Samacheer Kalvi Class 11 Maths Solutions Chapter 1 Sets, Relations and Functions Exercise 1.1

Question 1. Write the following in roaster form.
(i) \( \{x \in N : x^2 < 121 \text{ and } x \text{ is a prime}\} \)
(ii) The set of positive roots of the equation \( (x - 1) (x + 1) (x - 1 ) = 0 \)
(iii) \( \{x \in N : 4x + 9 < 52\} \)
(iv) \( \{x : \frac{x-4}{x+2} = 3, x \in R - \{-2\}\} \)
Answer:
(i) First, we find natural numbers \( x \) whose square is less than 121. This means \( x < 11 \). Next, we pick the prime numbers from this list. A prime number is a whole number greater than 1 that has only two divisors: 1 and itself.
The prime numbers less than 11 are 2, 3, 5, 7.
So, \( A = \{2, 3, 5, 7\} \)
(ii) We need to find the positive solutions for the given equation. We simplify the equation by combining terms.
\( (x - 1) (x + 1) (x - 1) = 0 \)
\( \implies (x + 1) (x - 1)^2 = 0 \)
This means either \( (x + 1) = 0 \) or \( (x - 1)^2 = 0 \)
\( \implies x = -1 \) or \( x = 1 \)
The positive root is \( x = 1 \). So, \( A = \{1\} \)
(iii) We need to find natural numbers \( x \) that satisfy the inequality. We solve the inequality step-by-step to find the possible values of \( x \).
\( 4x + 9 < 52 \)
\( \implies 4x + 9 - 9 < 52 - 9 \)
\( \implies 4x < 43 \)
\( \implies x < \frac{43}{4} \)
\( \implies x < 10.75 \)
Since \( x \) must be a natural number (positive whole numbers), the natural numbers less than 10.75 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
So, \( A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)
(iv) We need to solve the equation for \( x \). We are told that \( x \) cannot be -2, because that would make the denominator zero.
\( \frac{x-4}{x+2} = 3 \)
\( \implies x - 4 = 3(x + 2) \)
\( \implies x - 4 = 3x + 6 \)
\( \implies x - 3x = 6 + 4 \)
\( \implies -2x = 10 \)
\( \implies x = \frac{-10}{2} \)
\( \implies x = -5 \)
So, \( A = \{-5\} \)
In simple words: To write a set in roaster form, list all the elements of the set inside curly brackets, separated by commas. Make sure each element meets all the conditions given in the set builder form.

🎯 Exam Tip: Always pay attention to the domain of the variable (e.g., natural numbers, integers, real numbers) as it affects the elements included in the set. Also, ensure you apply all conditions (like primality or inequalities) when listing elements.

Question 2. Write the set \(\{-1, 1\}\) in set builder form.
Answer: To write this set in set builder form, we need to find a property that both -1 and 1 share, and no other number does. We can notice that these are the roots of the equation \( x^2 - 1 = 0 \).
\( A = \{x : x^2 - 1 = 0, x \in R\} \)
In simple words: Set builder form describes the properties that all elements in a set have. Here, the numbers are the solutions to \( x^2 - 1 = 0 \).

🎯 Exam Tip: When converting to set builder form, try to find a simple equation or inequality that generates exactly the given elements, and always specify the universal set for the variable (e.g., \( x \in R \) for real numbers).

Question 3. State whether the following sets are finite or infinite.
(i) \( \{x \in N : x \text{ is an even prime number}\} \)
(ii) \( \{x \in N: x \text{ is an odd prime number}\} \)
(iii) \( \{x \in Z : x \text{ is even and } < 10\} \)
Answer:
(i) First, we identify the even prime numbers. The only prime number that is even is 2. All other prime numbers are odd. Therefore, this set has only one element.
Let \( A = \{x \in N : x \text{ is an even prime number}\} \)
\( A = \{2\} \)
Since the set A has a definite, countable number of elements (just one), it is a finite set.
(ii) We list the odd prime numbers. Prime numbers are whole numbers greater than 1 that have only two divisors: 1 and themselves.
Let \( B = \{x \in N: x \text{ is an odd prime number}\} \)
\( B = \{3, 5, 7, 11, \ldots\} \)
There are infinitely many prime numbers, and all of them except 2 are odd. Therefore, this set has an unlimited number of elements, making it an infinite set.
(iii) We need to find integers \( x \) that are even and less than 10. Integers include positive and negative whole numbers and zero.
Let \( C = \{x \in Z : x \text{ is even and } < 10\} \)
\( C = \{\ldots, -8, -6, -4, -2, 0, 2, 4, 6, 8\} \)
This set includes all even integers counting downwards from 8, which goes on forever. Therefore, this set has an unlimited number of elements, making it an infinite set.
In simple words: A finite set has a specific, countable number of elements. An infinite set has an endless number of elements that cannot be counted.

