Question 3 of 12

Question 3 (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (a) $A=\{0,1,2,3\}$, $B=\{1,3,4,5\}$. (b) Three arbitrary sets $A,B,C$ within a universe $U$. (c) Three sets $A,B,C$ satisfying $A\cap C=B\cap C$ and $A\cup C=B\cup C$.

Find. (a) The elements of $(B-A)\times(A-B)$. (b) A proof of the set identity. (c) A proof that $A\cap C^c=B\cap C^c$.

Approach. (a) compute the two difference sets, then the Cartesian product; (b) rewrite $X-Y^c$ as $X\cap Y$ on both sides; (c) chase an arbitrary element $x\in A\cap C^c$ into $B\cap C^c$ and vice versa, using the two hypotheses.

1. (a) $(B-A)\times(A-B)$. $B-A=\{1,3,4,5\}-\{0,1,2,3\}=\{4,5\}$ (elements of $B$ not in $A$). $A-B=\{0,1,2,3\}-\{1,3,4,5\}=\{0,2\}$ (elements of $A$ not in $B$). The Cartesian product pairs every element of the first set with every element of the second: $$(B-A)\times(A-B)=\{4,5\}\times\{0,2\}=\{(4,0),(4,2),(5,0),(5,2)\}$$ $\boxed{\{(4,0),(4,2),(5,0),(5,2)\}}$
2. (b) Prove $(A\cap B)-C^c=(B\cap C)-A^c$. Use the identity $X-Y^c=X\cap(Y^c)^c=X\cap Y$ (subtracting a complement is the same as intersecting with the original set): $$(A\cap B)-C^c = (A\cap B)\cap C = A\cap B\cap C$$ $$(B\cap C)-A^c = (B\cap C)\cap A = A\cap B\cap C$$ Both sides reduce to the same three-way intersection $A\cap B\cap C$: $$\boxed{(A\cap B)-C^c\ =\ A\cap B\cap C\ =\ (B\cap C)-A^c}$$
3. (c) Prove $A\cap C^c=B\cap C^c$ (i.e. $A-C=B-C$) given $A\cap C=B\cap C$ and $A\cup C=B\cup C$. Take any $x\in A\cap C^c$, so $x\in A$ and $x\notin C$. Since $x\in A\subseteq A\cup C=B\cup C$, either $x\in B$ or $x\in C$; but $x\notin C$, so $x\in B$. Combined with $x\notin C$, this gives $x\in B\cap C^c$. This shows $A\cap C^c\subseteq B\cap C^c$. The argument is symmetric in $A,B$ (both hypotheses are symmetric under swapping $A\leftrightarrow B$), so the reverse containment $B\cap C^c\subseteq A\cap C^c$ holds by the same chase with $A$ and $B$ exchanged. Mutual containment gives equality: $$\boxed{A\cap C^c = B\cap C^c}$$