 So, we will continue our discussion of countable and uncountable sets. So, let us quickly recall that in the last class we proved a few things about countable sets namely that first of all a subset of a countable set is countable then union of a countable family of countable sets is countable and finite product of countable sets is countable and using that we also saw that some of the familiar sets like this set of all integers is countable set of all rational numbers is countable. So, now let us proceed with that I will now introduce one more notation let us say we have two sets a and b then we shall denote this let us say this is read as a is dominated by b a is dominated by b or if this means there exists f from a to b and f is injective injective means as you know 1 1 f may or may not be onto f may or may not be onto of course if f is onto also then we know that this means a is numerically equivalent to b. So, what it you can interpret this as number of elements in a is less than or equal to number of elements in b that is the that is roughly the meaning. Now, what can we say about this relation the properties of this relations first of all we can say this for a obviously a is dominated by a you can take f as an identity function then second thing we can say is that if let us say if a is dominated by b and let us say b is dominated by c then that a is dominated by c that is also clear. So, if f is a function going from a to b which is 1 1 and if g is a function going from b to c which is also 1 1 just consider g composed with f that will be a function from a to c also be 1 1. So, that is clear third property is this if a is dominated by b and b is also dominated by a then yes if somebody said something then what can I say about a and b a is equal to b that is then a is numerically equivalent to b but this last thing is not obvious. So, what it means is that suppose you have injective function going from a to b and similarly an injective function going from b to a those two functions cannot be same then there exists a bijective function from a to b. Now, that is something that is not obvious at all and in fact that is a very well known theorem in set theory it is a very famous theorem it is known as Schroder-Berstein theorem and what Schroder-Berstein theorem says is just what I said just now if you take two sets a and b and if there is a 1 1 function from a to b and if there is a 1 1 function from b to a then there exists a 1 1 and onto function from a to b that is Schroder-Berstein theorem. The proof of this is somewhat like this so we shall not discuss that here I mean those of you who are interested you can see a proof of it in Siemens book I mentioned Siemens book right in the beginning it is it is given there is one more source in which you can find a fairly good proof of this perhaps many of you have heard of Professor S. Kumeration quite famous for this MTTS program some of you may have attended that also. So, he is a professor in University of Hyderabad Mathematics Department University of Hyderabad you look at his home page in University of Hyderabad and that home page contains some popular articles and one of those articles is a proof of Schroder-Berstein theorem you can find it there or also there is an MTTS home page which also contains some popular mathematics articles there also you can find this proof alright now let us again go to one another famous theorem in this by the way can you see that these three relationship issues look these three properties of this relation they look something like a this is not an equivalence relation but this is something like a partial order again not exactly a partial order because in partial order we would have required it if this happens then we should have had a equal to b. So, it is not a equal to b we only got a is numerically equivalent to b but what we can say is that if it is a partial order see suppose you look at this equivalence relation among the family of all sets then that equivalence relation will split will partition the family of sets into equivalence classes and what are what are the members of equivalence classes those are the members which are numerically equivalent to each other. So, suppose you take the set of all that class all those classes then on that class this is a partial order because then the equivalence class containing this what this says is that equivalence class containing a it is same as equivalence class containing b then let us go little further. So, far we have not seen any example of an uncountable set to do that let us again go to a famous theorem it is known as the cantors theorem this theorem says the following let a be any set then consider the power set of a that is the set of all subsets of a then first of all a is dominated by this power set of a we shall use this notation 2 power a as mentioned earlier that 2 power a or script p of a these two notations are very commonly used for the set of all subsets of a let us use this and it is not equivalent that second part is more important first part is more or less trivial what does it