🎯 Exam Tip: To determine if a set is finite or infinite, try to list its elements. If you can count them all and reach an end, it's finite. If the list goes on forever, it's infinite. Always check the domain of numbers (N for natural, Z for integers, R for real) specified in the set definition.

Question 4. By taking suitable sets A, B, C, verify the following results.
(i) \( A \times (B \cap C) = (A \times B) \cap (A \times C) \)
(ii) \( A \times (B \cup C) = (A \times B) \cup (A \times C) \)
(iii) \( (A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A) \)
(iv) \( C - (B - A) = (C \cap A) \cup (C \cap B') \)
(v) \( (B - A) \cap C = (B \cap C) - A = B \cap (C - A) \)
(vi) \( (B - A) \cup C = (B \cup C) - (A - C) \)
Answer:
(i) To verify the distributive law of Cartesian product over intersection, we will choose simple sets.
Let \( A = \{1, 2, 5, 7\} \), \( B = \{8, 9\} \), \( C = \{8, 10\} \)
First, find \( B \cap C \):
\( B \cap C = \{8, 9\} \cap \{8, 10\} = \{8\} \)
Now, calculate \( A \times (B \cap C) \):
\( A \times (B \cap C) = \{1, 2, 5, 7\} \times \{8\} = \{(1, 8), (2, 8), (5, 8), (7, 8)\} \) ..... (1)
Next, calculate \( A \times B \):
\( A \times B = \{1, 2, 5, 7\} \times \{8, 9\} = \{(1, 8), (1, 9), (2, 8), (2, 9), (5, 8), (5, 9), (7, 8), (7, 9)\} \)
Then, calculate \( A \times C \):
\( A \times C = \{1, 2, 5, 7\} \times \{8, 10\} = \{(1, 8), (1, 10), (2, 8), (2, 10), (5, 8), (5, 10), (7, 8), (7, 10)\} \)
Finally, find the intersection of \( (A \times B) \) and \( (A \times C) \):
\( (A \times B) \cap (A \times C) = \{(1, 8), (2, 8), (5, 8), (7, 8)\} \) ..... (2)
Since (1) = (2), the result \( A \times (B \cap C) = (A \times B) \cap (A \times C) \) is verified.
(ii) To verify the distributive law of Cartesian product over union, we use the following sets.
Let \( A = \{1, 2\} \), \( B = \{3, 4\} \), \( C = \{4, 5\} \)
First, find \( B \cup C \):
\( B \cup C = \{3, 4\} \cup \{4, 5\} = \{3, 4, 5\} \)
Now, calculate \( A \times (B \cup C) \):
\( A \times (B \cup C) = \{1, 2\} \times \{3, 4, 5\} = \{(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)\} \) ..... (1)
Next, calculate \( A \times B \):
\( A \times B = \{1, 2\} \times \{3, 4\} = \{(1, 3), (1, 4), (2, 3), (2, 4)\} \)
Then, calculate \( A \times C \):
\( A \times C = \{1, 2\} \times \{4, 5\} = \{(1, 4), (1, 5), (2, 4), (2, 5)\} \)
Finally, find the union of \( (A \times B) \) and \( (A \times C) \):
\( (A \times B) \cup (A \times C) = \{(1, 3), (1, 4), (2, 3), (2, 4)\} \cup \{(1, 4), (1, 5), (2, 4), (2, 5)\} \)
\( (A \times B) \cup (A \times C) = \{(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)\} \) ..... (2)
Since (1) = (2), the result \( A \times (B \cup C) = (A \times B) \cup (A \times C) \) is verified.
(iii) To verify the intersection property, we use the following sets.
Let \( A = \{1, 2, 5, 7\} \), \( B = \{2, 7, 8, 9\} \)
First, calculate \( A \times B \):
\( A \times B = \{(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9), (7, 2), (7, 7), (7, 8), (7, 9)\} \)
Next, calculate \( B \times A \):
\( B \times A = \{(2, 1), (2, 2), (2, 5), (2, 7), (7, 1), (7, 2), (7, 5), (7, 7), (8, 1), (8, 2), (8, 5), (8, 7), (9, 1), (9, 2), (9, 5), (9, 7)\} \)
Now, find the intersection of \( (A \times B) \) and \( (B \times A) \):
\( L.H.S. = (A \times B) \cap (B \times A) = \{(2, 2), (2, 7), (7, 2), (7, 7)\} \) ..... (1)
Now for the right-hand side, first find \( A \cap B \):
\( A \cap B = \{1, 2, 5, 7\} \cap \{2, 7, 8, 9\} = \{2, 7\} \)
Then, find \( B \cap A \):
\( B \cap A = \{2, 7, 8, 9\} \cap \{1, 2, 5, 7\} = \{2, 7\} \)
Finally, calculate \( (A \cap B) \times (B \cap A) \):
\( R.H.S. = (A \cap B) \times (B \cap A) = \{2, 7\} \times \{2, 7\} = \{(2, 2), (2, 7), (7, 2), (7, 7)\} \) ..... (2)
Since (1) = (2), the result \( (A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A) \) is verified.
(iv) To verify the set difference identity, we use the following sets and visually represent them with a Venn diagram.
Let \( A = \{1, 2, 3\} \), \( B = \{2, 3, 4\} \), \( C = \{3, 4, 5\} \)

