04-BS-16 · December 2013

Question 2 of 12: Sets — Union, Intersection, Cartesian Product, Power Set

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examination, 04-BS-16 Discrete Mathematics, Dec 2013. Closed book, no aids, 3 hours, 12 questions of 10 marks each (100 marks); the exam instructs "answer 10 of 12" but every question is solved below as a complete study resource.

Reference texts: Rosen, Discrete Mathematics and Its Applications, 7th ed. (logic Ch.1, sets Ch.2, induction & pigeonhole Ch.5-6, relations Ch.9, counting Ch.6, discrete probability Ch.7, graphs Ch.10-11); Epp, Discrete Mathematics with Applications.

Question 2: Sets — Union, Intersection, Cartesian Product, Power Set (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. $A=\{1,2,3\}$ (elements $1,2,3$); $B=\{1,\{2\}\}$ (two elements: the number $1$ and the SET $\{2\}$ — not the number $2$); $C=\{\emptyset\}$ (one element: the empty set).

Find. The six set operations in part (a) and the four truth values in part (b).

(a) The key distinction. $B$'s elements are $1$ and $\{2\}$ (a set, not the integer $2$) — this governs every operation below. $$A\cup B=\boxed{\{1,2,3,\{2\}\}},\qquad A\cap B=\boxed{\{1\}}\ (\text{only }1\text{ is common; }2\ne\{2\}),\qquad A-B=\boxed{\{2,3\}}.$$
(a) Cartesian product and cardinality. $B\times C$ pairs every element of $B$ with the single element of $C$: $$B\times C = \boxed{\{(1,\emptyset),\ (\{2\},\emptyset)\}},\qquad |C|=\boxed{1}\ (\text{C has one element: }\emptyset).$$
(a) Power set of $B$. $B$ has $|B|=2$ elements, so $|\mathcal P(B)|=2^2=4$: $$\mathcal P(B)=\boxed{\{\emptyset,\ \{1\},\ \{\{2\}\},\ \{1,\{2\}\}\}}.$$
(b-i) $2\in B$? $B$'s elements are $1$ and $\{2\}$; the bare integer $2$ is not one of them. $\boxed{\text{False}}$.
(b-ii) $\emptyset\in C$? $C=\{\emptyset\}$ literally contains $\emptyset$ as its element. $\boxed{\text{True}}$.
(b-iii) $\emptyset\subseteq C$? The empty set is a subset of every set (vacuously — it has no elements that could fail to be in $C$). $\boxed{\text{True}}$.
(b-iv) $|A|=|B\cup C|$? $|A|=3$. $B\cup C=\{1,\{2\},\emptyset\}$, which has $3$ distinct elements, so $|B\cup C|=3$. $3=3$: $\boxed{\text{True}}$.

Final results — Question 2
Part	Result
$A\cup B$	$\{1,2,3,\{2\}\}$
$A\cap B$	$\{1\}$
$A-B$	$\{2,3\}$
$B\times C$	$\{(1,\emptyset),(\{2\},\emptyset)\}$
$\|C\|$	1
Power set of $B$	4 subsets — see boxed list
$2\in B$	False
$\emptyset\in C$	True
$\emptyset\subseteq C$	True
$\|A\|=\|B\cup C\|$	True (3 = 3)

Topic: Set membership vs. subset, sets-of-sets, Cartesian products, power sets.

Key relations: $|\mathcal P(S)|=2^{|S|}$; $\emptyset\subseteq S$ for every set $S$; $x\in\{x\}$ but a set containing $\{x\}$ does NOT contain $x$ itself unless $x$ is separately listed.

Why this works: $B=\{1,\{2\}\}$ is a set with a set as one of its two elements — treating $\{2\}$ as "the same as" $2$ is the single most common error on this question, and every part hinges on keeping that distinction (it changes $A\cap B$, $2\in B$, and the power-set listing).

Common pitfall: Confusing $\in$ (membership) with $\subseteq$ (subset) — $\emptyset\in C$ is true because $\emptyset$ is literally an element of $C$, while $\emptyset\subseteq C$ is true for the unrelated reason that the empty set is a subset of everything.

Source: Rosen, Discrete Mathematics and Its Applications, §2.1-2.2 (set operations, power sets, Cartesian products).

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