mean in terms of functions or definition it means there exists a 1 1 function going from a to power set of a, but there is no bijection there is no bijection between these two sets. So, let us go to the proof now first of all is it clear to you that this whole thing is trivial if a is an empty set if a is an empty set its power set will contain just one element and the set itself contains no element. So, there can be no bijection between the two and the 1 1 function will be a trivial function. So, that case let us forget about it. So, if a is empty it is trivial nothing to be true. So, assume a is non-empty. So, in the first part we need to show that there exists an injective function from a to its power set that is given any element let us say x in a we want to construct a function that is f going from its. So, suppose we take x in a this x f of x should be some subset of a f of x should be some subset of a now can you see that there is a very obvious choice for this then the singleton set containing the element x that is the most obvious function that one can think of. So, define f of x as singleton x. So, this is a function from a to its power set that function is 1 1 that is clear. So, f is 1 1 this part is proved. So, this part is proved now we will look at this. Now, here what is it that we have to prove that a is not numerically equal that means there cannot exist any bisection between these two sets. So, the way to proceed is fairly straight forward and that is assume that there exists a bisection and get a contradiction. So, suppose g from a to its power set of a is a bisection. In fact, we can show that there cannot exist any onto function from a to its power set of a, but that is all right. Now, we will construct by the way this proof is also given original proof is also given by Kantor. Kantor is a famous very famous German mathematician who has done several things in analysis and set field. You will hear this name again and again by the way Schroder and Bustin these two are also famous German mathematicians. So, suppose this is an onto function I will think of a set B as follows define B it is set of all x in a such that x does not belong to g x. Remember g x is a subset of a x is a function from sorry g is a function from a to its power set. So, every x in a g of x is a subset of a. So, given any x may or may not belong to g x. So, you pick up those x for which x does not belong to g x. I am not saying anything about whether this set is empty or not or whether such an x exist or not whatever is that. So, take all those x for which x does not belong to g x you call that set B. Now, this g is a bisection g is a bisection and this B is a subset of a. So, what follows from that there must exist some element in a such that g of that element is this. So, we can say that since in fact for this all that we require g is onto since g is onto I will say there exists suppose I call that element small b there exists small b in a such that g of this small b is equal to b that is fine. Now, we ask a question what can we say about this element B. Does this element B belongs to B of course, B is a subset of a and B belongs to a. So, B has to be either in B or outside B. Let us say what happens suppose suppose B belongs to B that means what this implies B does not belong to g of B right B does not belong to g of B but what is g of B. So, look at this what if what you have seen if B belongs to B then B does not belong to B that is a contradiction that is a contradiction. What is the other possibility suppose B does not belong to B suppose B does not belong to B but that will give that B belongs to B because that is how on the other hand that is on the other hand if small b does not belong to B which is nothing but g of B then the way in which we have defined B it means B must belong to B small b must belong to B then small b must belong to B. That means B belongs to small b belongs to B and small b does not belong to B both are leading to contradiction such a thing cannot happen B is an element of a and small b is an element of a big b is a subset of a. So, small b has to be either inside b or outside b and we are here we are seeing that both the statements are leading to a contradiction and again what is the source of this contradiction this we assume that there exists a bijection. So, that must be false. So, this is a contradiction. So, this completes the proof is this clear this is contrast original proof of this theorem. So, let me again come back to the statement of theorem that given any set a of course this is trivial part a is dominated by its power set, but a is not numerically equivalent to the power set of it a and its power set or there can be no bijection between any set a and its power set. Does it immediately give us an example of an uncountable set does this theorem immediately yes what is that to do with this theorem right that is right that is suppose we take a as the set of all natural numbers suppose I take a as the set of all natural numbers then n and 2 power n there can be no bijection there can be no bijection