Calculate the left-hand side \( C - (B - A) \):
First, find \( B - A \):
\( B - A = \{2, 3, 4\} - \{1, 2, 3\} = \{4\} \)
Now, find \( C - (B - A) \):
\( C - (B - A) = \{3, 4, 5\} - \{4\} = \{3, 5\} \) ..... (1)
Calculate the right-hand side \( (C \cap A) \cup (C \cap B') \):
First, find \( C \cap A \):
\( C \cap A = \{3, 4, 5\} \cap \{1, 2, 3\} = \{3\} \)
Next, find \( B' \). Assuming the universal set \( U \) for the numbers used is \( \{1, 2, 3, 4, 5\} \).
\( B' = U - B = \{1, 2, 3, 4, 5\} - \{2, 3, 4\} = \{1, 5\} \)
Now, find \( C \cap B' \):
\( C \cap B' = \{3, 4, 5\} \cap \{1, 5\} = \{5\} \)
Finally, find the union \( (C \cap A) \cup (C \cap B') \):
\( (C \cap A) \cup (C \cap B') = \{3\} \cup \{5\} = \{3, 5\} \) ..... (2)
Since (1) = (2), the result \( C - (B - A) = (C \cap A) \cup (C \cap B') \) is verified.
(v) To verify the set identity for differences and intersections, we use the following sets.
Let \( A = \{1, 2, 5, 7\} \), \( B = \{2, 7, 8, 9\} \), \( C = \{1, 5, 8, 10\} \)
Calculate the left-hand side \( (B - A) \cap C \):
First, find \( B - A \):
\( B - A = \{2, 7, 8, 9\} - \{1, 2, 5, 7\} = \{8, 9\} \)
Now, find \( (B - A) \cap C \):
\( (B - A) \cap C = \{8, 9\} \cap \{1, 5, 8, 10\} = \{8\} \) ..... (1)
Calculate the middle part \( (B \cap C) - A \):
First, find \( B \cap C \):
\( B \cap C = \{2, 7, 8, 9\} \cap \{1, 5, 8, 10\} = \{8\} \)
Now, find \( (B \cap C) - A \):
\( (B \cap C) - A = \{8\} - \{1, 2, 5, 7\} = \{8\} \) ..... (2)
Calculate the right-hand side \( B \cap (C - A) \):
First, find \( C - A \):
\( C - A = \{1, 5, 8, 10\} - \{1, 2, 5, 7\} = \{8, 10\} \)
Now, find \( B \cap (C - A) \):
\( B \cap (C - A) = \{2, 7, 8, 9\} \cap \{8, 10\} = \{8\} \) ..... (3)
Since (1) = (2) = (3), the result \( (B - A) \cap C = (B \cap C) - A = B \cap (C - A) \) is verified.
(vi) To verify this set identity involving union and difference, we use the following sets.
Let \( A = \{1, 2, 3, 4\} \), \( B = \{3, 4, 5, 6\} \), \( C = \{5, 6, 7, 8\} \)
Calculate the left-hand side \( (B - A) \cup C \):
First, find \( B - A \):
\( B - A = \{3, 4, 5, 6\} - \{1, 2, 3, 4\} = \{5, 6\} \)
Now, find \( (B - A) \cup C \):
\( (B - A) \cup C = \{5, 6\} \cup \{5, 6, 7, 8\} = \{5, 6, 7, 8\} \) ..... (1)
Calculate the right-hand side \( (B \cup C) - (A - C) \):
First, find \( B \cup C \):
\( B \cup C = \{3, 4, 5, 6\} \cup \{5, 6, 7, 8\} = \{3, 4, 5, 6, 7, 8\} \)
Next, find \( A - C \):
\( A - C = \{1, 2, 3, 4\} - \{5, 6, 7, 8\} = \{1, 2, 3, 4\} \)
Finally, find \( (B \cup C) - (A - C) \):
\( (B \cup C) - (A - C) = \{3, 4, 5, 6, 7, 8\} - \{1, 2, 3, 4\} = \{5, 6, 7, 8\} \) ..... (2)
Since (1) = (2), the result \( (B - A) \cup C = (B \cup C) - (A - C) \) is verified.
In simple words: To verify set identities, pick small, simple sets for A, B, and C. Then, work out each side of the equation separately, following the order of operations for set operations (like intersection, union, difference, Cartesian product) and compare the final sets. They should be identical.

