 Well, we were discussing finite and countable sets in the last class. Let us recall again a basic definition namely that we had said that two sets A and B are numerically equivalent and we had used this notation A is numerically equivalent to B if there is a bisection between A and B and then we had said that A is finite if A is numerically equivalent to some segment 1, 2, 3, n, etc. and A is countable if A is numerically equivalent to the that is countably infinite if it is numerically equivalent to n. Countable infinity is same as also as I said some book called denumerable and countable means finite or countably infinite and similarly of course uncountable there is something I did not mention last time uncountable this simply means not countable that means neither finite nor countably infinite or to be more precise it means that there is no bisection between that set and the set of all natural numbers that is that is the meaning of uncountable. Now, the whole question is how does one show that a certain set is a finite or countable or uncountable or anything like that and let us now see some techniques of that in this class. Let me again remind you that in the last class we showed that every infinite set has a countably infinite subset and using that we showed that every infinite set is numerically equivalent to a proper subset and so this is a property which distinguishes between finite and infinite sets. No finite set can be numerically equivalent to a proper subset of it whereas every infinite set is numerically equivalent to a proper subset of it. Before proceeding further let us first observe that this relation is basically an equivalence relation that is something that is fairly easy to see. So, we will see this thing that is first of all a is numerically equivalent to itself that is clear you can take the identity map. Secondly if a is numerically equivalent to b then b is numerically equivalent to a that is also easy to see if f is a bisection from a to b you can consider the it is inverse function f inverse that will be a map from b to e b to a and that will also be a bisection and similarly the so called transitivity if a is numerically equivalent to b and b is numerically equivalent to c then these two things should imply that a is numerically equivalent to c and how can this last thing we proved just consider that is if f is a map from a to b which is a bisection and if g is a map from b to c you consider g composed with f that will be a map from a to c and that will be a bisection alright. Now, this is of course a very straight forward thing to c but what is the great use of this it is the following that if you have shown already that some set is finite and if you know that there is a bisection between a and b it will give automatically that b is also finite similarly if a is countable then b also will be countable or if a is not countable uncountable then b also is uncountable. So, you can use the information about the sets that you already proved using this fact we will see how this can be done now to first of all let us get some working list of countable sets we already know of course that n is countable countable in this case countable infinity and then let us now to prove certain things about countable sets we need some properties of this set of all natural numbers n alright. Now, how does one prove properties of natural numbers n I mean if you want to proceed a very logically you have to start from the definition of n and how n how this set of all natural numbers n is constructed and standard way of that is following what is called this piano axioms it is called piano axioms this is a method of constructing the set of all natural numbers, but that will be very time consuming and we shall not go into that kind of destruction of natural numbers if you are interested I will give you some references where you can find these things, but there are two properties of natural numbers which we shall be using very often and let me list those properties. First is what is called well ordering principle the principle is very easy to state it simply says that if you take any non-empty subset of n then it has a least element so every non-empty subset of n has a least element least element means what that element is less not equal to every other element in that set of course using this property one can show that that least element is unique alright. The second property is what is called principle of mathematical induction of course this is a principle which you have used right from your school days to give various proofs. Now what does this principle say of course you would have used in some form let us let us see in what form we are going to use the principle says this suppose you have a subset of n suppose s is a subset of n is a subset of n and it satisfies these two properties what are the property the first property is that one belongs to s and the second property is that if some natural number m belongs to s then m plus 1 also belongs to s so that is the second property that is n belongs to s implies n plus 1 that is if n belongs to s then n plus 1 belongs to s so if you consider a set satisfying these two properties then what should happen that set s must be the set of all natural numbers then conclusion is this then s is equal to n any set satisfying these two properties must be the set of all natural numbers this is something you have used several times in proving so many things may not been this particular form what is not and we shall be using these two properties of natural numbers these two in in proving that several sets are countable since that is going to be used again and again I mentioned it again you can take this as an exercise actually these two principles are equivalent what