and it means that 2 power n is an uncountable set. So, this immediately gives that 2 power n is an uncountable set. So, we have got an example of an uncountable set in fact it can be shown that this 2 power n is actually numerically equivalent to real numbers this 2 power n is actually numerically equivalent to real numbers. Let me give you some idea about this how one shows that I will just give you some steps in this may not be the whole thing you have heard of this term binary sequence it is a sequence whose terms are 0 and 1 sequence is what we have seen the sequence is a function from the set of all natural numbers. So, any sequence whose terms are just 0 and 1 those are called binary sequences on the. So, what I want to say is that suppose you take the set of all binary sequences suppose we take the set of all binary sequences then that set is uncountable. So, set of all binary sequences is uncountable of course there are several ways of seeing this, but one way of seeing that is that we can say that this set is nothing but this set this set is nothing but this set of all binary sequences nothing but in the sense it is numerically equivalent and let us give some name to this suppose I call x is the set of all binary sequences let x denote the set of all binary sequences you want to say x is uncountable. So, I want to say this I will write this as a claim claim means this is something that I want to show claim is x is numerically equivalent to 2 power n I will take I will take a map from here to here. So, consider f from 2 power n to x that is given a subset of n given a subset of n I want to construct a binary sequence I want to construct a binary sequence. So, let a be a subset of n and define f of a define f of a you have all heard of what is meant by a characteristic function of a set of a subset. So, f of a I define it as a characteristic function of a what does this mean that if it is if it is 1 if 1 and if a number belongs to a that is remember this let me report this this chi suffix a of some natural number it is a function see it is a function from n to n it is a because it is a sequence it is a function from n and not n to n n to the set 0 1 n to the set binary. So, we will define n is equal to 1 if n belongs to a and 0 otherwise that is 0 if n does not belong to a for example, if a is a set of all even numbers then the corresponding sequence is 0 1 0 1 0 1 etcetera similarly we can So, is it clear that this characteristic function is basically a binary sequence characteristic function is a function it is a function going from n to see each characteristic function is a function going from n to this set 0 and 1 its values are 0 and 1. So, it is a binary sequence is this function 1 1 that is that is given a subset let us say if there are if the characteristic functions of two sets coincide that is when do you say something is 1 1 suppose f of a is equal to f of b that is same as saying that the characteristic functions coincide will it imply that a is equal to b that is clear is it on to that means given a binary sequence given a binary sequence can we construct a subset of n whose characteristic function is the given binary sequence that is again clear you take those n for which that that is you look at those n for which the value of the sequence is 1 collect those n and take that as a set a take that as a set a that will be a subset of n. So, this map which takes a to its characteristic function is is a bisection so f is a bisection that proves this right. So, f is a bisection we already seen that this is an uncountable set so x is an uncountable set we can also see one more proof of this that the set of all binary sequences is uncountable or that is again I will discuss this because it is also one of the well known methods of doing something is uncountable that is if a set is countable we already know that is infinite then you can arrange its elements in the form of a sequence right. So, suppose x is countable then we can write x as x 1 x 2 etcetera remember each of this x 1 x 2 is a sequence right each of this x 1 x 2 etcetera is a sequence. So, let us have some rotation for this for example, what is this sequence x 1 this sequence x 1 is I will denote it by x 1 as say x 1 1 x 1 2 x 1 3 etcetera remember each of this x 1 1 x 1 to either 0 or 1 each of this elements etcetera or either 0 or 1 so similarly so let us say ith sequence x i I can denote it as x i 1 x i 2 etcetera and to show that this set is suppose this were the case it will mean that you can list all binary sequences in this fashion and what we want to show is that that cannot be done. So, suppose I construct a sequence which is different from all these then it will mean that x is not countable right all right. Now, define binary sequence x as follows binary sequence x as follows I will use some other rotation because x 1 x I will call that sequence y define the binary sequence y as follows y is equal to y 1 y 2 etcetera and I should say what is y 1 what is y 2 what is y 1 what is y 2 etcetera what I say is as follows take y 1 you look at x 1 1 