🎯 Exam Tip: When choosing sets for verification, use small, distinct numbers to make calculations easier. Clearly show each step for both the left-hand side and the right-hand side to prove their equality. For Venn diagrams, accurately place each element in the correct region.

Question 5. Justify the trueness of the statement: “An element of a set can never be a subset”
Answer: The statement is true. An element is a single item belonging to a set. A subset, however, is a collection of elements that are all part of another set. For example, if we have a set \( A = \{1, 2, 3\} \), then 1 is an element of A, written as \( 1 \in A \). The set \( \{1\} \) is a subset of A, written as \( \{1\} \subseteq A \). An element itself, like 1, is not a set and therefore cannot be a subset. A set can be a subset of itself (e.g., \( A \subseteq A \)), but it cannot be a proper subset of itself because a proper subset must be strictly smaller and not equal to the original set.
In simple words: An element is a single item, like a number. A subset is a smaller group of items taken from a bigger group. You can't say a single item is a group, so an element cannot be a subset.

🎯 Exam Tip: Clearly understand the difference between an element (represented as \( x \in A \)) and a subset (represented as \( B \subseteq A \)). Remember that a set containing a single element, like \( \{x\} \), is a subset, but the element \( x \) itself is not.

Question 6. If \( n(P(A)) = 1024 \), \( n(A \cup B) = 15 \) and \( n(P(B)) = 32 \), find \( n(A \cap B) \).
Answer: We are given the number of elements in the power set of A and B, and the number of elements in the union of A and B. We need to find the number of elements in the intersection of A and B.
Given: \( n(P(A)) = 1024 \), \( n(A \cup B) = 15 \), \( n(P(B)) = 32 \)
We know that \( n(P(X)) = 2^{n(X)} \).
For set A:
\( n(P(A)) = 1024 \)
\( \implies 2^{n(A)} = 1024 \)
\( \implies 2^{n(A)} = 2^{10} \)
\( \implies n(A) = 10 \)
For set B:
\( n(P(B)) = 32 \)
\( \implies 2^{n(B)} = 32 \)
\( \implies 2^{n(B)} = 2^5 \)
\( \implies n(B) = 5 \)
Now, we use the formula for the number of elements in the union of two sets:
\( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
Substitute the known values into the formula:
\( 15 = 10 + 5 - n(A \cap B) \)
\( 15 = 15 - n(A \cap B) \)
\( \implies n(A \cap B) = 15 - 15 \)
\( \implies n(A \cap B) = 0 \)
This means that sets A and B have no common elements.
In simple words: First, use the power set formula to find how many elements are in set A and set B. Then, use the main set union formula to find how many elements they share. In this case, they share no elements.

