logical notation.

$\displaystyle x\in \cap (\mathcal{F}\cup \mathcal{G})$
$\displaystyle \equiv \forall A(A \in (\mathcal{F}\cup \mathcal{G})\Rightarrow x\in A)$
$\displaystyle \equiv \forall A( (A\in \mathcal{F}\cup A\in \mathcal{G})\Rightarrow x\in A)$
Where $\displaystyle P: A\in \mathcal{F}$
$\displaystyle Q:A\in \mathcal{G}$,
and $\displaystyle R: x\in A$

$\displaystyle (P\Rightarrow R) \wedge (Q\Rightarrow R)$
$\displaystyle \equiv \forall A ((A\in \mathcal{F}\Rightarrow x\in A) \wedge (A\in \mathcal{G}\Rightarrow x\in A))$
$\displaystyle \equiv \forall A (A\in \mathcal{F}\Rightarrow x\in A) \wedge \forall A (A\in \mathcal{G}\Rightarrow x\in A)$
$\displaystyle (\cap \mathcal{F}) \cap (\cap \mathcal{G})$

6. Exactly.

7. Originally Posted by novice
$\displaystyle \cap (\mathcal{F}\cup \mathcal{G})= \{A_1\cap B_1\cap A_2\cap B_2\cdots \cap A_n \cap B_n\}$
That's incorrect.

It should be:

$\displaystyle \cap (\mathcal{F}\cup \mathcal{G})= A_1\cap B_1\cap A_2\cap B_2\cdots \cap A_n \cap B_n$

/

Suppes is a great book. But watch out for one thing: He allows that there may exist urelements, so some of the first formulations in his book and some of the formulations of later theorems are somewhat different from many other set theory books. Also, for certain subjects, such as transfinite recursion, I find that Enderton's book provides a fuller and better step by step explanation. I find that they make a great combination; you can study them together - along side each other - to get a real good grasp.