Adding a condition to make a set theory identity true

• November 28th 2013, 11:17 AM
Ragnarok
Adding a condition to make a set theory identity true
Let A, B, and C be sets. The formula

$A-(B-C) = (A-B)\cup C$

is sometimes false. State an additional necessary and sufficient condition for it to be always true.

(The minus sign here means set difference.) It is pretty easy for me to come up with sufficient conditions (i.e., if $C\subseteq A$ and $B\cap C=\varnothing$, then the equation is true), but I'm not sure how to find a necessary condition.

Any element which is in both sets above must be in A. I think we could have three cases:

1. $x\in A$, $x\not\in B$, $x\in C$.
2. $x\in A$, $x\not\in B$, $x\not\in C$.
3. $x\in A$, $x\in B$, $x\in C$.

But I'm not really sure where to go from there.
• November 28th 2013, 12:02 PM
Plato
Re: Adding a condition to make a set theory identity true
Quote:

Originally Posted by Ragnarok
Let A, B, and C be sets. The formula
$A-(B-C) = (A-B)\cup C$
is sometimes false. State an additional necessary and sufficient condition for it to be always true.

\begin{align*} A\setminus (B\setminus C)&=A\cap (B\cap C^c)^c\\ &= A\cap(B^c\cup C)\\&=(A\cap B^c)\cup(A\cap C)\end{align*}

Now what if $C\subseteq A~?$ Does equality hold?

If $A=\{1,2,3\},~B=\{2,3,4\},~C=\{3,4,5\}$, does equality hold? What is the difference?
• November 29th 2013, 07:40 PM
Ragnarok
Re: Adding a condition to make a set theory identity true
Okay, I'm starting to see. It is a necessary condition that $C\subseteq A$ since if it is not then there is an element which is in the second set but not the first. I think this is a sufficient condition too? I'll go check…

Thank you!