 discuss set operations and laws of set operations now our starting point is you the universe and capital A capital B capital C etc which are subsets of you now we define several operations involving these sets ABC we define the following A union B the union of set A and set B is all x strictly speaking all x in you such that x belonging to A or x belongs to B now this is intuitively very clear this only means that given any two sets A and B I can consider a set which contains all the elements of A and all the elements of B needless to say that it will contain the elements which are both in A and B and this bigger set is denoted by A union B the symbol this symbol is the symbol of union of two sets so this is an operation on sets the next operation that we define is called the intersection of two sets we write intersection as this and B so this symbol is called the intersection and this is the set of all x belonging to you such that x belongs to A and x belongs to B at this point I would like to mention a diagrammatic representation of sets proposed by John when who lived between 1834 to 1923 the diagrammatic representation that he proposed is famous as when diagrams the idea behind when diagram is very simple it just says that when we are considering the universe you will write a rectangle or draw a rectangle and call that you our universal set and the subsets or the sets of the universal sets that we discuss that we discuss within this framework which are of course subsets of you so these sets that is ABC and all the others will be drawn as circles inside you where the interior of the circles may be shaded when I write this this means that a consists of the elements that are within this circle now suppose we take another set B in general it will have some common area shared with the circle corresponding to a let us call it B let us go to the next page and draw this situation one more time so here we have the universal set you or simply the universe I am representing a set the set A by the inner portion of a circle and B the inner portion of another circle if we want to know the set A union B this will consist of the shaded area which consists of the region that is covered by A and the region covered by B so this whole shaded area is A union B again we take A and B if we consider only the region that is within the interior of the circle corresponding to A and the interior of the circle corresponding to B then we get the set A intersection B now this pictorially gives us the sets A union B and A intersection B and these diagrams are very useful in getting an intuitive feeling of intuitive feeling of operations on sets the next operation that we discuss is called symmetric difference now let us check the definition of symmetric difference this is the third sub part of the first definition that we have started symmetric difference is denoted by ? so again we have two sets and I am writing a ? B this is the symmetric difference of the set A and the set B and this consists of all the elements in U such that X belongs to A union B and X does not belong to A intersection B now let us take the help of the Venn diagram to understand this operation suppose these are my sets A and B so I am looking for the region which is inside A union B and which is outside A intersection B let us recall what is A union B and what is A intersection B A union B is the shaded region that covers the whole of A and B and A intersection B is this part so if I superimpose these two diagrams then we will find that this part is A intersection B and therefore I have to cut this portion out I do not want elements there and all the other elements will be in the symmetric difference therefore the symmetric difference is a set which can be represented by the Venn diagram as the shaded region within A union B so this is whole of A union B leaving aside the portion A intersection B there are two more operations that I would like to discuss over here one is A set – B A set – this is the set of all X such that X belongs to A and X does not belong to B if we look at the Venn diagram corresponding to this operation then we see something like this again we have the universal set and we are taking two sets A and B within that a universal set and I want the elements which are in A but not in B therefore I said the portion of A which is outside A intersection B this shaded region corresponds to A set – B or simply A – B now if we inverted this situation and if we had taken B – A we would have gotten X belongs to you and X belongs to B and X does not belong to A now in Venn diagram that means the region which is inside B but not in A intersection B so if you have A here and B here this is the region corresponding to B – A this is B I am sorry this is B – A thus we can combine these two things and get a different description of symmetric difference of A and B the symmetric difference of A and B can also be thought of as A set – B union B set – A we have to remember this now let us look at some examples suppose our universal set is integers from 1 to 10 and let us consider the sets A, B, C as then A union B is 1, 2, 3, 4, 5, 6 and 7 so that is the integers which are both in A and B if you want to find out A intersection B that will take only the integers which are common to both A and B and we see that 3 is common to both A and B 4 is both common to A and B and same is 5 therefore it will be 3, 4 and 5 now if we take B intersection C B intersection C is only 7 because we see that the 7 is the only integer which is common to both B and C and then if we consider A intersection C we see that A is 1, 2, 3, 4, 5 and C is 7, 8, 9 therefore A intersection C does not have any element so it is the empty set ? we have defined empty set in the last lecture and here we see an example of the empty set now let us consider the symmetric difference between A and B so in order to do that I have to find out elements which are in A but not in B and the elements which are in B and not in A we can as well use the relation that we have obtained over here that is A-B union B-A if I do that and first let us consider A-B so let us compare A and B 1 is in A but not in B 2 is in A but not in B but 3, 4, 5 all are in A all are in B therefore A-B is going to be only 1 and 2 union now we see that 3, 4 and 5 are common to both A and B and in B we have 2 more terms or 2 more elements 6 and 7 which are not in A therefore those elements will appear in B-A and thus we have the set 1, 2 union 6 set 6, 7 and taking the union we have 1, 2, 6, 7 so this is a symmetric difference between A and B one more operation that we need to consider is said to be the complement. So we have a set A which is of course a subset of the universal set complement of A is denoted by A bar or A over line defined as complement of A is a set consisting of all x belonging to U such that x is not in A or in other words by using the operations that we have defined already the complement of A is U set-A that is the elements in U which are not in A now we will consider the laws of set theory I will note down several laws here many of them are quite intuitive and extremely clear. So I will not give the proofs in general but I will give sketches of proofs as we go on describing these laws only one or two laws we will check in details now the first one says that A complement complement is equal to A this is called the law of double complement it is not difficult to prove this because we can always say that suppose I have got an element x which is inside A complement of complement