Universal Set

1.7 Universal Set

When we study sets, we rarely work in complete isolation. Almost every discussion about sets takes place within some larger, well-understood collection of objects. For instance, when a number theorist talks about "the set of even numbers" or "the set of prime numbers," they are implicitly working inside the set of natural numbers (or integers, or real numbers). That larger, all-encompassing set — the one that contains every object under consideration in a given context — is called the universal set.

The universal set is denoted by the letter $U$. All other sets we discuss in that context are subsets of $U$. We typically label these subsets with capital letters like $A$, $B$, $C$, and so on.

Examples to fix the idea:

• If you are studying the system of numbers, a natural choice for $U$ is the set of real numbers $\mathbb{R}$. Then the set of natural numbers $\mathbb{N}$, the set of integers $\mathbb{Z}$, the set of rational numbers $\mathbb{Q}$ — all are subsets of $U$.
• In human population studies, the universal set could be the set of all people in the world. Then "the set of all left-handed people" or "the set of all people born in January" are subsets of $U$.
• For the set of all integers, you could choose $U = \mathbb{Q}$ (the rational numbers) or $U = \mathbb{R}$ (the real numbers). Both are valid universal sets because every integer is a rational number and every integer is a real number.

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Choosing a Universal Set

The textbook gives two explicit examples to show how we choose $U$ in practice.

Example 1: For the set of all integers, the universal set can be the set of rational numbers $\mathbb{Q}$, or the set of real numbers $\mathbb{R}$. Both work because $\mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$.

Example 2: In human population studies, the universal set consists of all the people in the world. Any subset — such as "people who speak Hindi" or "people over 6 feet tall" — is a subset of this $U$.

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Exercises from the Textbook (with Reasoning)

The textbook includes a set of exercises at the end of this section. These exercises test your understanding of subsets, the universal set, and the distinction between $\in$ (element of) and $\subset$ (subset of). We go through them systematically.

Exercise 1.3, Question 1: Fill in $\subset$ or $\not\subset$

(i) $\{2, 3, 4\} \_\_\_ \{1, 2, 3, 4, 5\}$

Every element of the first set (2, 3, 4) is also in the second set. So $\{2, 3, 4\} \subset \{1, 2, 3, 4, 5\}$.

(ii) $\{a, b, c\} \_\_\_ \{b, c, d\}$

The element $a$ is in the first set but not in the second. So $\{a, b, c\} \not\subset \{b, c, d\}$.

(iii) $\{x : x \text{ is a student of Class XI of your school}\} \_\_\_ \{x : x \text{ is a student of your school}\}$

Every Class XI student is a student of the school. So the first set is a subset of the second. Answer: $\subset$.

(iv) $\{x : x \text{ is a circle in the plane}\} \_\_\_ \{x : x \text{ is a circle in the same plane with radius 1 unit}\}$

The first set contains all circles (of any radius). The second set contains only circles of radius 1. A circle of radius 2, for example, is in the first set but not in the second. So the first set is not a subset of the second. Answer: $\not\subset$.

(v) $\{x : x \text{ is a triangle in a plane}\} \_\_\_ \{x : x \text{ is a rectangle in the plane}\}$

No triangle is a rectangle. The two sets have no common elements. So $\{x : x \text{ is a triangle}\} \not\subset \{x : x \text{ is a rectangle}\}$.

(vi) $\{x : x \text{ is an equilateral triangle in a plane}\} \_\_\_ \{x : x \text{ is a triangle in the same plane}\}$

Every equilateral triangle is a triangle. So the first set is a subset of the second. Answer: $\subset$.

(vii) $\{x : x \text{ is an even natural number}\} \_\_\_ \{x : x \text{ is an integer}\}$

Every even natural number (2, 4, 6, …) is an integer. So $\{x : x \text{ is an even natural number}\} \subset \{x : x \text{ is an integer}\}$.

Exercise 1.3, Question 2: True or False

(i) $\{a, b\} \not\subset \{b, c, a\}$

The set $\{a, b\}$ has elements $a$ and $b$. Both are in $\{b, c, a\}$. So $\{a, b\} \subset \{b, c, a\}$. The statement says $\{a, b\} \not\subset \{b, c, a\}$, which is false.

(ii) $\{a, e\} \subset \{x : x \text{ is a vowel in the English alphabet}\}$

The vowels are $a, e, i, o, u$. Both $a$ and $e$ are vowels. So $\{a, e\}$ is a subset. True.

(iii) $\{1, 2, 3\} \subset \{1, 3, 5\}$

The element 2 is in the first set but not in the second. So $\{1, 2, 3\} \not\subset \{1, 3, 5\}$. The statement is false.

(iv) $\{a\} \subset \{a, b, c\}$

The only element of $\{a\}$ is $a$, which is in $\{a, b, c\}$. So $\{a\} \subset \{a, b, c\}$. True.

