 Hello and welcome to the session. In this session we discussed the following question which says, check whether first set is a subset of second set or not. In the first part we have a set A equal to set of letters in the word here and set B is equal to set of letters in the word square. In the B part we have set C is equal to X such that X square minus 1 is equal to 0 where X belongs to N and set D is equal to X such that X is equal to Y plus 2 where Y belongs to W and Y is greater than equal to 0 and less than 4. Let us now define the subset of a set if every member of a set A is also a member of a second set B then the set A is called a subset of set B and this is written as A is a subset of B. This is the key idea that we use for this question. Let us proceed with the solution now. In the first part we have A is the set of letters of the word here. So this means set A consists of the elements H, E, R. Now though the letter E is occurring two times in the word here but it would be written only once in the set since the repetition of the elements is not allowed in the set. Then we have set B which is the set of the letters of the word sphere. So set B consists of the elements S, P, H, E, R. Again in the word sphere letter E appears two times but in the set we would write it only once since the repetition of elements is not allowed. Thus we have got two sets A and B. Now we have to check if the first set is a subset of the second set or not that is A is a subset of B or not. Now from the definition given in the key idea we have that if every member of a set A is also a member of the second set B then we say that A is a subset of B. Now as you can see here that every member of the set A is a member of the set B that is the elements H, E and R appear in the set B. So we say that A is a subset of B. Now let's see the second part of the question in which we have set C is equal to X such that X square minus 1 is equal to 0 where X belongs to the set of natural numbers. Now 1 belongs to the set of natural number N such that 1 square minus 1 is equal to 0. Therefore we can say that C is the single term set 1. Then we have set D is equal to X such that X is equal to Y plus 2 where Y belongs to W and Y is greater than equal to 0 and less than 4. As Y belongs to W and Y is greater than equal to 0 and less than 4 therefore we have that Y takes the values 0, 1, 2 and 3. Now X is equal to Y plus 2 so we will find the values of X for Y equal to 0, 1, 2 and 3. Now for Y equal to 0, 1, 2, 3 we obtain X equal to 2, 3, 4 and 5. Thus we get the set D is equal to the set containing the elements 2, 3, 4, 5. Now we have to check whether the set C which is single term 1 is a subset of set D containing the elements 2, 3, 4, 5. Now as you can see that 1 does not belong to the set D therefore C is not a subset of D. Thus we have obtained that A is a subset of the set B but set C is not a subset of set D. So this completes the session. Hope you have understood the solution of this question.