🎯 Exam Tip: Remember the fundamental formulas: \( n(P(X)) = 2^{n(X)} \) and \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \). Make sure to correctly convert the power set size back to the number of elements in the set by finding the correct power of 2.

Question 7. If \( n(A \cap B) = 3 \) and \( n(A \cup B) = 10 \), then find \( n(P(A \Delta B)) \).
Answer: We are given the number of elements in the intersection and union of two sets, and we need to find the number of elements in the power set of their symmetric difference. The symmetric difference, \( A \Delta B \), includes elements that are in A or B but not in their intersection.

We are given:
\( n(A \cap B) = 3 \)
\( n(A \cup B) = 10 \)
The number of elements in the symmetric difference \( A \Delta B \) is given by the formula:
\( n(A \Delta B) = n(A \cup B) - n(A \cap B) \)
Substitute the given values:
\( n(A \Delta B) = 10 - 3 = 7 \)
Now, we need to find the number of elements in the power set of \( A \Delta B \). The formula for the power set is \( n(P(X)) = 2^{n(X)} \).
\( n(P(A \Delta B)) = 2^{n(A \Delta B)} \)
\( n(P(A \Delta B)) = 2^7 \)
\( n(P(A \Delta B)) = 128 \)
The power set of \( A \Delta B \) will contain 128 subsets.
In simple words: The symmetric difference includes elements that are in either set A or set B, but not in both. Once you find how many elements are in this special difference, you can find the number of subsets (the power set) by raising 2 to that number.

🎯 Exam Tip: Remember the formula for symmetric difference: \( n(A \Delta B) = n(A \cup B) - n(A \cap B) \). Also, be fluent in calculating the power of 2 for different exponents to find the size of the power set.

Question 8. If A x A contains 16 elements and two of its elements are (1, 3) and (0, 2), find the elements of A.
Answer: We know that if \( n(A \times A) = m \), then \( n(A) = \sqrt{m} \). We are given information about the Cartesian product of set A with itself.
Given \( A \times A \) contains 16 elements.
So, \( n(A \times A) = 16 \)
This means \( n(A) = \sqrt{16} = 4 \). So, set A contains 4 elements.
We are also told that \( (1, 3) \) and \( (0, 2) \) are two elements of \( A \times A \).
This implies that the first components of these ordered pairs (1 and 0) must be elements of A. Similarly, the second components (3 and 2) must also be elements of A.
Thus, the elements of A must include 0, 1, 2, and 3.
Since we know \( n(A) = 4 \) and we have identified four distinct elements (0, 1, 2, 3), these must be all the elements of A.
Therefore, \( A = \{0, 1, 2, 3\} \)
In simple words: If A multiplied by A has 16 items, then A must have 4 items. Since (1,3) and (0,2) are in A x A, it means 1, 3, 0, and 2 must all be in A. As A only has 4 items, these are the only items in A.

🎯 Exam Tip: Remember that if \( (a, b) \) is an element of \( A \times A \), then \( a \) must be in \( A \) and \( b \) must be in \( A \). Also, \( n(A \times A) = (n(A))^2 \), which means \( n(A) = \sqrt{n(A \times A)} \).

