⛲ Example 0. Consider these three sets

*A*=the set of all even numbers, *B*={2,4,6}, *C*={2,3,4,6}

Here *B*⊂*A* since every element of *B* is also an even number, so is an element of *A*.

More formally,we could say *B*⊂*A* since if *x*∊*B*,then *x*∊*A*.

It is also true that *B*⊂*C*.

*C* is not a subset of *A*, since *C* contains an element, 3, that is not contained in *A*.

Let

D={} andE={1,2,3,4}. IsD⊂E? To showD⊄E, you must find at least one element of setDthat is not an element of setE. Because this cannot be done,D⊂Emust be true. Using the same reasoning. we can show that the empty set is a subset of every set, including itself.⛲ Question 1. Let

F,GandHbe three sets. IfF∊GandG⊂H, is it true thatF⊂H?. If not, give an example.

✍ Solution:

No. LetF={1},G={{1},2} andH={{1},2,3}. HereF∊GasF={1} andG⊂H. ButF⊄Has 1∊Fand 1∉H.

Note that an element of a set can never be a subset of itself.⛲ Ex1. In each of the following, determine whether the statement is true or false. If it is true, then prove it. If it is false, then give an example.

(i) Ifx∊JandJ∊K, thenx∊K.

(ii) IfJ⊂KandK∊L, thenJ∊L.

✍ Solution:

(i) False,

LetJ={2},K={{2},3}

Clearly, 2∊JandJ∊K, but 2∉K.

So,x∊JandJ∊Kneed not imply thatx∊K.

(ii) False, LetJ={2},K={2,3} andL={{2,3},4}

Clearly,J⊂KandK∊L, butJ∉L.

Thus,J⊂KBandK∊Lneed not imply thatJ∊L.

⛲ Q2. Let *M*={a,b,{c,d},e}. Which of the following statements is/are true?

(i) {c,d}∊*M* (ii) {{c,d}⊂*M*

✍ Solution:

Given, *M*={a,b,{c,d},e}

(i) Since a,b,{c,d} and e are elements of *M*.

∴ {c,d}∊*M*

Hence, it is a true statement.

(ii) As {c,d}∊*M* and {{c,d}} represents a set,which is a subset of *M*.

∴ {{c,d}}⊂*M*

Hence, it is a true statement.

⛲ Ex2. If *N*={3,{4,5},6}, then find which of the following statements are true.

(i) {4,5}⊂*N* (ii) {4,5}∊*N* (iii) Ø⊂*N* (iv) {3,6}⊂*N*

✍ Solution:

Given, *N*={3,{4,5},6}

Here, 3, {4,5}, 6 all are elements of *N*.

(i) {4,5}⊂*N*, which is not true. Since, element of any set is not a subset of any set and here {4,5} is an element of *N*.

(ii) {4,5}∊*N*, is a true statement.

(iii) It is always true that Ø⊂*N*

(iv) {3,6} makes a set,so it is a subset of *N* i.e., {3,6}⊂*N*

⛲ Q3. **is each either Element or Proper Subset? But not as Both!**

Let *P*={1,2,{3,4},5}. Which of the following statements are incorrect and why?

(i) {3,4}⊂*P*

(ii) {3,4}∊*P*

(iii) {{3,4}}⊂*P*

(iv) 1∊*P*

(v) 1⊂*P*

(vi) {1,2,5}⊂*P*

(vii) {1,2,5}∊*P*

(viii) {1,2,3}⊂*P*

(ix) Ø∊*P*

(x) Ø⊂*P*

(xi) {Ø}⊂*P*

✍ Solution:

*P*={1,2,{3,4},5}

(i) The statement {3,4}⊂*P* is incorrect because 3∊{3,4}; however, 3∉*P*.

(ii) The statement {3,4}∊*P* is correct because {3,4} is an element of *P*.

(iii) The statement {{3,4}}⊂*P* is correct because {3,4}∊{{3,4}} and {3,4}∊*P*.

(iv) The statement 1∊A is correct because 1 is an element of *P*.

(v) The statement 1∊*P* is incorrect because an element of a set can never be a subset of itself.

(vi) The statement {1,2,5}⊂*P* is correct because each element of {1,2,5} is also an element of *P*.

(vii) The statement {1,2,5}∊*P* is incorrect because {1,2,5} is not an element of *P*.

