 So, last class we were discussing accountability of a set. So, we said a set is set E is said to be countable if it can be put into a bijection with the set of natural numbers. In other words if you can write the set in a list and if you can exhaust writing all the elements of the set in a list it is a bijection and then that set is countable. So, if you put the elements of a set in a list and you exhaust the list the set is automatically countable. So, one thing so we one example we gave is that the set of all rationals if you take the set of all rationals in 0 1. So, we could write it as 0 1 1 by 2 1 by 3 2 by 3 1 by 4 3 by 4 dot dot dot right. So, this is a list. So, I have listed out all the rationals in the interval 0 1 right. So, every rationals you any rationals you can give me in the interval 0 1 is somewhere in this list right. Therefore, we said the set of all rationals in 0 1 is a countable set. Then we also stated a theorem without proof that countable union of countable sets is a countable set right. So, then we argued that rationals in any interval i i plus 1 is countable by the same argument and then the set of all rationals is after all the union of all these sets countable union of all these sets. Therefore, the set q is a countable set right which is which some of you may find some words surprising because they appear to be so many rationals right. So, but no according to our definition there are only as many rational numbers as there are natural numbers because there is a bijection right. Is that clear to everybody? So, there are other examples of countable sets there is have you heard of algebraic numbers algebraic numbers. So, algebraic numbers are those real numbers which are roots which are 0's of polynomials with integer coefficients. If I have any nth degree polynomial any degree polynomial with integer coefficients the set of all roots right the set of all real roots of those polynomials are called algebraic numbers and algebraic numbers are countable. For example, square root of 2 is an algebraic number it is not rational, but it is algebraic why because it is a root to the equation x square minus 2 equal to 0 right. But you can show that pi is not pi e these are all not algebraic they are all called transcendental numbers. So, you can show that algebraic numbers are countable the reason algebraic numbers are countable is there are only countable many polynomials with integer coefficients which you can again show. So, it has to have only countable many roots alright. Another very interesting set of countable another interesting countable set is what is known as the set of computable numbers. The computable numbers are the set of all real numbers which can be computed to any degree of precision with a terminating computer program. So, this it turns out that. So, algebraic numbers are a larger set than irrational numbers, but still countable and computable numbers are even larger set than algebraic numbers, but computable numbers are also a countable set. So, these if you want you can read Wikipedia or math world on these topics this is just for your additional information. So, we have identified several countable sets right now. So, now, we move on to the topic of uncountable sets right. So, here we are going to say that there are certain infinite sets which are truly larger in cardinality than natural numbers or rational numbers alright. So, these are called uncountable sets. So, the definition is as follows definition a set F is uncountable we do not say uncountably infinite it is understood that it is infinite right it is it is just uncountable uncountable F it has cardinality strictly larger than that of N. So, the definition says a set F is uncountable if its cardinality is strictly bigger than that of the natural numbers the cardinality of natural numbers. Now, what does it mean to say that a set has strictly larger cardinality than some other set we gave a definition remember. So, it just means that there is there exists an injection from F to N right. So, I E there exists an injective function from F or to N, but no such bijective function exists this is the definition we gave alright you can find an injection from F to N, but no bijection right no bijection can possibly exist between F and N. N to F did I make a mistake I think this is correct no. So, if you know. So, what I am saying here is that F has a cardinality strictly bigger. So, they cannot be a bijection that is clear right an injection means that. So, should I say N F here N F here is that fine then the bijection is fine. So, it is the other way around. So, you are saying that this is how it should be. So, we are saying that with every natural number yeah you can associate I think this is correct right. So, I think this is correct I think the earlier version has seen correct yeah no it is also have to correct it. So, this is sorry. So, if this is correct everybody with this. So, there is an injection from N to F which means that if I have this is my N and that is my F this is my F then for every natural number I can associate. So, there exists an injective function like that right. So, for every natural number which means there are at least as