That is, how to represent set $A$ in terms of D, B, C given the equation above? Or is it possible?
N.B. To represent a set is to define this set, i.e. $A = \cdots$, not $A \subseteq \cdots$.
Asked
Active
Viewed 116 times
-1
-
Here is a diagram to accompany answers you've been already given. You can see that $D$ (the shaded area) does not depend in any way on elements in $A\cap C\setminus B$. (This is one of the areas in the picture.) – Martin Sleziak Nov 01 '12 at 09:31
1 Answers
1
If $D\cap C=\varnothing$ and $B\subseteq D\cup C$, the solutions are the sets $A$ such that $D\setminus B\subseteq A\subseteq D\cup C$. Otherwise, there is no solution.
To see this, picture $B$ as the upper half of a square, $C$ as its left half and $D$ as its right half. Then $A$ must contain the lower-right quarter and must not meet the top-left quarter.
Edit: There exists a unique solution $A$ if and only if $D\cap C=\varnothing$, $B\subseteq D\cup C$ (existence) and $D\setminus B=D\cup C$ (uniqueness). These conditions are equivalent to $B=C=\varnothing$, then the unique solution is $A=D$ (obviously).
Did
- 279,727
-
Again, please provide an equation, instead of the inclusion relation among sets. p.s. By the equation given, your 'if' is surely valid. – MeadowMuffins Nov 01 '12 at 09:39
-
1I can give you an equation, but not without lying to you, and I don't want to do that in public. – Harald Hanche-Olsen Nov 01 '12 at 10:26
-
Meadow: Again is inaccurate here (and rude) since you modified your question after I posted my answer. Anyway, please read what I wrote: the solutions are the sets A such that... means that every such set A is a solution and that no other set A is--so, whether you like it or not, this is the answer. (Just as the real numbers $x$ solving $\sqrt{x^2-2x^3+x^4}=x-x^2$ are those such that $0\leqslant x\leqslant1$, all of them and none other.) – Did Nov 01 '12 at 10:57
-
Meadow: Somehow in the same vein, I am rather baffled that you still did not see fit to accept this answer. – Did Nov 01 '12 at 11:14