if x 1 1 is 0 you take y 1 as 1 and if it is 1 you take y 1 as 0 right. So, take y 1 as 1 if x 1 1 is 0 and 0 if x 1 1 is equal to 1 similarly you take y 2 same way look at this x 2 2 second sequence x 2 second sequence x 2 x 2 1 x 2 2 etcetera look at this entry x 2 2 here if x 2 2 is 1 you take y 2 as 0 if it is 0 you take y 2 as 1. So, you take y 2 as 1 if x 2 2 is 0 and 0 if x 2 2 is equal to 1 and now it is clear how to go about it proceed in this way take the general entry y n as follows y n is equal to 1 if x n n is 0 that is x n n will be somewhere here suppose this is x n may be x n n and 0 if x n n is equal to 1 then y is a binary sequence y is a binary sequence, but since y if y is a binary sequence it must be 1 of this x 1 x 2 x n right, but you can see that it cannot be x 1 because y 1 is different from x 1 1 it cannot be x 2 because y 2 is different from x 2 2 it cannot be x n because y n is different from x n n. So, y cannot be any of this and still it is a binary sequence right. So, this is a contradiction and this contradiction we got because of what because we assume that x is countable and wrote the elements of x in this form x 1 x 2 that cannot be done this is also fairly standard technique of proving that a set is not countable and this method is known as diagonal method diagonal method, diagonal method of proof and you can see the reason why it is called diagonal method that is you are arranging the elements in something like in the form of a matrix and looking at the diagonal entries and then constructing a new sequence which differs from each of the diagonal entries that is why it is called diagonal method or diagonal process is this clear. Now, you all know that every real number has a decimal expansion can be expressed in terms of its decimal expansion right you also know that it can also be expressed using binary expansion right decimal expansion is just one choice it can also express by using binary numbers just 0 and 1. So, what you can say is that this set of all binary sequences is nothing but the set of all real numbers you take any real number and take its binary expansion that is a binary sequence right. So, for each binary so each real number you can associate a binary sequence which is nothing but its binary expansion right and similarly if you are given any binary sequence you can associate a real number with that only problem is that which of the entries you take as an integer part and you which of this you take as a fractional part that will be a problem to decide with, but let us say we take only those numbers lying between 0 and 1. Let us just take the numbers lying between 0 and 1 right then this problem will not be there there is no integer part. So, you can say that all so suppose you are given any binary sequence like that you let us say some sequence 1 0 1 you can take that number at 0 point 1 0 1 0 etcetera that is the binary expansion of the given number. So, in other words this sequence x you can say that it is numerically equivalent to x this sequence x is a numerically equivalent to x. So, what does that prove that the interval 0 1 is uncountable it is it is numerically equivalent to the set of all binary sequences which we have already showed to be uncountable. So, that is uncountable all right. Now, we have put in the last class that a subset of a countable set is countable does it also follow from that immediately that if A is a set and if it has an uncountable subset then the A itself must be uncountable right. Suppose a set contains an uncountable subset then the whole set itself must be uncountable right it is basically the same it is basically the same statement said in a different method different language. So, now if this is uncountable that will mean that R is uncountable this is uncountable it means that R is also uncountable. In fact you can show something more if we were simply to say that R is uncountable then this is enough pick up some subset that is uncountable once show that, but we can say something more it is the following I will use that you as an exercise you take any open interval of R then you can show that that open interval is numerically equivalent to the whole of R that is in particular I will just give you this to you as an exercise show that and if you can do this for this interval you can do it for any interval show that what does it mean that show that there exists a bijection from the open interval 0 to 1 to the whole whole of R element exercise try to try to find such a function on your own and once you can do this you can show that there is nothing particular about 0 and 1 you can take any open interval and that any open interval is numerically equivalent to R and in particular any two open intervals are numerically equivalent to each other. But once you show this it will it will mean that R is numerically equivalent to 2 power n that is clear right we already shown that this is numerically equivalent to x and x is equal to 2 power n and this is numerically equivalent to R. So, comparing all this you can say that R is numerically equivalent to