many elements in F as there are N, but there is no bijection which means there is something that will always be left behind here right the surjective property does not hold right. No matter what mapping I do I can never ensure that everything here is covered the core domain and the range will never be the same right. So, is this is the definition clear to everybody all right. So, it turns out that there are several examples of uncountable sets. So, I will write them down. So, the R the set of all real numbers is uncountable. So, is the set of all well the set of all irrational numbers R minus Q right. So, this is the all irrational numbers is also uncountable. So, any interval like A B is uncountable 0 1 is uncountable is an uncountably infinite set. The most elementary example in a mathematical sense is the set of all binary strings. If you take the set 0 1 power infinity right this is the set of all infinite binary strings right. So, you write out all possible infinite binary strings 0 1 1 0 0 0 0 right every string is infinite and all possible combinations are allowed right. So, 0 1 power N you understand right it has is the set of all N long binary strings right there are 2 power N of them right. This of course is an infinite set because you are considering infinite sequences of sets infinite binary sequences set of all infinite binary sequences. However, so this is infinite clearly clearly an infinite set it is actually an uncountably infinite set. You can never put this in a list it is not very clear now, but it you can prove it this is the most elementary example in some sense of an uncountable set. Another example is 2 power N. The set of all subsets of natural numbers. So, if I give you a set S 2 power S is the power set right set of all possible subsets of S right. This is the power set of the natural numbers and basically it includes the set of all possible subsets of the natural numbers. So, elements of 2 power N will be. So, a given element of 2 power N will be some subset of natural numbers and all possible subsets of natural numbers are included in this set right examples. So, there may be some finite subsets 23 14 something right that may be 23 14 48 right that is a subset of 2 that is a element of 2 power N being a therefore, it is a subset of rational numbers right, but it also contains infinite subsets right the set of all prime numbers is contained here the set of all even numbers is contained here the set of all multiples of 17 is contained here right. So, if you collect the set of all those subsets of natural numbers it is clearly an infinite collection it is in it is actually an uncountable collection. So, of course, we have to prove this right. The most elementary proof is this the uncountability of 0 1 power infinity the set of all infinite binary strings right. So, this theorem was proved by Cantor using a very celebrated argument known as Cantor's diagonal argument it is a very elegant argument right. So, if you theorem 0 1 power infinity is an uncountable set how do you prove that. So, in order to prove this what do you have to do you have to prove that there exists a injection from N to this set, but no bijection exists right. Can you prove that an injection exists from N to this can somebody give me an injection from N to F. So, for example, so injection it is easy to produce an injection F from N to F or F to N to F into let me call this let me call this S or something N to S the set I am calling S. So, an N to F for example, of an injection is. So, with every natural number you associate that string which has 1 at that point 1 at that index right. For example, you can take let us say F of 1 is equal to 1 0 0 dot dot dot right actually this is wrong notation I should actually just write 1 binary string right 1 0 0 0 dot dot dot F 2 is equal to 0 1 0 0 dot dot and so on right F 3 is equal to 0 0 1 0 0 right 0 0. There are many such injections you can if you want to flip the 0s and once you are welcome to do it you just need to produce 1 injection. So, what have I done here if you give me a natural number N I will consider that sequence which has 0s everywhere except at that index N is that clear. So, that is an injection is very easy right. So, now I have to prove that there is no bijection right that is the technical part right. So, this is where Cantor's diagonal argument comes in right in order to prove no bijection exists what do you think you should do. So, I have to prove that. So, I cannot say that oh I am not able to find a bijection right. So, I have found an injection, but I cannot prove that no bijection exists by saying oh I am not able to find it right that is not a proof right. You have to say that you have to prove that no bijection can possibly exists right. How do you prove something that cannot possibly exist? Assume that it exists and end up with the contradiction right. In order to prove that no bijection exists assume the contrary which means that I am assuming there exists a listing of 0 1 power infinity which exhaust the set 0 1 power infinity all right. So, for example, in particular suppose the following listing exhausts 0 1 power infinity ok. So, I am going to assume that some. So, I am