Question 1: Find the smallest set such that

Answer:

Smallest set

Question 2: Let . Verify the following identities:

(i) (ii)

(iii) (iv)

(v) (vi)

Answer:

(i) Given

Hence

(ii) Given

Hence

(iii) Given

Hence

(iv)

Hence

(v) Given

Hence

(vi) Given

Hence

Question 3: If is the universal set and , , then prove that: (i) (ii)

Answer:

(i) Given , and

Hence

(ii) Given , and

Hence

Question 4: For any two sets and , prove that

(i) (ii) (iii)

Answer:

(i)

For all

or

(Definition of union of sets)

(ii)

Fora all

and (definition of intersection of sets)

(iii)

Let

Let

and

and (as )

Question 5: For any two sets and , show that the following statements are equivalent:

(i) (iii)

Answer:

In order to show that the four statements are equivalent, we need to show that , , and

i) Let us assume

Let , since

Now and

Since , each element of is an element of

ii) Let

Therefore every element of is also an element of .

Therefore

iii) Let us assume

and

Therefore since every element of is in

iv) Let

if also

Therefore . All elements of are elements of also.

Question 6: For three sets and , show that

(i) need not imply (ii)

Answer:

(i) need not imply

Let and

Therefore

Hence

But we can clearly see that

(ii)

Given

Let

but (By definition of )

and (Since )

Thus . This is true for all

Therefore

Question 7: For any two sets, prove that: (i) (ii)

Answer:

(i)

[Since union is distributive over intersection ]

[ Since ]

[ Since as union of two sets is biggest than each of the individual sets ]

Hence . Hence Proved.

(ii)

[ Since intersection is distributive on union ]

[Since ]

Question 8: Find sets and such that and are non-empty sets and

Answer:

Consider and

Therefore

Also

Hence there are no common elements between all three sets.

Question 9: For any two sets and , prove that .

Answer:

Given

for

There are no common elements between and

Thus and

Question 10: If and are sets, then prove that and are pair wise disjoint.

Answer:

We need to show:

Let us first show

Let but

This is true for all

Hence

Similarly, let us prove

but

Therefore

Now prove

If but

If but

We can see that there is no intersection of and . Hence

Question 11: Using properties of sets, show that for any two sets and , .

Answer:

To prove:

Now (using associative property)

[ Since and ]

[Since ]

Question 12: For any two sets of and , prove that: (i) (ii)

Answer:

(i) To prove:

Let [ Since ]

Therefore and

[Since ]

or

But

Thus

This is true for all

Therefore

(ii) To prove

Let [ Since ]

[Since ]

[Since ]

Thus

This is true for all . Therefore

Question 13: Is it true that for any sets and , Justify your answer.

Answer:

False

Let

or

or

Therefore

Again let but or

Therefore

Thus is not necessarily a subset of

Question 14: Show, that for any sets and , (i) (ii)

Answer:

(i) We know that and

… … … … … i)

Let

and

and but

[ Since and are not possible simultaneously]

Therefore … … … … … ii)

From i) and ii) we get

(ii) Let

or or

or

Therefore … … … … … i)

Let or

or and

or

Therefore … … … … … ii)

From i) and ii) we get

Question 15: Each set , contains elements and each set , contains elements and . If each element of belongs to exactly of the and to exactly of , find the value of .

Answer:

It is given that each set contains elements

Therefore

But it is given that each element of belongs to exactly of that

Number of distinct elements in … … … … … i)

It is also given that each set contains elements and

Therefore

Also each element of belongs to exactly of … … … … … ii)

Therefore from i) and ii) we get