MHB Set Theory Proofs: A, B, and C - Solving for Set Equality and Complements

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The discussion focuses on proving set equality and complements through three specific problems involving arbitrary sets A, B, and C. For part (a), the user demonstrates that if the intersections and unions of sets A with B and C are equal, then B must equal C, although they need clarification on expressing the implications clearly. In part (b), the user correctly identifies that if the difference between sets A and B is equal, then A must equal B. The main challenge lies in part (c), where the user seeks assistance in proving that if the intersections of the sets are equal and their union equals the universal set U, then the symmetric difference of A, B, and C equals U. A suggested approach involves using the characteristic function to establish the proof.
MikeLandry
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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
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then
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. Suppose then that
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then
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and thus
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. Contradiction.

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

b)
AB=ABc where Bc is the complement of B.

Now if AB then (x)[xABc or xBAc]for
 
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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
 
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
 
MikeLandry said:
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 $A\cap B=A\cap C=B\cap C$.
 
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