Solving Equations of Sets in P(E)

  • Thread starter geoffrey159
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To be honest, I've always seen this notation in topology and measure theory, but yes, it seems to be a bad idea to use a notation with E in it when you already defined it in the statement...Sorry about this, I didn't know it wasn't an international notation !.To be honest, I've always seen this notation in topology and measure theory, but yes, it seems to be a bad idea to use a notation with E in it when you already defined it in the statement...No problem, it's just confusing when the notation is not consistent. But your summary is correct and clear.
  • #1
geoffrey159
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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 ?
 
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  • #2
geoffrey159 said:

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.
 
  • #3
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
geoffrey159 said:
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.
 
Last edited:
  • #5
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
geoffrey159 said:
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.
 
  • #7
Ray Vickson said:
I have no idea what ##C_E (A)## means.
##C_E(A) = \{ x \in E : x\notin A\}##
 
  • #9
Ray Vickson said:
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.
 
  • #10
Sorry about this, I didn't know it wasn't an international notation !
 

1. What is P(E) in the context of solving equations of sets?

P(E) refers to the power set of a set E, which is the set of all possible subsets of E. In other words, P(E) includes all the distinct combinations of elements that can be formed from the elements in set E.

2. How do you represent equations of sets using P(E)?

Equations of sets can be represented using set notation and operations such as union, intersection, and complement. For example, the equation A ∪ B = C can be written as P(A) ∪ P(B) = P(C), where P represents the power set operation.

3. What is the purpose of solving equations of sets using P(E)?

Solving equations of sets using P(E) allows us to find the solutions for complex sets by breaking them down into simpler sets and using set operations. It also helps in visualizing and understanding the relationships between different sets.

4. Can P(E) be used to solve equations with more than two sets?

Yes, P(E) can be used to solve equations with any number of sets. The same principles of set operations and breaking down complex sets into simpler ones can be applied to solve equations with multiple sets.

5. Are there any limitations to using P(E) for solving equations of sets?

While P(E) is a powerful tool for solving equations of sets, it may not be suitable for solving very large or infinite sets. In such cases, other methods such as Venn diagrams or algebraic techniques may be more efficient.

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