Chapter 1.3 - Subsets

In the previous section, we saw the empty set, finite set and equal sets. In this section, we will see subsets

Some basics about subsets can be written in 14 steps:
1. Consider the following two sets:
X = {x : x is a student in the school}
Y = {y : y is a student in class 11}
2. It is clear that, all elements in Y will be present in X.
• We say that: Y is a subset of X.
   ♦ Using symbols, this is written as: Y ⊂ X
   ♦ The symbol ‘⊂’ stands for ‘is a subset of’ or ‘is contained in’.
3. We can write the definition:

Definition 4 A set A is said to be a subset of set B if every element of A is contained in B.

4. We already know that:
If a is an element of A, we can write a ∈ A.
• But since A is a subset of B, the element a will be present in B also. So we can write: a ∈ B
5. That means,
If a ∈ A implies a ∈ B then, A ⊂ B.
• The word implies can be replaced by the symbol ‘⇒ ’. So we can write:
If a ∈ A ⇒ a ∈ B then, A ⊂ B.
6. There are two situations where A can be a subset of B
(i) Every element in A is present in B. Further, B has some extra elements also.
(ii) Every element in A is present in B. There are no extra elements in B.
   ♦ This means B has the exact same elements as A.
7. Let us consider the second situation carefully:
• If B has the exact same elements as A, we can write A ⊂ B alright.
   ♦ But we can write B ⊂ A also.
• That means, if two sets A and B have the exact same elements, we can write both:
A ⊂ B and B ⊂ A.
8. Recall that, if two sets have the exact same elements, they are equal sets.
• So we can write:
If A ⊂ B and B ⊂ A, the sets A and B are equal.
• Using symbols, we can write:
A ⊂ B and B ⊂ A ⇒ A = B
9. Conversely, we can write:
If two sets A and B are equal, then A ⊂ B and B ⊂ C.
10. A theorem and it’s converse can be written using the symbol ‘⇔’.
• This symbol indicates two way implication. So we get: A ⊂ B and B ⊂ A ⇔ A = B
   ♦ The forward direction indicates what we wrote in (8).
   ♦ The reverse direction indicates what we wrote in (9).
11. What we saw in steps (7) to (10) is the analysis of the situation in 6(ii).
• What if the situation is as in 6(i) ?
   ♦ In such a situation, we say two things:
         ✰ A is a proper subset of B.
         ✰ B is a superset of A.
12. Let us see some examples:
Example 1:
• Consider the following two sets:
A = {x : x is a divisor of 56}
B = {x : x is prime divisor of 56}
• Obviously, all elements in B will be present in A.
   ♦ So we can write: B ⊂ A
Example 2:
• Consider the following two sets:
A = {1, 3, 5}
B = {x : x is an odd number less than 6}
• The set B in roster form is {1, 3, 5}.
• So we can write:
   ♦ All elements of A are present in B.
         ✰ So A ⊂ B
   ♦ All elements of B are present in A.
         ✰ So B ⊂ A
• Since A ⊂ B and B ⊂ A, we can write A = B.
Example 3:
• Consider the following two sets:
A = {a, l, o, y}
B = {a, e, i, o, u}
• We see that, A is not a subset of B.
   ♦ We write: A ⊄ B
• We see that, B is not a subset of A.
   ♦ We write: B ⊄ A
13. From the above steps, it is clear that, every set A is a subset of itself
• Also, since the empty set ɸ has no elements, ɸ is a subset of every set
14. If a set A has only one element, we call it a singleton set
• For example, {a} is a singleton set

Solved example 1.18
Consider the sets
ɸ, A = { 1, 3 }, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}.
Insert the symbol ⊂ or ⊄ between each of the following pair of sets:
(i) ɸ . . . B  (ii) A . . . B  (iii) A . . . C  (iv) B . . . C
Solution:
(i) Since is a subset of every set, we can write: ɸ ⊂ B
(ii) 3 is not present in B. So we write: A ⊄ B
(iii) Both 1 and 3 are present in C. So we write: A ⊂ C
(iv) All three elements 1, 5 and 9 are present in C. So we write B ⊂ C

Solved example 1.19
Let A, B and C be three sets. If A ∈ B and B ⊂ C, is it true that A ⊂ C?. If not, give an example.
Solution:
1. Let A = {1, 2}
• Given that, set A is an element of B. So we can write: B = {3, {1, 2}, 4}
• Given that B is a subset of C. So we can write: C = {3, {1, 2}, 4, 5}
2. Here we are inclined to think that, A is a subset of C
• But if we write a subset of C, it will be: {{1, 2}}
   ♦ {{1, 2}} is different from {1, 2}
   ♦ That is: {{1, 2}} is different from A.
• So the answer is: No, A cannot be a subset of C.
3. Another way to analyze this situation can be written in 3 steps:
(i) If A is to be a subset of C, all elements of A should be present in C as separate elements.
(ii) But in C, the elements 1 and 2 are present together as a set. Not as separate elements.
(iii) So the answer is: No, A cannot be a subset of C.

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In the next section, we will see subsets of the set R

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