it means is that well ordering principle implies this and principle of mathematical induction implies this so I will give that you as an exercise it is an interesting exercise to try show that I will show that one is equivalent to two that means if you assume well ordering principle you can prove this on the other hand if you assume principle of mathematical induction then you can prove well ordering principle I will not discuss that proof here it is a try it on your own if you get held up we shall discuss here interesting exercise on its own now how we are going to use this as I said we shall use it to show that several sets are several familiar sets are countable to begin with let us just start with this by the way why is this called a well ordering principle the reason is the following suppose you let you take a nonempty set let us say a a is a nonempty set then it has a least element suppose you call that element let us say a 1 you remove that element a 1 then a minus a 1 either it is the whole of a or it is still nonempty if that is the case that remaining set also has a least element you can call it a 2 then again do the same thing look at a 1 a 2 either that is the whole of a or a minus a 1 a 2 is again nonempty if it is nonempty then that also has a least element you can call it a 3 so this way you can order the elements of a you can order the elements of a that is why it is called well ordering principle and this is in fact the principle which we shall be using in some of the proofs so in fact in this particular proof which let us what is what does this theorem say it is that every subset of a not not every subset of n is countable we already know that n itself is countable so let us see how this can be proved so let a be a subset of n if a is empty there is nothing to be proved right so if a is empty then a is finite nothing to be proved finite means countable because our definition of countable means finite or countable infinite so assume a is nonempty once you say a is nonempty I can invoke this I can invoke this so if it is a nonempty it has a least element then a has a least element I do exactly the thing that I mentioned just few minutes before that least element I will call a 1 so that least element a has a least element suppose I call that element a 1 now consider a minus a 1 consider a minus this is equal to a 1 right then again there are two possibilities either a minus a 1 is empty or nonempty right so a minus a 1 is empty or a minus a 1 is nonempty if this happens if a minus a 1 is empty what does it mean it means a is just singleton a 1 it has only one element it is a finite set all right so this implies that a is singleton a 1 and is finite what does this imply what will this mean if it is nonempty you again apply the same principle here by this that will have a least element and I will decide to call that least element a 2 so if a minus a 1 is nonempty it has a least element it has a least element and suppose I decide to call that least element a 2 what is to be done after that you consider set a minus a 1 a 2 so consider a minus a 1 a 2 consider a minus a 1 a 2 all right now I will not write everything in the book let us consider a minus a 1 a 2 what are the possibilities here either this is finite or infinite if it is finite we again just say then a is just a 1 a 2 it is a finite set consisting of two elements if not again consider least element call that element a 3 and proceed proceed so this this tells you a technique of how to proceed with this suppose we go on like this what are the possibilities there are two possibilities that is after n steps this may stop then that is a minus a 1 a 2 a n will be empty in which case a is same as a 1 a 2 a n right so then it is a finite set it is it is a 1 a correspondence with 1 2 n if this does not end that means at every stage n a minus a 1 a 2 a n is nonempty you can consider least element of that set call it a n plus 1 and proceed in which case a will be equal to a 1 a 2 n infinite set and that will be in bisection with n so it is countably infinite that is clear so in either case either this procedure if this procedure ends after a finite number of steps it will be a finite set consisting of n elements otherwise it is an infinite set and that infinite set is in 1 1 correspondence with the set of all natural numbers that will be a countably infinite set so it is either a finite or countably infinite which is basically that it is countable that is clear all right now let us come back to this principle which I had said once you prove that certain set is countable anything which is in 1 1 correspondence with that will also be countable so using that fact we can prove the following I will call it as a crawler now that we have proved that every subset of n is countable does it follow from here that every subset of a countable set is countable yeah that is what I said just if you prove something for one set you can say the same thing for any other set which is in 1 1 correspondence with that which is numerically equivalent so I will say a corollary of that is every subset of a countable set is countable now in order to show that a set is countable we have seen that there are two ways one is that you can show that it is in 1 1 correspondence with the set of all natural numbers let me state that here I will only consider the case of countably infinite sets if a set is finite I mean there is nothing much to be proved so consider the set a then I want to say that the following things are equivalent the following statements are equivalent in fact this is something we are going to use very often so there is again a very standard notation used for this it is this t f a the following are equivalent we will just use this t f a to show that the