Elementary Set theory: Consider any three arbitrary sets A, B, and C. 
Consider any three arbitrary sets $A$, $B$ and $C$.

*

*Show that $C \cap A = C \cap B$ and $C \cup A = C \cup B$, then $A = B$.

*Show that if $A − B = B − A$, then $A = B$.

*Show that if $A\cap B = A\cap C = B \cap C$ and $A\cup B \cup C = U$, then $A\oplus B \oplus C = U$.


My attempts:

*

*My logic behind it is that I can prove this by showing that $A$ is a subset of $B$ and $B$ is a subset of $A$:  $x \in  C\cap A$ implying $x  \in A$ and $x  \in  C$. But now, i get a little confused about the other side. I can't just say $x  \in  C\cap  B$. But if you look at the $C \cup  A = C \cup  B$, you can say that since $x \in  A$, would that imply $x \in  B$.


*I went about it the same way as 1, trying to state that $A$ is a subset of $B$ and $B$ is a subset of $A$. Changing $A-B$ to $A  \cap\lnot B $, but that means $x \in  A$, and $x \in \lnot B$. Can $x$ be an element of $B$ and $\lnot B$?
 A: Hint for 2: If $A\setminus B=B\setminus A$ then as you noted correctly, $A\cap B'=B\cap A'$. Let $x\in A$, we have two choices: 


*

*$x\in B$

*$x\in B'$


If $x\in B$ then $A\subset B$.
If $x\in B'$ then $x\in A\cap B'=B\cap A'$ and from that we have $x\in B\cap A' $. Since $x\in A$ so $x\in B$ and again $A\subset B$.
A: Hints:
$1.$  two cases: if $x\in C$ then use the first equality: $x\in A \Leftrightarrow x\in B$, and if $x\notin C$ then use the second equality and also $x\in A \Leftrightarrow x\in B$
$2.$ Suppose there's  $x\in A$ and $x\notin B$ then $x\in A-B=B-A$ so $x\in B$. Contradiction.
A: *

*Suppose that $ x \in A $. Now either $ x \in C $ or $ x \notin C $. 
a.If $ x \in C $, then $ x \in C \cap A $ and so $ x \in C \cap B $, which implies $ x \in B $. 
b.Now if $ x \notin C $ then $ x \in C \cup A = C \cup B $ and yet $ x \notin C $ so $ x \in B $ (because otherwise it couldn't be in $ C \cup B $.

*Suppose $ x \in A - B $. Then $ x \notin B $. However, $ x \in B - A $, which implies $ x \in B $. This is a contradiction. Thus, $ A - B = \emptyset $. Similarly, $ B - A = \emptyset $. Therefore, $ A = B $ because this implies that there's nothing in A that's not in B, and vice versa.

*Suppose $ x \in A $. Then $ x \in B $ or $ x \notin B $.
a. If $ x \in B $, then $ x \in A \cap B = B \cap C $. Thus, $ x \in C $.
b. If $ x \notin B $, then $ x \notin A \cap B = A \cap C $. Thus, $ x \notin C. $
This effectively shows that if $ A \subset C $. We could do exactly the same for all the other sets and get $ A = B = C $. This implies that $ A \oplus B \oplus C = (A \oplus B) \oplus C = \emptyset \oplus C = C $. Note that $ C = A \cup B \cup C = U $.
A: For $1.)$, $A=(C\cup A)\cap A=(C\cup B)\cap A=(C\cap A)\cup(B\cap A)=(C\cap B)\cup(B\cap A)=(C\cup A)\cap B=(C\cup B)\cap B=B$
$2.)$  $A-B=A\cap \bar B$, then $A=A\cap(B\cup \bar B)=(A\cap B)\cup(A\cap \bar B)=(A\cap B)\cup(B\cap \bar A)=B\cap(A\cup \bar A)=B$