going to say that a 1 1 a 1 2 a 1 3 dot dot dot dot a 2 1 a 2 2 a 2 3 dot dot dot. So, 1 right a n 1 a n 2 a n 3 dot dot dot dot dot dot dot dot dot. So, what am I doing here? So, this is my first binary string. So, this is each of this a i j's are 0 or 1. So, a i j is the j th bit of the i th string. So, you understand what I have done here right. So, each of this could be 0 1 1 0 blah blah blah and so on. So, I am assuming that this listing completes totally exhaust the set 0 1 power infinity that is the assumption and then I have to end up with the contradiction right. So, this is clear now. So, this is this is cantors proof all right. So, here is where the master stroke comes in. So, you have to prove that now this is cannot. So, I have assumed that this is a complete listing of the set 0 1 power infinity. Now, I am going to produce a binary string infinite binary string that cannot be contained in this listing. So, if I assume that this listing is exhausting the set and then I produce a string that cannot possibly be contained in this list, then I have ended up with the contradiction correct. Now, so cantor streak involves considering these guys the diagonal guys. So, you consider a i i. So, you take if this is a 0 or 1 you put that you take that right you just take the all diagonal elements. So, you consider the string. So, you consider the string a 1 1 a 2 2 a 3 3 dot dot dot except you flip these bits. So, if this guy is a 1 you write down a 0 if this guy is a 0 you write down a 1 this guy is a 0 you write down a 1 and so on right. So, you are taking all the diagonal bits and flipping complimenting this is a valid infinite binary string. Now, the key is in understanding that this string cannot possibly be contained in this list. How do you prove that suppose this string was contained in some mth position, then the mth bit will be wrong for any m correct. Suppose this suppose I claim that this is my 20th string in this list the problem is a 2020 will be complimented right. So, I have made everything wrong in the diagonal I have complimented everything right. So, this this string cannot possibly be in this list correct it is a very elegant proof right. There is nothing very mathematically sophisticated about it right, but it is a very elegant proof very elementary, but elegant proof that. So, this string consider this string blah blah blah this string is not contained in the this thing above you see why right. So, if it were contained in the mth position the mth bit is wrong. So, it cannot be contained in any of the positions here right. So, this is a contradiction right. So, it is a contradiction therefore, we conclude that no no bijection can possibly exist. See, I have not considered a specific bijection here where I have not tried a specific listing. I am saying if at all there exists a listing of the set that listing cannot contain that string. So, no listing of the set can possibly exhaust the set right understand this argument. So, there is exist no bijection, but there exists an injection. So, this set is an uncountably infinite set which one. No, I only no see this I was produce a specific injection, but I cannot try see I cannot say that there exists no injection by trying to by just failing to find bijection that is not a proof right. Maybe you are just looking at the wrong bijection. No, no, no see this is an injection right I am saying that there exists no bijection right. This does not lead to a bijection is not enough you have to prove that no bijection can possibly exist between the set and natural numbers. So, I have assume that let us say some bijection exist and arrived at a contradiction. So, again right. So, in order to prove that something exist you produce it, but in order to prove that some bijection cannot possibly exist you cannot just say I did not manage to find it right. You have to prove that no such bijection can possibly exist correct good. So, this is this is the famous canter diagonal argument it is a very elegant argument to prove that the set of all infinite binary string is an uncountable set. Any questions in this listing? We have taken all those rows as like with infinite number of bits right. They are all infinite by these things. Yes sir. If I am taking say for example the bit would just 1 or say 1 0. Sorry. So, no each of this is a bit. So, yes. So, infinite number of all those listings contains infinite number of elements. So, if I am saying if I am asking you where is the bit stream 1 0. There is no 1 0 right. So, all elements of this are infinite strings. So, 1 0 is simply a finite string. That also contains in this. No all elements are infinite strings. Each element of s is an infinite binary string. You can have all 0s, but it is an infinite string. So far. So, now using this the reason I said this is the mathematically speaking this is the fundamental example of an uncountable set. Because you know there is a elementary proof of this. Now, using this you can prove that the rest of the guys are in fact these guys are all uncountable right. So, what am I going to do? So, in order to prove that some other set is uncountable what do I need to do? So, if I find a bijection between this and this