following are equal since we are going to use it again and again we will use this standard notation and it is fairly common it is used fairly often also and what is that first I want to say that a is countably infinite then there exists a subset b of n and a map f from b to a that is on all right and similarly we can say that there exists a subset let us say c of n and a map so we are called g from a to c map g from a to c that is 1 1 what does it mean that these three are equivalent that is and what is the practical use of this theorem that is if I want to show that a set is countable we have seen earlier that by definition I should say that there exists a 1 1 correspondence from a to n but what this says is that in fact one can show something less for example it is enough to show that there is a map from a to some subset of natural numbers which is 1 1 right that is enough there is we can show g is there is a map g from a to c not not dc this will mean complex number a to c that is 1 1 or you can find a subset of natural numbers and find a map from that subset to a which is onto that is sufficient that is sufficient all right now first of all it is clear that 1 implies 2 and 1 implies 3 you can just take that subset as the set of all natural numbers itself you can just take this b is equal to n and then there is a map from n to a which is 1 1 and onto and so in particular onto similarly here you can take this csn you can take this csn so real issue is to prove that 2 implies 1 and 3 implies 1 to show that see right let me just write 1 implies 2 1 implies 3 obvious clear right so the issue is to prove 2 implies 1 and 3 implies 1 all right so we need to prove 2 implies 1 and 3 implies 1 since 3 implies 1 is easy I will take that first what does this say that there exists a subset c of n there exists a subset c of n and a map g from a to c that is 1 1 right that is there exists a subset c of n and a map g from c to a that is 1 1 only thing is that this map may not be onto only thing is that this map g from c to a may not be onto but does not matter see instead of I can I can consider no I think I not c to a a to c a to but what we can say is that even if it is not onto onto n we can consider this g of a we can consider this g of a then a to look at the map a to g of a look at the map a to g of a we already know that it is 1 1 and since we are not we are taking only the elements of g of a that is onto we are taking only elements from a to g of a that is onto. So, a and so what it means is that a is numerically equivalent to g of a right is numerically equivalent to g of a all right is this g of a subset of n right its subset of n now look at this theorem every subset of n is countable we have already shown every subset of n is countable. So, in particular this g of a is countable and hence countably infinite because it is not finite. So, it is infinite so it is countably infinite that shows that a it is countably infinite fine in a similar way you can show that 2 implies 1 because what we can think of is the map which takes f from b to a that is is this clear first of all 3 implies 1 it is enough to just consider a subset of a natural numbers and show that there is a map from a to c that is 1 1 and similarly it is enough to consider a subset of b of n and show that there exists a map b to a that is onto right. See what you can do is that you can see the map again similarly as we have followed here the map from b to a is onto it is not 1 1, but what we can do even if it is not 1 1 from b to a I can think of a subset b of a and a map from b of a which is both onto as well as 1 1 and in that subset is countable and how to consider that map from b to a which is 1 1 and onto either you can take either you can use axiom of choice or you can use well ordering principle that is you can take that is you look at an element a consider it is inverse image in b that will be some non-empty set if that is a subset of just 1 element there is no problem if it contain many more elements you choose the least element of that and do it for each element in a that way you will get a subset of b which is in 1 1 correspondence with a that will be a subset of natural number I have given you the main idea you can develop the proof with this. Now, we just consider one more extension of this instead of considering a subset b of n that is what I want to do is that I want to replace this n by any countable infinite set right instead of taking n I can say that there exist subset b of some countable set and a map b from f from b to a which is onto. So, again by the principle which we thought of just now this is similarly I can consider any other subset from a to any subset c which is instead of taking subset c of n I can take any countable set and say that there exist some map from a to c that is 1 1 that will be enough that is enough right and this is a principle that we are going to use to show that several well known sets are countable or not countable. In fact, this will be used to say that they are countable. So, first of all let us just take an illustration of this I want to say that n cross n is countable and of course, since it is easy to say it is infinite we shall show that it is countable infinite n cross n is countable infinite I will use this 3 if I am able to say find a subset of natural numbers and a map. So, suppose I call this a is n cross n if I am able to find a subset of natural numbers and a map g from this set a to that subset c which is 1 1 that is enough right what are the elements of n cross n yes ordered pairs of the natural numbers ordered pairs of. So, n cross n this is the set of all elements of the form m comma n where m and n both are natural numbers. So, I consider a map let us say g from a in fact I will take