Sets — solutions

Solutions to the practice problems at the end of Sets.

Problem 1: Set-builder

$\{0, 3, 6, 9\}$ . The multiples of three among $\{0, 1, 2, \ldots, 10\}$ .
$\{0, 2, 6, 12, 20\}$ . The values of $n^2 - n$ for $n = 1, 2, 3, 4, 5$ .
$\{(1, 2), (2, 1)\}$ . The pairs with $a \neq b$ out of the four pairs in $\{1, 2\} \times \{1, 2\}$ .

Problem 2: Operations by hand

$A \cup B = \{1, 2, 3, 4, 5, 6, 7\}$ .
$A \cap B = \{4, 5\}$ .
$A \setminus B = \{1, 2, 3\}$ .
$(A \cup B) \cap C = \{1, 2, 3, 4, 5, 6, 7\} \cap \{2, 4, 6, 8\} = \{2, 4, 6\}$ .
$A \cap B \cap C = \{4, 5\} \cap \{2, 4, 6, 8\} = \{4\}$ .
By inclusion–exclusion with $|A| = 5$ , $|B| = 4$ , $|C| = 4$ , $|A \cap B| = 2$ , $|A \cap C| = 2$ , $|B \cap C| = 2$ , and $|A \cap B \cap C| = 1$ , we get $|A \cup B \cup C| = 5 + 4 + 4 - 2 - 2 - 2 + 1 = 8$ . The union is $\{1, 2, 3, 4, 5, 6, 7, 8\}$ , which checks out.

Problem 3: Inclusion and exclusion

Let $M$ be the set of students taking math and $C$ the set taking computer science. We have $|M| = 35$ , $|C| = 28$ , and $|M \cap C| = 15$ . Then

$|M \cup C| = |M| + |C| - |M \cap C| = 35 + 28 - 15 = 48.$

So 48 students take at least one of the two courses, and $60 - 48 = 12$ take neither.

Problem 4: Subsets of a small set

The eight subsets of $\{a, b, c\}$ are $\emptyset$ , $\{a\}$ , $\{b\}$ , $\{c\}$ , $\{a, b\}$ , $\{a, c\}$ , $\{b, c\}$ , and $\{a, b, c\}$ . In general, a set with $n$ elements has $2^n$ subsets.

Four of these contain $a$ : $\{a\}$ , $\{a, b\}$ , $\{a, c\}$ , and $\{a, b, c\}$ . In general, exactly half of the subsets of a nonempty set contain any one chosen element, because fixing that element to be “in” or “out” splits the subsets into two equal classes.

Problem 5: Prove a set identity

We show that $(A \cup B) \setminus C \subseteq (A \setminus C) \cup (B \setminus C)$ and vice versa.

Forward direction. Let $x \in (A \cup B) \setminus C$ . Then $x \in A \cup B$ and $x \notin C$ , so $x \in A$ or $x \in B$ , and in either case $x \notin C$ . If $x \in A$ , then $x \in A \setminus C$ ; if $x \in B$ , then $x \in B \setminus C$ . Either way, $x \in (A \setminus C) \cup (B \setminus C)$ .

Reverse direction. Let $x \in (A \setminus C) \cup (B \setminus C)$ . Then $x \in A \setminus C$ or $x \in B \setminus C$ . In the first case, $x \in A$ and $x \notin C$ , so $x \in A \cup B$ and $x \notin C$ , so $x \in (A \cup B) \setminus C$ . The second case is symmetric.

Both inclusions hold, so the two sets are equal.

Problem 6: Common letters

def common_letters(s, t):
    return {c.lower() for c in s if c.isalpha()} & {c.lower() for c in t if c.isalpha()}

The two comprehensions build sets of lowercased letters from each string, and & intersects them.

Problem 7: Jaccard by hand

The word sets are $A = \{\text{a, rose, is}\}$ and $B = \{\text{roses, are, red, blue}\}$ . The intersection is $\emptyset$ (no word appears in both), and the union has seven distinct words, so

$J(A, B) = \frac{|A \cap B|}{|A \cup B|} = \frac{0}{7} = 0.$

The score is zero even though the two sentences are intuitively related: “rose” and “roses” count as different words under this tokenization. Reducing words to a common root (stemming or lemmatization) before computing the sets would raise the score.