QUESTION
If A and B are two different nonempty sets, how many distinct sets can be formed with these sets using as many unions,intersections,complements and parentheses as desired.
EDIT: lol, I was wondering about the "n" number of sets case. I kinda missed the question there.
MY ANSWER,
which I don't know if it is right.
This is my first answer and I found this question hard. It's wordy because I want to know if I'm thinking on it right.
I don't think $2^{2^n - 1} $ is right. This was a previous answer.
I think the use of "unions,intersections,complements and parentheses as desired" means distinct spaces on a venn diagram.
When I say 'space' I mean distinct set. On a venn diagram this is an area you can fill in on its own and is different from all other spaces.
1 set has 2 spaces, everything in it and the complement of it (ie. U). Total $2^{1}$ spaces
2 sets have 4 spaces, namely, $(A\cup B)', A\cap B, A\cap B', A'\cap B $ . Total $2^{2}$spaces
3 sets have ... Total $2^{3}$ spaces
n sets have ... Total $2^{n}$ spaces
The next thing to notice is that any combination of n things is also $2^{n}$. For example, when there is one set there are two spaces, (everything inside, everything outside.) how many ways can we combine these two spaces?
1)our set alone
2)everything outside our set
3)everything outside our set and our set
4) nothing inside or outside our set
So:
one set has 2 spaces, and 4 different combinations of spaces.
2 sets have 4 spaces, and 16 different combination of spaces.(you could draw a venn diagram)
3 sets have 8 spaces, and 256 different combinations of spaces.
All we need now is a formula that goes from our set number to our spaces number to our combinations number.
Spaces ---> combinations
2 ---> 4
4 ---> 16
8 ---> 256
After playing with it, if we let n=1,2,3,4... we can use $2^{2^{n}}$.
Number of sets-->Spaces ===> combinations
1---> 2 ===> 4 or $2^{2^{1}}$
2 ---> 4 ===> 16 or $2^{2^{2}}$
3 ---> 8 ===> 256 or $2^{2^{3}}$
n ---> 2^n ===> $2^{2^{n}}$
I would like to know if this is correct.