this set c as n itself. So, I will take a map a g from a to n as follows define g from a to n as follows g of this m comma n g of this m comma n I will take this as say 2 to the power m into 3 to the power n. So, this is a map from a to n that is ok is it 1 1 see how does one show that something is 1 1 see suppose you take some other m 1 n 1 then g of m 1 n 1 is 2 to the power m 3 to the power m 1 3 to the power n 1. So, if let us say we consider g of m 1 n 1 this will be 2 to the power m 1 multiplied by 3 to the power n 1 suppose these 2 are equal g of m n is g of m 1 n 1 then that means 2 to the power m into 3 to the power n is same as 2 to the power m 1 into 3 to the power n 1 will it follow from that that m is n is same as m 1 n 1 that is suppose these 2 are equal then that will imply that 2 to the power m into 3 to the power n is same as 2 to the power m 1 into 3 to the power n 1. So, what I can say from here is that this will imply that 2 to the power m minus m 1 is same as 3 to the power n 1 minus n, but some power of 2 is equal to the some other some power of 3. So, when can this happen this only one way that this can happen that is that must be 1 that is if this is to happen then this must be equal to 1 that means m minus m 1 is 0 and n 1 minus n is 0 which is same as saying that m is equal to m 1 n is equal to n 1. So, does it prove whatever we wanted to say we have found a map from n cross n into n which is 1 1. So, that is does it show that n cross n is countable that is that was our 3 that was our 3. What if I take n cross n cross n I can say I can take a map to 2 to the power m 3 to the power n. So, let us something like 5 to the power something and what is the most general statement I can take a finite number of times countable product finite number of times the product of that is n cross n n cross n finite number of times in other words product of n cross n cross n taken finite number of times is a countable set. And all that what is what is so peculiar about this 2 and 3 it is just that they are primes it just that they are primes you can just choose a finite number of primes and construct a map like that. So, a product of a quotation product of n taken with itself a finite number of times is a countable set. In case the infinite number of products. Yes. Infinite number of products. No, we will come to that later. As far as finite. So, Cartesian product of a finite number of copies of n is countable. So, this is a theorem here this corollary is that I can replace n by any other countable set. In other words if I what I want to say is that suppose if a 1 a 2 let us say a k are countable sets are countable sets a 1 cross a 2 cross a k is countable. In other word if you take a finite family of countable sets then and then take the Cartesian product then that is countable. Does it follow from here all that we are doing is replacing n by any countable set each of this a 1 a 2 a k is equivalent to n and so replacing n by all that. Let us look at the union and in union we can take even a countably infinite family. So, let us say that let a i i belonging to let us say n be a family of countable sets a be their union a b equal to union of a i i belonging to n then a is countable or to put it in words if you take a countable family of countable sets then their union is countable that is union of a countable family of countable sets is countable that is what we say in words. The proof is very simple we can view this union a as a subset of n cross n we can view this union as a subset that is in other words we can establish a 1 1 correspondence between a and a subset of n cross n and since we have shown that n cross n is countable every subset of it is countable and hence that is countable. And how does one show that let us look at the set a 1 we know that a 1 is countable. So, it is finite or countably infinite. So, it is either finite or it is in verbal correspondence with n. So, suppose it is finite I can write its element as let us say a 1 1 a 1 2 etcetera. If it is finite it will stop somewhere if it is infinite it will continue it will be in 1 1 correspondence with n and similarly for a 2 I can write elements of a 2 as a 2 1 a 2 2 etcetera. If it is finite it will stop at some stage a 2 n otherwise it will continue and do it for any i. So, a i will be a i 1 a i 2 etcetera does it more or less finish the proof. Now, a is union of all this a i's and so I can write a as nothing but the set of all elements of the form a i's a because if you look at take all of them all elements can be expressed as a i's. And can you see an obvious map from this into this set n cross n. For example, I can say that you take a map f from a 2 n cross n what is the map a i's a goes to i comma j a i's a goes to i comma j. So, that map is 1 1. So, a is in 1 1 correspondence with some subset of n cross n and hence a is countable. We do not have to show that it is on to we already seen that it is essentially we are using this principle 3. So, let us again recall what did we prove so far that if you take any now let us go to some familiar sets go to this set z is the set of all integers. What are the elements of this they are all integers. So, what are so they are let us say the first is set of all natural numbers. So, what are the extra elements other than this d 0 this I will say this union 0 union minus 1 minus 2 etcetera. This is any way a countable set we know already this is a finite set what about this this is a in 1 1 correspondence with n it is a countable if it is. So, this is a finite union of countable sets this is a finite union of countable sets. So, it is countable. So, now we have got how did we begin we wanted to discuss how to show us that certain set is countable or finite. So, after proving these theorems