for example, then I know that this set cannot be in bijection with n why? Because if this set were to be in bijection then the composed function will be a bijection with 0 1 power infinity and that is a contradiction correct. So, I have to find. So, the moment I find a bijection with this set I am done or in fact even if I find an injection one of the direction saying that the set of interest has at least as many elements as this set I am done. Even in the injection is in the one direction from if I find an injection from here to another set I am done. So, in order to for example, if you want to prove that the interval 0 1 let us say the closed interval 0 1. So, if I take. So, I want to prove that this is uncountable what do I do? Yeah you can do that right. So, this guy. So, this includes all these numbers between 0 and 1 boundaries included. So, you write out the binary expansion of this right just like I mean. So, instead of normally we write down decimal expansion, but you write out binary expansions right. So, you the binary expansion of any number here will be like 0.0 1 1 0 something right some binary string some infinite binary string right and that is that looks like a bijection between. So, if I have x let us say some x is in 0 1 and write x in binary right. So, I will have x is equal to 0 point a 1 a 2 dot dot dot right where these are all binary bits basically right a 1 a 2 a 3 dot dot dot. Now, you can identify this infinite binary expansion with an element of this infinite string of infinite these strings right. So, basically you can see that they see appears to be one to one correspondence between the set the binary expansion and that infinite binary strings right. So, that is essentially the core of the proof to establish that the open interval 0 1 is an uncountable set there is one glitch here there is one tiny glitch here the glitch is the following. So, when you write a binary expansion. So, let us say you write the decimal expansion suppose I give you the number 0.0999999 for ever you will write that as that is actually equal to 0.1 right. So, similarly if you have a number like 0.010111111 something like that this number is actually equal to what this number is actually equal to 0.011000000 these two are actually the same real number right. They are actually numerically equal. So, if you write the value of this is equal to the value of this just like 0.99999 is equal to 1 right these two are actually equal. So, it is not really true that. So, but on the other hand if you look at these guys as elements of this infinite they are different strings right. So, there are two different infinite binary strings corresponding to the same real number in 0 1. So, that is a little bit of a sticking point right. So, this problem occurs only for those numbers which have terminating decimal binary expansions terminating binary expansions. So, these numbers are called dyadic rational numbers. The numbers which have terminating binary expansions are after all their rational numbers. In fact, they are called dyadic rational numbers they are of the form a by 2 power b right. So, but dyadic rational numbers are in fact countable because rational numbers are countable. So, there is just a problem with a small countable subsets of this bijection goes off track only in this small dyadic set right. So, you have to find a clever way of handling that and that proof is given in the notes. So, you have to just for these dyadic rational numbers of which there are countably many you have to have some way of handling it. So, that is the essence of the proof. So, I will not complete the proof, but I have given the essential argument right. The diagonal argument I am not sure it works here no the main see the problem is that this because the 2 different real numbers corresponding to the same infinite binary string you have a you mean you have a slightly smaller set actually. The set 0 1 if you expand it is a slightly smaller set, but it is only smaller by a countable set of dyadic. So, that is that is the essential proof. So, the proof is given in the notes the proper proof the complete proof even taking care of this dyadic is given in the notes. So, now that this we know is uncountable essentially because you can write out binary expansion and that corresponds to 0 1 power infinity correct. Now, this is uncountable then can we prove that r is uncountable you have to find some bijection between 0 1 and r and you can find there are many bijections between 0 1 and r that is not a big deal. So, there is you have to prove that no. So, you have to say that if you have a union of sets in which at least one element is uncountable then you have to prove a lemma saying that that set is uncountable it is correct, but. So, we have proved that countable union of countable sets is countable, but I have not said that if one of the sets is uncountable the overall set is uncountable it is true, but you have to prove it right. If you are going from fundamentals you have to prove that a better idea is to simply get a bijection from 0 1 to r actually I think I have a bijection in my notes here. So, if you want to have. So, if you want to prove that r is uncountable let us