(v) $\{a\} \in \{a, b, c\}$

The set $\{a\}$ is not an element of $\{a, b, c\}$; the elements of $\{a, b, c\}$ are $a$, $b$, and $c$ (individual letters, not sets). So $\{a\} \notin \{a, b, c\}$. The statement is false.

(vi) $\{x : x \text{ is an even natural number less than 6}\} \subset \{x : x \text{ is a natural number which divides 36}\}$

The first set is $\{2, 4\}$ (even natural numbers less than 6). The second set is $\{1, 2, 3, 4, 6, 9, 12, 18, 36\}$ (all natural numbers that divide 36). Both 2 and 4 are in the second set. So $\{2, 4\} \subset \{1, 2, 3, 4, 6, 9, 12, 18, 36\}$. True.

Exercise 1.3, Question 3: Let $A = \{1, 2, \{3, 4\}, 5\}$

This is a tricky set because one of its elements is itself a set: $\{3, 4\}$. We must be careful with the difference between $\in$ and $\subset$.

(i) $\{3, 4\} \subset A$ — Incorrect. The elements of $A$ are $1$, $2$, $\{3, 4\}$, and $5$. The set $\{3, 4\}$ is an element of $A$, not a subset. For $\{3, 4\}$ to be a subset of $A$, every element of $\{3, 4\}$ (i.e., 3 and 4) would have to be in $A$. But 3 and 4 are not elements of $A$ (only the set $\{3, 4\}$ is). So $\{3, 4\} \not\subset A$.

(ii) $\{3, 4\} \in A$ — Correct. The set $\{3, 4\}$ is explicitly listed as an element of $A$.

(iii) $\{\{3, 4\}\} \subset A$ — Correct. The set $\{\{3, 4\}\}$ has one element: the set $\{3, 4\}$. Since $\{3, 4\}$ is an element of $A$, the singleton set $\{\{3, 4\}\}$ is a subset of $A$.

(iv) $1 \in A$ — Correct. 1 is an element of $A$.

(v) $1 \subset A$ — Incorrect. The symbol $\subset$ is used between sets. $1$ is not a set (it is a number), so $1 \subset A$ is meaningless. Even if we interpret it as $\{1\} \subset A$, that would be correct, but the statement as written is incorrect.

(vi) $\{1, 2, 5\} \subset A$ — Correct. All elements 1, 2, 5 are in $A$.

(vii) $\{1, 2, 5\} \in A$ — Incorrect. The set $\{1, 2, 5\}$ is not listed as an element of $A$. The elements are 1, 2, $\{3, 4\}$, and 5.

(viii) $\{1, 2, 3\} \subset A$ — Incorrect. The element 3 is not in $A$ (only the set $\{3, 4\}$ is). So $\{1, 2, 3\} \not\subset A$.

(ix) $\phi \in A$ — Incorrect. The empty set $\phi$ is not listed as an element of $A$.

(x) $\phi \subset A$ — Correct. The empty set is a subset of every set.

(xi) $\{\phi\} \subset A$ — Incorrect. The set $\{\phi\}$ has one element: $\phi$. Since $\phi$ is not an element of $A$, $\{\phi\}$ is not a subset of $A$.

Exercise 1.3, Question 4: Write all subsets

(i) $\{a\}$: Subsets are $\phi$ and $\{a\}$.

(ii) $\{a, b\}$: Subsets are $\phi$, $\{a\}$, $\{b\}$, $\{a, b\}$.

(iii) $\{1, 2, 3\}$: Subsets are $\phi$, $\{1\}$, $\{2\}$, $\{3\}$, $\{1, 2\}$, $\{1, 3\}$, $\{2, 3\}$, $\{1, 2, 3\}$.

(iv) $\phi$: The only subset is $\phi$ itself.

Exercise 1.3, Question 5: Write as intervals

(i) $\{x : x \in \mathbb{R}, -4 < x \leq 6\}$: This is $(-4, 6]$.

(ii) $\{x : x \in \mathbb{R}, -12 < x < -10\}$: This is $(-12, -10)$.

(iii) $\{x : x \in \mathbb{R}, 0 \leq x < 7\}$: This is $[0, 7)$.

(iv) $\{x : x \in \mathbb{R}, 3 \leq x \leq 4\}$: This is $[3, 4]$.

Exercise 1.3, Question 6: Write intervals in set-builder form

(i) $(-3, 0)$: $\{x : x \in \mathbb{R}, -3 < x < 0\}$

(ii) $[6, 12]$: $\{x : x \in \mathbb{R}, 6 \leq x \leq 12\}$

(iii) $(6, 12]$: $\{x : x \in \mathbb{R}, 6 < x \leq 12\}$ …