Question 9. Let A and B be two sets such that \( n(A) = 3 \) and \( n(B) = 2 \). If \( (x, 1), (y, 2), (z, 1) \) are in \( A \times B \), find A and B, where x, y, z are distinct elements.
Answer: We are given the number of elements in set A and set B, and some elements of their Cartesian product \( A \times B \). We need to identify the actual sets A and B.
Given: \( n(A) = 3 \) and \( n(B) = 2 \).
The elements \( (x, 1), (y, 2), (z, 1) \) are in \( A \times B \).
In a Cartesian product \( A \times B \), the first component of each ordered pair always belongs to set A, and the second component always belongs to set B.
From the given elements:
First components are \( x, y, z \). Since these are distinct elements and \( n(A) = 3 \), set A must be \( \{x, y, z\} \).
Second components are 1, 2, 1. The distinct second components are 1 and 2. Since \( n(B) = 2 \), set B must be \( \{1, 2\} \).
Therefore, \( A = \{x, y, z\} \) and \( B = \{1, 2\} \).
In simple words: The first numbers in the pairs like \( (x, 1) \) belong to set A, and the second numbers like \( (x, \underline{1}) \) belong to set B. Since A has 3 different items, \( x, y, z \) must be those items. Since B has 2 different items, 1 and 2 must be those items.

🎯 Exam Tip: Remember the definition of the Cartesian product \( A \times B \): it's the set of all ordered pairs \( (a, b) \) where \( a \in A \) and \( b \in B \). The number of elements \( n(A \times B) = n(A) \times n(B) \). Use the given elements to deduce the members of A and B, ensuring the count matches \( n(A) \) and \( n(B) \).

Question 10. If A × A has 16 elements, S = \( \{ (a, b) \in A \times A: a < b \} \); (-1, 2) and (0, 1) are two elements of S, find the remaining elements of S.
Answer: We are given information about the Cartesian product \( A \times A \), a subset S of \( A \times A \), and two elements of S. We need to find the other elements of S.
Given \( A \times A \) has 16 elements.
This means \( n(A \times A) = 16 \). Therefore, \( n(A) = \sqrt{16} = 4 \). Set A contains 4 elements.
We are given that \( (-1, 2) \) and \( (0, 1) \) are two elements of S.
The definition of S is \( \{ (a, b) \in A \times A: a < b \} \). This means for any pair \( (a, b) \) in S, \( a \) must be less than \( b \). Both given elements satisfy this condition: \( -1 < 2 \) and \( 0 < 1 \).
Since \( (-1, 2) \in A \times A \), it implies \( -1 \in A \) and \( 2 \in A \).
Since \( (0, 1) \in A \times A \), it implies \( 0 \in A \) and \( 1 \in A \).
Combining these, the elements \( -1, 0, 1, 2 \) must all be in set A. Since \( n(A) = 4 \), these are exactly the elements of A.
So, \( A = \{-1, 0, 1, 2\} \).
Now we need to find all elements \( (a, b) \in A \times A \) such that \( a < b \).
Let's list all possible ordered pairs where the first element is from A, the second element is from A, and the first is less than the second:
Possible values for \( a \) are \( -1, 0, 1 \). (If \( a=2 \), there is no \( b \in A \) such that \( 2 < b \)).
If \( a = -1 \), then \( b \) can be \( 0, 1, 2 \) (since \( -1 < 0, -1 < 1, -1 < 2 \)).
Pairs: \( (-1, 0), (-1, 1), (-1, 2) \)
If \( a = 0 \), then \( b \) can be \( 1, 2 \) (since \( 0 < 1, 0 < 2 \)).
Pairs: \( (0, 1), (0, 2) \)
If \( a = 1 \), then \( b \) can be \( 2 \) (since \( 1 < 2 \)).
Pair: \( (1, 2) \)
So, the set S is \( \{ (-1, 0), (-1, 1), (-1, 2), (0, 1), (0, 2), (1, 2) \} \).
The given elements of S are \( (-1, 2) \) and \( (0, 1) \).
The remaining elements of S are \( \{ (-1, 0), (-1, 1), (0, 2), (1, 2) \} \).
In simple words: First, figure out what numbers are in set A. Since A x A has 16 pairs, A must have 4 numbers. The given pairs like (-1, 2) tell us that -1, 2, 0, and 1 are all in A. So, A is \( \{-1, 0, 1, 2\} \). Then, list all possible pairs from A x A where the first number is smaller than the second. Remove the two pairs already given to find the remaining ones.

🎯 Exam Tip: Always deduce the elements of the base set (A, in this case) first by using the size of the Cartesian product and the given elements. Then, apply the condition defining the subset (like \( a < b \)) to systematically list all its elements before identifying the "remaining" ones.