we have now found one more technique. So, what were the techniques that we showed so far either you can show that it is in 1 1 correspondence with n or with some subset of n or with some set which you already shown to be countable and now we have got one more technique namely suppose you are able to show that it is a union of countable sets either finite union or countable union of countable sets then that is also countable. So, this is now one more technique to show that some set is countable you can show that it is a union of countable sets either finite union or a countable. In this case we have shown that it is a finite union of countable sets. So, z is countable so z is countable. So, z minus equal to 0 that is also countable. So, now suppose I take the product cross product z cross this that is also countable set. Now, can you see that obviously this is look at the set set of all rational numbers set of all rational numbers now what is that set by definition it is the set of all elements of the form p p by q where p q are integers and q dot 0 that describes all rational numbers that describes all rational numbers. Does it follow that this is countable because we have seen that this there is obviously the relationship between this and this set there is only one problem here this representation p by q is not unique 2 by 3 is same as 4 by 6 that is same as 8 by 12 etcetera. But that is not a problem we can see that once so that means you have a map which goes from here to here I can say that I can take the map p q goes to sorry p q goes to p by q that. So, map from here this to q this to q that map is on 2 but may not be 1 1 that map is on 2 but may not be 1 1 because as I said just now this 1 3 2 4 they will all go to the same rational number. But we have seen that that is not a problem you look at the statement 2 of our theorem if you have a map which is from a countable set which is on 2 but not necessarily 1 1 still this set is countable in other words we can q we can say that q is in 1 1 correspondence with some subset of this q is in 1 1 correspondence with some subset of this and since we already shown that this is countable all its subsets are countable. So, the set of all rational numbers are countable is clear now before proceeding further I will give you one exercise. Let us to take that consider exercise we need a definition of what is called an algebraic number have you heard of this earlier what is been an algebraic number does not matter we will define it a number is set to an algebraic number if it is a root of a polynomial with integer coefficients. So, it can be a real number or a complex number so it is called so let us say that z is a complex number it is set of so z is called an algebraic number if z is a root of a polynomial if z is a root of a polynomial with integer coefficients with integer coefficients which is same as saying that means polynomials means what polynomials is a polynomial is a function of the form a naught plus a 1 z plus a 2 z etcetera all that now what we what is it all this a naught a 1 a 2 plus b integers. So, what we can say is that if there exists let us say a naught a 1 a n belonging to z of course all of them should not be 0. So, let us just say a n is not 0 that a n is not 0 and what should happen is that this z be a root of the polynomial that means a naught plus a 1 z plus etcetera that is a naught plus a 1 z plus a n z to the power n this should be 0 is a root of a of course if you take all of them to be 0 then it will mean that every numbers will mean. So, we do not want it that is why we take non-zero polynomial and one way of ensuring that a polynomial is non-zero is that at least one coefficient is non-zero we shall ensure that a n is not 0. So, this covers everything now how do we how do we know that certain number is algebraic or not. Obviously, all integers are algebraic numbers we can just we can just take the polynomial for example, if you take the number 2 I can consider polynomial z minus 2 it is a root of that. If I take root 2 I can take z square minus 2 and it is a root of that it is a root of that. So, root 2 is an algebraic number now to decide whether a particular number is algebraic or not is not very easy for example, in case of root 2 it is fine, but suppose I give a number like e or pi you cannot say easily whether it is a algebraic or not. In fact it is not they are not algebraic numbers, but that is fairly difficult to prove. Now, what is the problem? The problem is this show that this set of all algebraic numbers is countable that is the problem show that show that the set of all algebraic numbers is countable. I give you a hint I will not give the full solution I will just give you a hint what are the ways of showing that a set is countable? The most recent way that we saw is you can show that it is a union of countable sets or union of finite sets whatever. So, I will say that you try to show that this is a countable union of finite sets you can show that it is a countable union of finite sets. So, now we have seen so many examples of the sets which are countable which are countable that is finite or countable in fit. Now, this may raise a question whether there exist uncountable sets or not whether there do exist uncountable sets or not and that is something that we shall answer by now how does one show that certain set is not countable? By definition you have to show that there is no one one correspondence between that and the set of all natural numbers that is one way of showing and then once you show that some set is not countable then any set which is numerically equivalent to that is also not countable how to go about showing that we shall we shall discuss that in the next class. We will stop with this for today.