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Set Theory Proofs

MikeLandry

New member
Mar 7, 2013
2
I have gotten to this point with a and b but do i am totally lost with c. Any help would be much appreciated

Consider any three arbitrary sets A, B and C.
(a) Show that if A ∩ B = A∩ C and A ∪ B = A ∪ C, then B = C.
(b) Show that if A − B = B − A, then A = B.
(c) Show that if A∩B = A∩C = B ∩C and A∪B ∪C = U, then A⊕B ⊕C = U.

For the three proofs so far i have

a) So A intersects C = A intersects B and A union B= A union C.

Let
then
. Suppose then that
then
and thus
. Contradiction.

Similarly, let
then
. Suppose that
then
and so
. Contradiction

b)

AB=ABc where Bc is the complement of B.

Now if AB then (x)[xABc or xBAc]











for
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Hi MikeLandry,

Welcome to MHB! :)

I think you have the right idea for part one, but I would alter a couple of things. To show two sets, $B$ and $C$ are equal you need to show $B \subseteq C$ and $C \subseteq B$. Put another way $x \in B \implies x \in C$ and $x \in C \implies x \in B$. I think you already showed both of those things by the contrapositive but you didn't write what you showed implies.

Jameson
 

MikeLandry

New member
Mar 7, 2013
2
Thank you very much for your quick reply. I feel confident with my solutions for questions a and b but any insite on how to solve part c would be greatly appreciated
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I feel confident with my solutions for questions a and b but any insite on how to solve part c would be greatly appreciated
An elegant way (but not the only one), is to use the characteristic function. Being $U$ an universal set and $M\subset U$ the characteristic function $1_M:U\to \{0,1\}$ is defined by: $$1_M(x)=\left \{ \begin{matrix} 1 & \mbox{ if }& x\in M \\0 & \mbox{if}& x\not\in M\end{matrix}\right.$$ Using the properties $$\begin{aligned}&M_1=M_2\Leftrightarrow1_{M_1}=1_{M_2}\\&1_{M\cup N}=1_M+1_N-1_M\cdot 1_N\\&1_{M\oplus N}=1_M+1_N-2\cdot1_M\cdot 1_N\end{aligned}$$ and the hypothesis $A\cup B\cup C=U$ (that is, $1_{A\cup B\cup C}=1_U$) you'll easily verify that $A\oplus B\oplus C=U$ iff: $$1_A\cdot 1_B+1_A\cdot 1_C+1_B\cdot 1_C-3\cdot 1_A\cdot 1_B\cdot 1_C=0$$ Now, use the hypothesis $A\cap B=A\cap C=B\cap C$.