MHB Saint's question from Yahoo Answers regarding set theory

[Chris L T521]

Here is the question:

Show that for any three sets $A$; $B$; $C$, we have $A-(B-C) = (A-B) \cup (A\cap C)$.

Here is a link to the question:

Show that for any three sets A; B; C , we have: A - (B -C) = (A-B) U (A ? C)? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Chris L T521]

Hello Saint,

We will need to use four facts. If $A,B,C\subseteq X$ are three sets, then

1. $(A^c)^c = A$
2. $A-B=A\cap B^c$
3. $(A\cap B)^c = A^c\cup B^c$
4. $A\cap(B\cup C) = (A\cap B)\cup(A\cap C)$

Hence, if $A,B,C\subseteq X$, we see that

\[\begin{aligned}A-(B-C) &= A\cap(B-C)^c\\ &= A\cap(B\cap C^c)^c\\ &= A\cap(B^c\cup (C^c)^c)\\ &= A\cap(B^c\cup C) \\ &= (A\cap B^c)\cup(A\cap C) \\ &= (A-B)\cup(A\cap C).\end{aligned}\]

Therefore, $A-(B-C) = (A-B)\cup(A\cap C)$.

I hope this makes sense!