Solving Equations of Sets in P(E)

[geoffrey159]

Homework Statement

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. $X - A = B$

Homework Equations

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 ?

 

[Ray Vickson]

Homework Statement

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. $X - A = B$

Homework Equations

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 ?

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.

 

[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.

 

[Ray Vickson]

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.

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.

 

[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 ?

 

[Ray Vickson]

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 ?

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

 

[geoffrey159]

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

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

 

[Ray Vickson]

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

When did the notation change? I have always seen, for the complement in the whole set $E$, either $A^c$, $\bar{A}$ or $A^{\prime}$, and for the complement of $A$ in $B$ just $B-A$ or $B\backslash A$. See, eg.,
http://www.rapidtables.com/math/symbols/Set_Symbols.htm or
http://www.mathwords.com/s/set_subtraction.htm .

 

[Mark44]

When did the notation change? I have always seen, for the complement in the whole set $E$, either $A^c$, $\bar{A}$ or $A^{\prime}$, and for the complement of $A$ in $B$ just $B-A$ or $B\backslash A$. See, eg.,
http://www.rapidtables.com/math/symbols/Set_Symbols.htm or
http://www.mathwords.com/s/set_subtraction.htm .

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.

 

[geoffrey159]

Sorry about this, I didn't know it wasn't an international notation !