The Empty Set, Finite and Infinite Sets, Equal Sets

Two sets are equal when they contain exactly the same elements; the order of listing does not matter, and repeated elements are written only once. Options A and D contain 1, 2, and 3 in a different order. Option B also represents {1, 2, 3}, because repetition does not create a new element. Option C contains the additional element 4, so it is not equal.

Taking square numbers as positive natural-number squares, list the squares below 20: 1² = 1, 2² = 4, 3² = 9, and 4² = 16. The next square is 5² = 25, which is not less than 20. Therefore A = {1, 4, 9, 16}, containing four distinct elements. It is finite, so option B is correct; 3 misses one square, 5 includes an excluded value, and the set is not infinite.

The equation x = x + 1 cannot be true for any number. Subtracting x from both sides gives 0 = 1, which is a contradiction. Therefore, no natural number satisfies the defining condition of A. A set with no member is called the empty set, so A = ∅. It is finite as well, but the requested type is empty set.

Equal sets contain exactly the same elements. The sets {a, b} and {b, a} both contain a and b; only the order of writing has changed. In set notation, order is irrelevant. Option A has 3 versus 4, option C has an element versus no element, and option D has an extra 6. Thus only option B shows equal sets.

The natural numbers divisible by 3 are 3, 6, 9, 12, 15, and so on. For every such number, adding 3 gives another natural number that is also divisible by 3. Since this process never ends and no upper bound is specified, the set has infinitely many elements. Therefore, A is an infinite set.

In the usual school convention, natural numbers are the positive counting numbers 1, 2, 3, and so on; some curricula also include 0, but neither convention includes negative numbers. Consequently, there is no negative natural number. Since no element satisfies the condition, A has no members and is the empty set, written A = ∅.

The natural numbers less than 4 are 1, 2, and 3, so A = {1, 2, 3}. In set notation, listing an element twice does not create a second element; therefore B = {1, 2, 3, 3} represents the same set as {1, 2, 3}. Since A and B have exactly the same members, A = B. Thus option A is correct; the other statements contradict the sets’ actual contents.

A finite non-empty set must contain at least one element and must have a limited total number of elements. The set {2, 4, 6} contains exactly three elements, so it is finite and non-empty. Option A is empty, option B contains endlessly many natural numbers greater than 10, and option D is also infinite. Therefore, option C is correct.

The symbol ∅ denotes the empty set, which contains no elements. However, A = {∅} is not the empty set itself; its outer braces show that the empty set is included as one element of A. Therefore, A is a singleton set and its cardinality is 1. Always distinguish between ∅ and {∅}.

To determine the set, first solve the equation x² = 4. Its integer solutions are x = 2 and x = −2. However, the condition also says that x must be a natural number. Since −2 is not a natural number, it must be excluded. Therefore, the only permissible element is 2, so A = {2}. This is a finite set containing exactly one element.

The equation x² = 2 has real solutions x = √2 and x = −√2, but neither solution is a natural number. In particular, √2 is irrational and is not a member of the natural numbers. Since no natural number satisfies the stated condition, the set contains no elements and is therefore empty.

Sets are equal when they contain exactly the same distinct elements; the order of elements does not matter, and repetition is ignored. Thus {1, 2} = {2, 1}, {a, a, b} = {a, b}, and {3, 4, 5} = {5, 4, 3}. But {0} has one element, whereas ∅ has no elements, so they are unequal.

The natural numbers less than 100 are 1, 2, 3, ..., 99, using the usual school convention that natural numbers begin with 1. This list has a fixed number of members, namely 99, and it ends at 99. A set with a limited, countable number of elements is called a finite set.

The set consists of 101, 102, 103, and so on. For every natural number in this set, a larger natural number can be found that is also greater than 100. Thus the list has no final member and continues without end. Therefore, A is an infinite set, even though it has a lower bound.

The letters appearing in LEVEL are L, E, V, E, and L. In a set, repeated elements are written and counted only once. Therefore, the set is A = {L, E, V}. It has three distinct elements, so the cardinality of A is 3, not 5, because the repeated L and E are not counted again.

A standard traffic light uses a fixed set of colors: red, yellow (or amber), and green. Whether the colors are named in English or Hindi does not change the number of members. Since the collection contains a limited, fixed number of colors, it is a finite set. It is neither empty nor infinite.

The condition requires x to be a natural number strictly greater than 2 and strictly less than 3. The endpoints 2 and 3 are excluded because both inequalities are strict. There is no natural number between two consecutive natural numbers, 2 and 3. Hence no value satisfies the condition, so A = ∅.

The set contains natural numbers that are less than or equal to 5. Under the common convention N = {1, 2, 3, ...}, its elements are {1, 2, 3, 4, 5}; if 0 is included in N, the set is {0, 1, 2, 3, 4, 5}. In either convention, there are only finitely many elements, so B is a finite set. The exact convention about including 0 does not change the answer.

The conditions require x to be a natural number, odd, and greater than 10. The elements therefore begin as 11, 13, 15, 17, 19, and so on. There is always another odd natural number greater than the previous one, so the list never ends. Hence C has infinitely many elements and is an infinite set. Therefore, option C is correct.

The governing concept is equality of sets: two sets are equal when they contain exactly the same elements, regardless of how they are described. The prime factorisation of 12 is 2 × 2 × 3, so its distinct prime factors are 2 and 3; repeated factors are written only once in a set. Hence B = {2, 3} = A, making option A correct. Option B ignores set equality, while C and D contradict the listed finite elements.

Solving x² = 9 gives x = 3 or x = −3. However, the definition also requires x to belong to N, the set of natural numbers. Under the standard school convention, 3 is a natural number but −3 is not. Therefore only 3 satisfies both conditions, so P = {3}, a singleton set. Option B is correct.

In a set, the order of elements is irrelevant and repeated listing of an element does not create a new element. Thus the pairs in A, B, and D are equal. But {0} contains the number 0, whereas {∅} contains the empty set as its sole element. Since 0 and ∅ are different objects, pair C is not equal.

The natural numbers are 1, 2, 3, 4, 5, and so on. Because 4 and 5 are consecutive natural numbers, there is no natural number strictly between them. Hence A contains no elements and is written as A = ∅. A set with no elements is called the empty set, so option A is correct.

The set A is explicitly listed as {0, 2, 4, 6, 8}. Its distinct elements can be counted: there are exactly five of them. Because the counting ends and no additional elements are implied by the notation, A has a limited number of elements. A set with a limited number of elements is called a finite set. Therefore, option C is correct.

The positive multiples of 7 are 7, 14, 21, 28, 35, and so on. Multiplying 7 by every positive natural number produces another member, and there is no greatest positive natural number. Consequently, the list of positive multiples never ends. Therefore A has infinitely many elements and is an infinite set.