# Equations of sets

1. Nov 30, 2015

### geoffrey159

1. The problem statement, all variables and given/known data
Let $A,B \in {\cal P}(E)$. Solve in ${\cal P}(E)$ the following equations:
1. $X\cup A = B$
2. $X\cap A = B$
3. $X - A = B$

2. Relevant equations

3. The attempt at a solution
1. We have $A\cup B = (A\cup X)\cup A = A\cup X = B$. So $A\subset B$ and the solution cannot be less than $C_B(A)$. So $X = C_B(A) \cup G,\ G\in{\cal P}(A)$
2. We have $A\cap B = A\cap X = B$. So $B\subset A$ and the solution cannot be less than $B$. So $X = B \cup G,\ G\in{\cal P}(C_E(A))$
3. We have $A\cap B = \emptyset$ and $A\cup B = A\cup X$. So the solution can't be less than $B$. Finally, $X = B\cup G,\ G\in{\cal P}(A)$

Is it correct ?

2. Nov 30, 2015

### Ray Vickson

These equations are impossible if there are no restrictions on $A$ and $B$. In other words, it is easy to give examples of $A, B$ in which the equations cannot possibly hold, no matter how you try to choose $X$.

Draw some Venn diagrams and see this for yourself.

3. Nov 30, 2015

### geoffrey159

Would you mind sharing your counterexamples for the 3 different cases? Assuming $A\subset B$ for q1, $B\subset A$ for q2, $A\cap B =\emptyset$ for q3.

4. Nov 30, 2015

### Ray Vickson

No, I said that without some restrictions on $A$ and $B$ you can find cases where they are impossible.

Naturally, if you make some additional assumptions about $A$ and $B$ you CAN find solutions. It is just that as originally stated (with no restrictions on $A,B$) it may not always be possible to have any $X$. In other words, you either copied down the question incorrectly, or you were given a trick question.

Last edited: Nov 30, 2015
5. Nov 30, 2015

### geoffrey159

Oh I see what you mean, I have been sloppy in my proof. In each cases, I assumed $X$ was a solution. With that assumption, I found restrictions for each cases, which are:
1. $A\subset B$
2. $B\subset A$
3. $A\cap B=\emptyset$.
So that outside these restrictions, there aren't any solution. Then I drew a diagram to find a minimal solution $X_{\text{min}}$ which are:
1. $C_B(A)$
2. $B$
3. $B$
Finally, I tried to find a 'maximal' solution $X = X_{\text{min}} \cup G$. In each case, it has to be true that $G$ belongs to
1. ${\cal P}(A)$
2. ${\cal P}(C_E(A))$
3. ${\cal P}(A)$
Is it ok then ?

6. Nov 30, 2015

### Ray Vickson

I have no idea what $C_E (A)$ means.

7. Nov 30, 2015

### geoffrey159

$C_E(A) = \{ x \in E : x\notin A\}$

8. Nov 30, 2015

### Ray Vickson

9. Nov 30, 2015

### Staff: Mentor

Yes, I am more familiar with $B - A$ or $B \backslash A$. I'm not sure why the author of the text in use in this thread felt the need to come up with new notation when there was existing notation that was clearer.

10. Dec 1, 2015