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Consider any three arbitrary sets $A$, $B$ and $C$.

  1. Show that $C \cap A = C \cap B$ and $C \cup A = C \cup B$, then $A = B$.
  2. Show that if $A − B = B − A$, then $A = B$.
  3. 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:

  1. 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 $x \in B$.

  2. 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 $\lnot B$?

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For $1$, as $C\cap A=C\cap B$, then if $x\in C\cap A\implies x\in C\cap B$ – Aang Mar 8 '13 at 7:55
You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. – Zev Chonoles Mar 8 '13 at 7:57
Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be closed, see meta. – Martin Sleziak Mar 8 '13 at 8:10
What means $A\oplus B$ for the sets? – user63181 Mar 8 '13 at 8:18
Thanks everyone! I owe you guys one. – MatthewL Mar 8 '13 at 8:27
up vote 2 down vote accepted
  1. 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 $.

  2. 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.

  3. 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 $.

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Thankyou! This was very helpful, completely makes sense now. The empty set part on #2 is a bit off, but that is okay, just needs some time to settle in. Thanks again. – MatthewL Mar 8 '13 at 8:24
for number 3, why does A ⊂ C ? – user65735 Mar 8 '13 at 8:40

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.

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Thankyou Sbr! I try to stick away from the full on answers, it really is quite helpful to actually be given a fork rather than spoon fed. Thanks for the quick reply! Your post was very helpful – MatthewL Mar 8 '13 at 8:30

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 $A\subset B$.

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Thankyou Babak! Your notation is a little wiered, and your approach feels different, and i see what you're saying. Thankyou Babak! I got it now. Thanks a ton – MatthewL Mar 8 '13 at 8:36
Nice! Thank you, yes! :+) – amWhy Mar 9 '13 at 0:05

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 $A=A\cap(B\cup \bar B)=(A\cap B)\cup(A\cap \bar B)=(A\cap B)\cup(B\cap \bar A)=B\cap(A\cup \bar A)=B$

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Oh thankyou! I thought there was a different way of going around it. Thanks. – MatthewL Mar 8 '13 at 8:22

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