see I think I have an example here. So, if you r is uncountable because I think this is right tan of pi x minus pi over 2 this is what the T s have come up with I think and x belonging in 0 1. So, if you consider this mapping this actually is a bijection from 0 1 to all of r. So, it is just verify this. So, this is if you vary x from 0 this is a bijection from this interval to all of real numbers. So, then you prove that r is uncountable you just have to produce 1 bijection. So, good r is uncountable. So, we are done with this is what we fundamentally derived and then we did that then we did that. So, how do you prove that irrationals are uncountable? So, I want to prove this. So, I want to prove that r minus q is uncountable. So, he is saying that if you remove a countable set from an uncountable set you get a uncountable set it is true, but you have to prove it. So, here is the thing. So, you write like this you write it as r minus q union q right after all real numbers are irrationals union rationals correct. Now, I want to prove that this is uncountable. Suppose that it is countable suppose irrationals are countable then that union another countable set will be a countable set contradiction right. So, this guy is uncountable correct. So, we are done with that also. So, there are. So, what do you know now? So, there are so many more irrational numbers then there are rational numbers although they are both distributed densely on the real line right. So, in any tiny little interval you find so many irrational numbers and so many rational numbers right, but according to our cardinality definition there are so many more irrational numbers right. So, what is the only thing that remains this guy right yes you have question. So, you have to create a bijection between this and any of these guys actually you can create a bijection with this. So, after all if you take any subset of natural numbers it either. So, for each number natural number it either leaves it out or takes it. So, any subset of natural numbers will either leave out a natural number or take that natural number. If it leaves out that natural number put a 0 if it takes it put a 1 right. So, for every string here I can I have you identify unique subset here for every subset here I can identify your unique string here correct. So, for example, if I am looking at so if I am looking at the set of all let us say odd numbers which is an element of this what is the string that identifies with here 1 0 1 0 1 0 1 0 right it just picks all the odd numbers right. If I want the subset of all let us say prime numbers I have to consider the string that has once in all the prime indices and 0s everywhere else correct. So, that is a natural bijection between this and this right. So, 2 power n is a uncountably infinite set. So, this is all clear are there any questions of this point no no no I am saying that no no no. So, I am saying that suppose this is countable I know this is countable if this were to be countable the union will be countable, but I know that r is not countable. So, this is a countable. Therefore, this is uncountable no I want to prove that this is uncountable right. So, I want to prove that this is uncountable. So, that I proved it I have assumed that this is a countable and ended up with a count addiction right I do not need to prove anything more here right I know this is countable. So, if this were to be countable the union will be a countable set which I know is not the case because I proved r is uncountable. So, this has to be uncountable yeah correct. Now, you mentioned odd numbers no no no see the power set of n contains as its elements the subsets of n. So, the only difference see 2 power n is a odd ball here it was odd one out because it is not just a set it is actually a set of subsets. So, the elements of this are themselves subsets of natural numbers. So, while it is true that the odd numbers are countable I am saying that the power set of natural numbers which contains as its elements all the possible subsets of natural numbers it is an uncountable collection of subsets. So, for example, if you want to write down see this 2 power n see the elements of 2 power n are themselves sets I hope this is clear right. So, I am just saying that if you give me an element of 2 power n it will be a subset of the natural numbers correct. For that subset you write down 0 whenever that index is not included and write down 1 whenever that index is included. So, that gives the bijection between this and this right therefore, this is an uncountable set right. So, the reason I mentioned odd numbers is. So, if you are considering. So, it is just one particular subset of 2 power n one particular element of 2 power n right the particular element of 2 power n will have may be finite some of them will be finite some of them will be infinite. Let us say there are let us say one such example will be the odd numbers the element is the set of all odd numbers. So, that corresponds to the string of 1 0 1 0 1 0 1 0 here and that is the bijection. So, whatever subset of n you give me I can create I can get the corresponding binary string right. So, that is the bijection alright. So, I will stop here.