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

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

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