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Sets - Empty or Null Set-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition: A set which does not contain any element is called the empty set or the null set or the void set. It is denoted by the symbol ∅\emptyset or empty braces {}\{\}.

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Cardinality: The number of elements in an empty set is zero. If A=∅A = \emptyset, then n(A)=0n(A) = 0.

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Universal Subset: The empty set ∅\emptyset is a subset of every set AA. Mathematically, ∅⊆A\emptyset \subseteq A for any set AA.

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Power Set of Empty Set: The power set of an empty set is not empty. Since A=∅A = \emptyset has 00 elements, its power set P(A)P(A) has 20=12^0 = 1 element. Thus, P(∅)={∅}P(\emptyset) = \{\emptyset\}.

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Distinction from Zero Set: The set {0}\{0\} is NOT an empty set because it contains one element, which is the number 00. Similarly, {∅}\{\emptyset\} is a singleton set containing the empty set as an element.

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Property of Intersection and Union: For any set AA, the intersection with a null set is always a null set (A∩∅=∅A \cap \emptyset = \emptyset), and the union with a null set is the set itself (A∪∅=AA \cup \emptyset = A).

📐Formulae

A={x:x≠x}=∅A = \{x : x \neq x\} = \emptyset

n(∅)=0n(\emptyset) = 0

∅⊆A for every set A\emptyset \subseteq A \text{ for every set } A

P(∅)={∅}P(\emptyset) = \{\emptyset\}

n(P(∅))=20=1n(P(\emptyset)) = 2^0 = 1

A∩∅=∅A \cap \emptyset = \emptyset

A∪∅=AA \cup \emptyset = A

💡Examples

Problem 1:

Examine if the set A={x:x∈R,x2+1=0}A = \{x : x \in \mathbb{R}, x^2 + 1 = 0\} is a null set.

Solution:

We are given the condition x2+1=0x^2 + 1 = 0. Solving for xx: x2=−1x^2 = -1 x=−1x = \sqrt{-1} Since the square of any real number is always non-negative (x2≥0x^2 \geq 0), there is no real number xx such that x2=−1x^2 = -1. Therefore, no element satisfies the condition x∈Rx \in \mathbb{R}. Hence, A=∅A = \emptyset.

Explanation:

In the set-builder form, if the conditions imposed on the variable cannot be satisfied by any member of the specified universal set (here, Real numbers), the set is empty.

Problem 2:

Let B={x:x is an even prime number greater than 2}B = \{x : x \text{ is an even prime number greater than } 2\}. Is BB an empty set?

Solution:

The set of prime numbers is P={2,3,5,7,11,… }P = \{2, 3, 5, 7, 11, \dots\}. The only even prime number is 22. Since we are looking for even prime numbers strictly greater than 22, there are no such numbers in the set of primes. Thus, B=∅B = \emptyset.

Explanation:

Since 2 is the unique even prime, the intersection of the set of even numbers and the set of primes greater than 2 is empty.

Problem 3:

Find the number of elements in P(P(∅))P(P(\emptyset)).

Solution:

Step 1: Find the power set of the empty set. A=∅  ⟹  P(A)={∅}A = \emptyset \implies P(A) = \{\emptyset\} Step 2: Find the power set of P(A)P(A). Let B=P(A)={∅}B = P(A) = \{\emptyset\}. The number of elements in BB is n(B)=1n(B) = 1. The power set P(B)P(B) will have 2n(B)=21=22^{n(B)} = 2^1 = 2 elements. P(P(∅))={∅,{∅}}P(P(\emptyset)) = \{\emptyset, \{\emptyset\}\} Therefore, n(P(P(∅)))=2n(P(P(\emptyset))) = 2.

Explanation:

This demonstrates that while the empty set has no elements, its power set has one element, and the power set of that has two elements, following the 2n2^n rule.