How would I prove that $(A-B)\cup (B-A)=(A\cup B)-(A\cap B)$ for arbitrary sets $A$ and $B$?
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$\begingroup$ What have you tried so far? Are you aware that saying $X=Y$ means $X\subseteq Y$ and $Y\subseteq X$ for sets $X$ and $Y$? $\endgroup$– Kevin LongOct 2, 2017 at 4:44
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$\begingroup$ By showing both: 1) every element of the LHS is an element of the RHS, and 2) every element of the RHS is an element of the LHS. $\endgroup$– David MitraOct 2, 2017 at 4:45
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$\begingroup$ Maybe a visual would help $\endgroup$– gen-ℤ ready to perishOct 2, 2017 at 4:50
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$\begingroup$ Draw a Venn diagram, with intersection A,B.If they do not intersect, what then? $\endgroup$– Peter SzilasOct 2, 2017 at 5:52
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2 Answers
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\begin{align} (A-B) \cup (B-A) &= (A \cap \bar{B}) \cup (B \cap \bar{A}) \\ &= [(A\cap \bar{B} )\cup \emptyset] \cup [(B \cap \bar{A}) \cup \emptyset]\\ &= [(A\cap \bar{B} )\cup (B \cap \bar{B})] \cup [(B \cap \bar{A}) \cup (A \cap \bar{A})] \\ &= [\bar{B}\cap(A\cup B)] \cup [\bar{A} \cap (A\cup B)] \\ &= (A\cup B) \cap (\bar{A} \cup \bar{B}) \\ &= (A\cup B) \cap \overline{(A \cap B)} \\ &= (A\cup B) - (A \cap B) \end{align}
answered Oct 2, 2017 at 5:02
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1$\begingroup$ Seems legit. Maybe the
\alignenvironment would tidy things up? $\endgroup$ Oct 2, 2017 at 5:10
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Consider the follow 4 sets.
$A\setminus B \subset A$.
$B\setminus A \subset B$
$A\cup B \supset A$ and $A\cup B \supset B$.
$A\cap B \subset A$ and $A\cap B \subset B$.
If $x \in A$ then $x\not \in B\setminus A; x\in A\cup B$. We don't know about the rest.
If $x \not \in A$ then $x \not \in A \setminus B$ and $x\not \in A\cap B$. We don't know about the rest.
If $x \in B$ then $x\not \in B\setminus A; x\in A\cup B$. We don't know about the rest.
If $x \not \in B$ then $x \not \in B \setminus A$ and $x\not \in A\cap B$. We don't know about the rest.
1) If $x \in A$ and $x \in B$ then $x\not \in B\setminus A; x\in A\cup B;x\not \in A\setminus B; x\in A\cap B$
So $x \not \in A\setminus B \cup B \setminus A$ and $x \not \in (A\cup B)\setminus (A\cap B)$.
2) If $x \in A$ and $x \not \in B$ then $x\not \in B\setminus A; x\in A\cup B;x \in A\setminus B; x\not\in A\cap B$
So $x \in A\setminus B \cup B \setminus A$ and $x \in (A\cup B)\setminus (A\cap B)$.
3) If $x \not \in A$ and $x \in B$ then $x \in B\setminus A; x\in A\cup B;x\not \in A\setminus B; x\not\in A\cap B$
So $x \in A\setminus B \cup B \setminus A$ and $x \in (A\cup B)\setminus (A\cap B)$.
4) If $x \not \in A$ and $x \not \in B$ .... well then $x$ is not in ANY of the sets.
So.... $ A\setminus B \cup B \setminus A$ and $x \in (A\cup B)\setminus (A\cap B)$ have precisely the same elements.
So they are equal.