(viii) The statement {1,2,3}⊂*P* is incorrect because 3∊{1,2,3}; however, 3∉*P*.

(ix) The statement Ø∊*P* is incorrect because Ø is not an element of *P*.

(x) The statement Ø⊂*P* is correct because Ø is a subset of every set.

(xi) The statement {Ø}⊂*P* is incorrect because Ø∊{Ø}; however, Ø∊*P*.

⛲ Q4. Let *Q*={1,2,{3,4},5}. Which of the following statements are incorrect and why?

(i) {3,4}⊂*Q* (ii) {3,4}∊*Q* (iii) {{3,4}}⊂*Q* (iv) 1∊*Q* (v) 1⊂*Q* (vi) {1,2,5}⊂*Q* (vii) {1,2,5}∊*Q* (viii) Ø⊂*Q* (ix) Ø∊*Q* (x) Ø⊂*Q*

✍ Solution:

We have, *Q*={1,2,{3,4},5}

(i) Since, {3,4} is a member of set *Q*.

∴ {3,4}∊*Q*

Hence, {3,4}}⊂*Q* is incorrect.

(ii) Since, {3,4} is a member of set *Q*. Hence, {3,4}∊*Q* is correct.

(iii) Since, {3,4} is a member of set *Q*. So,{{3,4}} is a subset of *Q*.

Hence, {{3,4}}⊂*Q* is correct.

(iv) Since, 1 is a member of *Q*. Hence, 1∊*Q* is correct.

(v) Since, 1 is a member of set *Q*. Hence, 1⊂*Q* is incorrect.

(vi) Since, 1, 2, 5 are members of set *Q*. So,{1,2,5} is a subset of set *Q*.

Hence, {1,2,5}⊂*Q* is correct.

(vii) Since, 1, 2 and 5 are members of set *Q*. So,{1,2,5} is a subset of *Q*.

Hence, {1,2,5}∊*Q* is incorrect.

(viii) Since, Ø is subset of every set. Hence, Ø⊂*Q* is correct.

(ix) Since, Ø is not a member of set *Q*. Hence, Ø∊*Q* is incorrect.

(x) Since, Ø is not a member of set *Q*. Hence,⊂*Q* is incorrect.

⛲ Ex3: **is each either Element or Proper Subset? But not as Both!**

Determine whether the following are true or false.

a) 3∊{3,4,5}

b) {3}∊{3,4,5}

c) {3}∊{{3},{4},{5}}

d) {3}⊂{3,4,5}

e) 3⊂{3,4,5}

f) {}⊂{3,4,5}

✍ Solution:

a) 3∊{3,4,5} is a true statement because 3 is an element of the set {3,4,5}.

b) {3}∊{3,4,5} is a false statement because {3} is a set, and the set {3} is not an element of the set {3,4,5}.

c) {3}∊{{3},{4},{5}} is a true statement because {3} is an element in the set. The elements of the set {{3},{4},{5}} are themselves sets.

d) {{3}⊂{3,4,5} is a true statement because every element of the first set is an element of the second set.

e) 3⊂{3,4,5} is a false statement because the 3 is not in braces, so it is not a set and thus cannot be a proper subset. The 3 is an element of the set as indicated in part (a).

f) {}⊂{3,4,5} is a true statement because the empty set is a proper subset of every set.

⛲ Ex4. Examine whether the following statements are true or false:

(I) {a,b}⊄{b,c,a}

(ii) {a,e}⊂{x:x is a vowel in the English alphabet}

(iii) {1,2,3}⊂{1,3,5}

(iv) {a}⊂{a,b,c}

(v) {a}∊(a,b,c)

(vi) {*x*:*x* is an even natural number less than 6}⊂{*x*:*x* is a natural number which divides 36}

✍ Solution:

(i) False. Each element of {a,b} is also an element of {b,c,a}.

(ii) True. a,e are two vowels of the English alphabet.

(iii) False. 2∊{1,2,3}; however, 2∉{1,3,5}

(iv) True. Each element of {a} is also an element of {a,b,c}.

(v) False. The elements of {a,b,c} are a,b,c. Therefore, {a}⊂{a,b,c}

(vi) True. {*x*:*x* is an even natural number less than 6}={2,4}

{*x*:*x* is a natural number which divides 36}={1,2,3,4,6,9,12,18,36}