the power set of n all right now I will just make a few comments about what are known as cardinal numbers and then we shall close this discussion about the countable uncountable sets etcetera cardinal numbers. I think I have mentioned it earlier also if a is a finite set let us take set if a is a finite set cardinal by the cardinal number is nothing but a symbol which we associate with every set we call it the cardinal number of that set we call it a cardinal number of that set. So, suppose a is a set let us say the suppose a is a finite set we shall say cardinal number of a is 0 if a is empty and this is equal to n this is equal to n if if a is numerically equivalent to this j suffix n we have said it a is finite we have said it a is finite that means a is either empty or a is numerically equivalent remember what was j suffix n it was this set 1 2 3 etcetera up to n segment consisting of first natural numbers. In other words this n cardinal number of at for finite set is nothing but the number of elements in that set is nothing but the number of elements in that set then cardinal number of this set of all natural numbers that is divided by this symbol it is called a left not it is it is read as a left not a left suffix 0 a left not and then cardinal number associated with this set with any of these sets r 0 1 or 2 power n etcetera whatever I will just take this set 2 power n that is usually taken as the symbol small c and that is called cardinality of the continuum that is called cardinality of the continuum. Now if you take various cardinal numbers then we define relationship between them so suppose suppose a and b are cardinal numbers of course one thing is clear a equal to b will happen in which case a equal to b means those are associated with two sets which are numerically equivalent to each other that is a equal to b means let us write it in full form it means there exists sets a and b such that cardinal cardinality of a is small a cardinality of b is small b and a is numerically equivalent to b. So if the two sets are numerically equivalent the cardinal number associated with them is the same. Let us now also see what is the meaning of this a less than or equal to b again it means that there exists sets a b such that cardinality of a is small a cardinality of b is small b and a is dominated by b that means there exists sets a and b such and an injective function f going from a to b which is 1 1 and the corresponding numbers are small a and small b and similarly one will also like to define what is meant by a strictly less than b this again as usual we will say that a is less than or equal to b and a not equal to b in terms of sets what does it mean it means that there exists sets a and b such that cardinality of b a is small a cardinality of b b small b and a is dominated by b but not numerically equivalent to b and in fact there exist no two sets with this property which are numerically equivalent to b that is the meaning of saying that a is strictly less than b. Now if you look at the set of all cardinal numbers can you see that this is now a partial order for example we can see this property that is of course a is not equal to a that is clear if it is not equal to a then second thing is that a is not equal to b and b is not equal to c that implies a is equal to c and lastly is not equal to b and b is not equal to a. Now this time, I can say this implies a is equal to b and then this last thing follows from the Schroder-Berstein's theorem. Because a is less than or equal to b and b is less than or equal to a means what? That means there exist two sets this a and b. Satisfying is probability that cardinality of a is small a, cardinality of b is small b and a is dominated by b and b is also dominated by a. Then in view of Schroder-Berstein's theorem a and b must be numerically equivalent which is same as saying that small a is equal to small b. So now among the set of all cardinal numbers this relation less than or equal to is a partial order. This relation is a partial order in those numbers. Now let us take one more definition in this. Suppose a is a cardinal number a. This is called infinite cardinal. If this number a f naught is less than or equal to a it is called infinite cardinal. Now coming back to this relationship for example a f naught itself is an infinite cardinal. c is an infinite cardinal. Any number bigger than or equal to c is also any cardinal number bigger than or equal to c is also infinite cardinal. One more thing that one should notice here is that in view of Cantor's theorem for any cardinal number a, a is less than 2 power a. And what is meant by 2 power a? If this number a is associated with a set a cardinal number of a then 2 power a is the cardinality of the power set of a. 2 power a is the cardinality of the power set of a because that is what follows from the various definitions that we have seen so far. Now we can see the relationship between whatever the cardinal numbers that we have seen. So if you try to arrange all of them in certain particular order the smallest cardinal number is of