this implies that x is not in A complement now if x is not in A complement this automatically in implies that x is inside A the reason is that A complement consists of exactly those elements which are not in A so if something is not not in A then it is of course in A therefore we see that any element in A complement complement is going to be in A therefore A complement complement is a subset of A on the other hand if I have an x inside A this will mean that x definitely is not inside A complement but from the definition of complement I can take again a complement because x is not in A complement and I can consider A complement as a set with which I am starting then I take complement of that and then x has to be inside complement of A complement and therefore A must be a subset of A complement complement and combining these two results we see that A complement complement is equal to A which is called the law of double complement the next result is the De Morgan's law which states that A union B complement is equal to A complement intersection B complement and A intersection B complement equal to A complement union B complement is called De Morgan's law now De Morgan's law needs a proof so let us start with the proof of De Morgan's law let us write De Morgan's law again it states that A union B complement is equal to A complement intersection B complement and A intersection B complement equal to A complement union B complement first we prove this part in order to prove this equality we start by considering an element x inside A union B complement which implies that x is not in A union B now let us draw a small Venn diagram over here and see what happens so here we have got of course this is the universal set u here we have got A and B and here somebody tells me that x is in A union B complement I know that A union B covers the interiors of the circles corresponding to A and B therefore A intersection B definitely is a region outside A union B so this shaded region is A union B complement so I am saying that A x belongs to A union B complement implies that x does not belong to A union B but what does it mean it means that x cannot be inside A neither can be x inside B therefore this means that x is not in A and x is not in B now we know that if somebody tells me that x is not in A this means definitely x is inside A complement and x not in B means definitely x is inside B complement now this means that x is inside A complement intersection B complement because this is the definition of intersection that if I have an element which is inside let us say set C and also inside set D then that element is inside C intersection D and here those two sets are A complement and B complement therefore x belongs to A complement and x belongs to B complement means x belongs to A complement intersection B complement now we have proved here that x belonging to A complement union sorry x belonging to A union B complement implies x belongs to A complement union sorry x belongs to A complement intersection B complement therefore we can say that A union B complement is a subset of A complement intersection B complement let us write this equation as the equation 1 next we start from the last line of the previous chain of arguments so we start from the point that x is an element of A complement intersection B complement suppose this happens then this implies that x is an element of A complement and x is an element of B complement this means x is not in A and x is not in B now we again refer back to the Venn diagram that I have drawn a while back we see that if I have a scenario where x is not in A and x not in B that means x will be in the region outside exactly outside A union B therefore I can safely say that x does not belong to A union B which means that x belongs to A union B complement therefore we can write that A complement intersection B complement is a subset of A union B complement let us denote this by 2 here we see that we are considering 2 sets that one is A union B complement and the other is A complement intersection B complement and we have proved that the set in the left hand side is a subset of the set in the right hand side and the right hand side is a subset of the set left hand side because of these 2 subset equality relations and therefore we can safely say that these 2 sets are equal and this is the first part of De Morgan's theorem now let us try to prove the second part of the De Morgan's theorem here our starting point is x is inside A intersection B complement now this means that x is not in A intersection B now x is not in A intersection B means that x is not in A or x not in B let us see that Venn diagram corresponding to the situation this is A this is B so let us write the sets AB and this is A intersection B now I am told that x is not in A intersection B so that means x is not here this means that x cannot be both in A and B because A if x is in A and B both then it is inside A intersection B so that means either one is correct either x is not in A and in B or x is in B but not in A or x is neither therefore x can be anywhere over here and that is expressed by this but this means that x belongs to A complement or x belongs to B complement which means that x belongs to A complement union B complement therefore like before we can write that A intersection B complement is a subset of A complement union B complement let us write this as 3 we can start again from the last line taking x to be belonging to A complement union B complement this means that x belongs to A complement or x belongs to B complement which in turn means that x is not in A or x not in B which in turn means x not in A intersection B which in turn means x belongs to A intersection B complement which gives us A complement union B complement is a subset of A intersection B whole complement call it 4 combining 3 and 4 we have A intersection B complement equal to A complement union B complement this completes the proof of de Morgan's law now let us quickly list all the other laws which are used very frequently A union B equal to B union A and A intersection B equal to B intersection A this is called commutative law for A union B union C is equal to A union B union C and A intersection B intersection C is equal to A intersection B intersection C these together is called associative law 5 A union B intersection C is equal to A union B intersection A union C and A intersection B union C is equal to A intersection B union A intersection C these two together is called distributive law 6 now A union A is equal to A and A intersection A is equal to A this is the idempotent law 7 A union ? equal to A that is of course if I take union of A to the element which has no to the set which has no element and the result is A similarly if I take A and take the intersection with the universe then also I will get A this is called identity law 8 A union A complement is U and A intersection A complement is ? this is called inverse law lastly we end today's lecture by a law which is called absorption law and which is useful very often which goes like this A union A intersection B is A and A intersection A B is also A this is called absorption law these laws more or less covers the most important properties of set operations which are used to manipulate sets we have proved some initial laws and later on we have given a complete proof of de Morgan's law which is not very easy to see and the rest of the other laws are quite intuitive and I leave it as exercise for the participants we stop the lecture today here thank you