My work

Let $x$ be any arbitrary element of $A-B$ $$A-B=\{x: x\in A,\; x\notin B\}$$ $$=\{x: x\in A,\; x\in B'\}$$ $$=\{x:x\in A\cap B\}$$

How do I proceed further?

  • $\begingroup$ How did you get $x\in A\cap B$? $\endgroup$ – Michael Burr Mar 29 '17 at 14:12
  • $\begingroup$ HINt: $A-B=A\cap B'$ $\endgroup$ – Mojtaba Mar 29 '17 at 14:12
  • $\begingroup$ Take A-B=A, now you need to prove that there is no common element between A and B. Assume the contrary and can try to get a contradiction? $\endgroup$ – danny Mar 29 '17 at 14:12
  • $\begingroup$ what part of the statement do you want to show in your attempt? $\endgroup$ – supinf Mar 29 '17 at 14:13
  • 1
    $\begingroup$ $\{x: x\in A,\; x\in B'\}=\{x:x\in A\cap B'\}$ $\endgroup$ – Mojtaba Mar 29 '17 at 14:20


${\color{Red} {\text{to proof}}}$: $A\cap B=\emptyset$:

  1. We can use Symmetric Difference of Equal Sets and $(A\setminus B)\cap B=\emptyset$ in fact: $$ \begin{align} (A\cap B)\Delta \emptyset&= ((A \cap B) \setminus \emptyset) \cup (\emptyset \setminus (A \cap B)) \\ &=((A \cap B) \setminus \emptyset) \cup \emptyset \\ &=((A \cap B) \setminus \emptyset)\\ &=(A \cap B) \setminus ((A\setminus B)\cap B) \\ &=(A \cap B) \setminus (A\cap B) \text{ (because we have that } A \setminus B= A) \\ &=\emptyset \\ &{\color{Red} \Longrightarrow} A \cap B=\emptyset \end{align}$$

  1. We can use $(A\setminus B)\cap B=\emptyset$ and " $A \setminus B= A\cap C \to A\cap B\cap C=\emptyset$ " in fact: $$\begin{align}A \cap B\cap C &= (A\cap C) \cap B \\ &= (A \setminus B) \cap B \text{ (because we have that }A \setminus B= A \cap C)\\ &=\emptyset \\ &{\color{Red} \Longrightarrow}A\cap B=\emptyset \text{ (because we have that } C=A) \end{align}$$
  1. Let $A-B=A$.

Assume for the sake of contradiction that $x\in A \cap B$. Then, in particular, $x\in A=A-B$, so $x\notin B$, which is a contradiction with $x\in A \cap B$.

  1. Let $A\cap B=\emptyset$.

a) If $x\in A$ then, since $A\cap B=\emptyset$, $x\notin B$, so $x\in A-B$. Thus $A\subset A-B$.

b) If $x\in A-B$ then, by definition, $x\in A$, so $A-B\subset A$.

In conclusion, $ A-B=A$.



$A\backslash B = A\backslash(A\cap B)$



For the forward direction.

  • To prove a set is empty, the typical approach is by contradiction.

  • Assume for contradiction that $A\cap B\not=\emptyset$ and $A\setminus B=A$, so let $x\in A\cap B$ ($x$ exists because $A\cap B$ is not empty).

  • Can you prove that $x\in A$, but $x\not\in A\setminus B$? Does this give a contradiction?

For the backwards direction,

  • Assume that $A\cap B=\emptyset$.

  • Since $A\setminus B\subseteq A$, all you need to show is that if $x\in A$, then $x\in A\setminus B$.

  • Since $x\in A$, $x\not\in B$ as $A\cap B=\emptyset$, so $x\in A\setminus B$.

You could also do the backwards direction using the contrapositive.

  • Suppose that $A\setminus B\not=A$.

  • Since $A\setminus B\subseteq A$, there is some $x\in A$ such that $x\not\in A\setminus B$.

  • Prove that $x\in B$ and get that $A\cap B\not=\emptyset$.


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