course 0. Then 0 that is less than or equal to 1 that is less than or equal to 2 etcetera. And this is less than or equal to n that takes into account all finite cardinal numbers. Those are same as the 0 and other natural numbers. Then all these finite cardinal numbers those are strictly less than a left dot. What it means that given any finite set you can have an injective map going from that set to n but no onto map there is no bijection. So that is the same as seen all these finite cardinals are strictly less than a left dot. And then by what we have observed there a left dot is strictly less than 2 power a left dot. But 2 power a left dot is same as c that is what we have seen here r is numerically equal to 2 power n cardinal number associated with n is a left dot that associated with r is correlated with r. So which is same as saying that is 2 power a left not it is same as c. Then c will be again strictly less than 2 power c. What is 2 power c? It is you take the set of all subsets of r and whatever the cardinal number you will associate with that that will be 2 power c. And similarly you can go on. So 2 power c will be less than 2 power c sorry 2 power 2 power c etcetera. Then that will be less than 2 power 2 power 2 power c that way you can go on. Now this leads to one very natural question. Now this is fine a left not is less than 2 power less than 2 power c that is c is less than 2 power c that is also fine. The question is do there exist any cardinal numbers lying between these two? Do there exist any cardinal numbers lying between these two? For example between a left not and c and between c and 2 power c. So let me make that question precise that is of course that question is trivial if you take here. If it is a finite cardinal then you can easily find the numbers lying between n and 2 power n that is trivial. But it is not clear about the infinite cardinals. So let me just say that let a be an infinite cardinal and infinite cardinal does there exist a cardinal number b such that a is strictly less than b and b is strictly less than 2 power a. Of course this is the question. In fact it will be easier to understand this question if you formulate it in terms of sets. Suppose you are given an infinite set. We have shown that every infinite set contains a countably infinite subset. So the cardinality of an infinite set will be bigger than or bigger than or equal to a left not. So if you take any infinite sets its cardinal number will be an infinite cardinal. So suppose you take any infinite set and you take its power set. Then cardinality of that set will be a cardinal number of the power set is 2 power a. So asking whether there exists a b with this probability means given an infinite set and its power set. Can you find a subset which is bigger than or equal to the given set that means this. Can you find a b such that a is dominated by b but not equal to b and b is dominated by 2 power a but not equivalent to 2 power a. That is what it means in terms of set theory. And as such nobody knows the answer to this question till now. This is an open question in set theory. The answer is not known. And there are certain special cases of this. They have some very special name. One of them is called continuum hypothesis. It is called continuum hypothesis. Continuum hypothesis assumes negative answer for this when a is a left not. In other words continuum hypothesis says that there is no cardinal line between alpha not and c. Or in terms of set there is no set which is strictly bigger than set of all natural numbers and strictly less than set of all real numbers. That is continuum hypothesis. And similarly there is another thing which is called generalized continuum hypothesis. Let me simplify generalize continuum hypothesis. Generalize continuum hypothesis says the answer to be known for this whole question. And what is known about this continuum hypothesis is that using the other standard axioms of the set theory you cannot prove or disprove continuum hypothesis. Now this is something you may be find it difficult to understand. There is more to do with logic. What it means is that if you assume that continuum hypothesis is true that is quite consistent with all other axioms of set theory. And you can develop certain kind of set theory assuming that continuum hypothesis is true. At the same time if you assume that continuum hypothesis is false then that is also consistent with all the other axioms of the set theory. And you can develop some other kind of model of set theory assuming all other axioms and assuming that continuum hypothesis is false. Now this is expressed by saying that continuum hypothesis is independent of all the other axioms of set theory. That is using other axioms of set theory you cannot prove or disprove continuum hypothesis. So that is the answer known about discontinuum hypothesis. But even that is not known about this generalized continuum hypothesis. So with this I will close this discussion of countable and countable and finite sets for the time being. From the next class we will go to the next topic.