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I am trying to prove that if $A – C = B – C$ and $A\cap C = B\cap C$ then $A = B$. I have tried using Venn Diagrams as a proof technique, but we are not able to use proof by Venn Diagrams.

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    Hint: For all sets U and V, we have $U = (U - V) \cup (U \cap V)$. – 2017-02-14

5 Answers 5

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By symmetry of the question, it suffices to show $A\subset B$. (The proof will work for $B\subset A$.) Let $x\in A$; we need to show $x\in B$.

Step 1: Suppose $x\in C$. Then $x$ is in the intersection $A\cap C$. Conclude with the hypothesis.

Step 2: Suppose $x\notin C$. Then $x\in A-C$. Again we can conclude easily.

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    I was 15 seconds too late, but I was pleased to see your name on the answer. I'm pretty sure you were my Geometry TA in my first year at EPFL. Is that possible? :) – 2017-02-14
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    Ahah maybe ;). (cf mon profil) :P C'est ton vrai prénom John? – 2017-02-14
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    Non, c'est Giovanni (Tessinois). J'ai commencé en 2011! – 2017-02-14
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Let $x\in A$. We have two cases:

  • $x\in C$. Then $x\in A\cap C=B\cap C$, hence $x\in B$.
  • $x\notin C$. Then $x\in A-C=B-C$, hence $x\in B$

Same goes for the opposite direction.

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$A= (A \cap C) \cup (A \cap C^C) = (A \cap C) \cup (A-C) = (B \cap C) \cup (B -C) = (B \cap C) \cup (B \cap C^C)=B$

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    $C^C$? What it is? – 2017-02-15
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    @mle $C^C$ is the set of objects *not* in $C$. – 2017-02-15
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In a set theory proof of $A=B$, you must prove two statements:

  1. If $x\in A$, then $x\in B$. (This proves $A\subseteq B$.)

  2. If $x\in B$, then $x\in A$. (This proves $B\subseteq A$.)

Let's try to prove that if $x\in A$, then $x\in B$. So, we assume that $x\in A$, and our goal is to conclude that $x\in B$. Now, there are two cases, either $x\in C$ or $x\not\in C$. We can consider each case separately.

If $x\in C$, then since $x\in A$ was assumed, $x\in A\cap C$. Since $A\cap C=B\cap C$, then $x\in B\cap C$. Hence $x\in B$ and $x\in C$. Therefore, $x\in B$.

If $x\not\in C$, then since $x\in A$ was assumed, $x\in A-C$, but since $A-C=B-C$, $x\in B-C$. Therefore, $x\in B$ and $x\not\in C$. Therefore, $x\in B$.

Since either case leads to $x\in B$, we know that $x\in A$ implies that $x\in B$. Hence $A\subseteq B$.

Now, you must prove the other direction.

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Let $x \in X$.

If $x \in A$ and $x \in C$ then $x \in A\cap C = B \cap C$ so $x \in A$.

if $x \in A$ and $x \not \in C$ then $x \in A \setminus C = B \setminus C$ so $x \in B$.

if $x \not \in A$ and $x \in C$ then $x \not \in A\cap C=B\cap C$. So either $x \not \in B$ or $x \not \in C$. But $x \in C$ so $x \not \in B$.

If $x \not \in A$ and $x \not \in C$ then $x \not \in A \setminus C$. So if $x \in B$ then $x \not \in C$ so $x \in B \setminus C = A \setminus C$. That's not possible so $x \not \in B$.

So $x \in A \iff x \in B$. So $A = B$

.... OR ..... $(D \setminus E)\cup (D \cap E) = \{x|x \in D; x \not \in E\} \cup \{x|x \in D; x \in E\} = \{x|x \in D; $ and either $x \in E$ or $x \not \in E\}=$ .... but all $x$ are either in $E$ or not in $E$ ... $= \{x|x \in D \} = D$.

Therefore $A = (A\setminus C) \cup (A\cap C) = (B\setminus C) \cup (B\cap C) = B$.

......

or

If $x \in A$ then if $x \in C$ $x \in A\cap C = B\cap C$ so $x \in B$.

If $x \not \in C$ then $x \in A \setminus C = B\setminus C$ so $x \in B$.

So $x \in A\implies x \in B$ so $A \subset B$.

Do the same to show $B \subset A$. So $A \subset B$ and $B \subset A$. So $A = B$.