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Let $A$ and $B$ be sets. Suppose that there is some set $C$ such that $A \cap C = B \cap C$ and $A \cup C = B \cup C$. Show that $A = B$.

My sketch is the following. Suppose first that sets are non-empty and instead that $A$ were not equal to $B$. Then there must exist an element in either $A$ or $B$ that is not in the other. In particular, choose a in $A\setminus B$. Then this a must be in $A \cap C = B \cap C$. But this contradicts the fact that a was not in $B$.

In fact, I am interested in the logic of my steps. Any comment would be helpful.

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    " In particular, choose a in A\B. Then this a must be in A ∩ C = B ∩ C" Um, the exact opposite if $a \in A\setminus B$ (which could be empty by the way) then we know for *certain* that $a \not \in B$ so $a \not \in B \cap C$. – 2017-01-28

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$a \in A\backslash B $ does not imply that $a \in A \cap C$. For example, if $A = \{ 1, 2, 3 \}$, $B= \{ 3, 4, 5 \}$, $C=\{1, 3, 5 \}$.

Rather, it would be helpful to prove that $x \in A $ implies $x \in B$ and vice versa.

If $x \in A$, then $x \in A \cup C$, so $x \in B \cup C$, which is equivalent to $x \in B$ or $x \in C$.

If $x \in B$, we are done. If $x \in C$, then $x \in A \cap C$, so $x \in B \cap C$. This implies $x \in B$. Therefore, we can conclude $A \subset B$. Can you prove the converse?(Just interchange the role of $A$ and $B$.)

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    Thank you. Your answer makes a lot of sense. – 2017-01-30
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1) $A$ and $B$ could be empty.

2) $A \setminus B$ could be empty even if $A$ and $B$ are not. (Although $A\setminus B$ and $B \setminus A$ can not both be empty if $A \ne B$ [even if one or the other of $A$ or $B$ is empty.])

3) If $a \in A\setminus B$ then $a \not \in B$ so $a \not \in B\cap C=A \cap C$. I have utterly no idea why you would say such a thing.

But if $a \in A \setminus B$ then $a \not \in B$ so $a \not \in B \cap C = A \cap C$. As $a \in A$ then $a \not \in A \cap C$ means $a \not \in C$. But $a \in A \cup C = B \cup C$. So $a \in B$ or $a \in C$. But we showed neither of those are possible.

So $A \setminus B = \emptyset$ and so $A \subset B$.

Likewise the exact same argument shows that $B \setminus A$ is empty. So $B \subset A$.

But I'd do it directly in the exact manner that bellcircle did.

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    I say 3) for the following reason. If a is an element of the set A. It is also an element of the set (A and B). However, the question gives us that ( A and B) is equal to ( B and C). This is were the contradiction comes from. Does that follow ? – 2017-01-30
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    Oh, I guess so. But that's reaching a somewhat complicated conclusion without explanation, while overlooking more obvious contradictions. You're right that that does work. To integrate yous and mine I'd say: if a in A/B then a in AUC=BC – 2017-01-30
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    I'd say: a in A/B means a in AUC=BUC but a not in B so a in C, so a in C $\cap $A=B $\cap $C so a in B which is a contradiction. So A/B is empty and so A subset B. Likewise B/A is empty by same argument. That'll work. – 2017-01-30