 Good afternoon everybody. So, last time we were discussing several properties of the knowledge operator. Some of them are listed here. The fact that when a player i knows a then he then a is in fact obtained then the fact that if i knows a and a is a subset of b then he then i also then i also knows b and that the knowledge operator is essentially self aware. So, which means that when i knows a, i also knows that i knows a and so on. And knowing a and b is the same as knowing the intersection of a and b. So, knowing a and knowing b is the same as knowing a and b. And then finally that when i, when if you take k i of a complement that means if you take the event that player i does not know a then player i knows that he does not know a. Then from there we went into a characterization of sets that can be that are common knowledge in a certain state of the world. So, for this what we did was we defined a graph g. We said let g be this graph which is defined on the states of the world. And there is an age between any two states omega and omega dash if there is at least one player, if there exists one player who cannot distinguish between omega and omega dash. So, omega dash belongs to a phi of omega for at least for the for one player for a player i. So, this does not need to be true for all players it just has to be there has to be just one player who cannot distinguish between these two then there is an age. And then we defined what is called a connected component in this. A connected component essentially is a subset of y such that there is a path between any two omega and omega dash in that subset but not from a vertex in c from in that subset to a vertex not in that subset. So, there is no age connecting a vertex in this subset to a vertex not in the subset. So, this is what is called was called a connected component. And our next what we ended with was this theorem basically said that an event an event a is common knowledge in the state of the world in state of the world omega star if and only if that event is a is a superset of the connected component of omega star. So, every state of the world omega is is part of some connected component. Let c of omega be that connected component and an event a is common knowledge in omega star if and only if a is a superset of c of omega star the connected component containing omega star. So, what we will do today is actually we will prove this and alongside we will also see the structure of we will understand how the what kind of what the the structure of sets that emerges common knowledge. So, first let us understand this suppose this is this is y and now suppose this is the partition of a of some player. So, suppose this is this is my event this blue shaded one this is this here is my event all right. And now what is the what are the states of the world in which so I have described here the partition of player i. So, let us let us call this the partition of player i ok. Now what are the states of the world in which player i knows this event let us call this event a what are the states of the world in which player i knows a. So, those are actually this these right the yellow highlighted ones are the states of the world in which player i knows a ok. Now so, k i of a therefore, is the yellow region right because it is those states of the world in which player i knows is those states of the world in which if you look at the element of the partition corresponding to those states of the world that element is completely contained in a ok. Let us take another example I will just draw another set here let us look at this green set what are the this is this is some other event let us call this event say let us say event b this is event b now tell me what is k i of b is this k i of b is this red shaded region right it is the states of the world in which player i knows b ok. Now what do you observe here what you observe is that for take whatever event you want k i of that event take any event a k i of that event is always an element of the partition of a player right. So, k i of for any event a this is k i of a is is always a union of elements is always a union of elements from f i right. So, there could be more than one such element so, if my a is large enough there could be more than one such element. So, it is always a union of elements of f i. Now what does this mean essentially if you take any any if you take any any event a it the the k i corresponding to that event the k i of a is going to be formed by taking taking the union of elements of the partition of player i ok. The elements of the partition of player i which intersect a but are not completely enclosed in a are not included in k i of a right. So, in particular you look at something like this this this here is a is an element of player a i's partition it intersects a but it is not completely enclosed in a right it is not a subset of a. So, therefore, it is not an it is not it though these events are therefore, not in k i of a right likewise it again intersects b but is not a subset of b right. So, therefore, it is these events are also not in k i of b right. So, what this means is that if you have elements of player i's partition which intersect an event but are not completely which are but are not a subset of it then they are not in player i's knowledge of knowledge of that event ok. So, they are not in k i of that event all right. So, now what is this mean? So, let us look let us look at again now k i of a complement. Now, what is k i of a complement then k i of a is necessarily containing those elements of player i's partition which are completely enclosed in a ok. So, k i of a complement will include what? Yeah the union of all the other elements of this partition right. So, in particular see suppose. So, here I said the yellow region is k i of a the red region here was k i of b. So, what is k i of a complement? k i of a complement is everything other than the yellow region right. So, k i of a complement is everything other than the yellow region. Now, if this is everything other than the yellow region ok. Now, what does that mean? That means that itself is a union of elements of partitions of player a or a player i right because it is it is a union of elements of f i that are not a subset of a clear. So, all the elements of f i which are not a subset of a are r k i of a complement ok. Their union is k i of a complement ok. So, therefore, now what is k i of k i of a complement? What is this? This is itself now a so, if you since this is itself a union of elements of player i's partition it is therefore, it is you take any omega in k i of a complement right. The f i corresponding to that omega is completely contained in k i of a complement by definition right. So, for any so, this is itself equal to k i of a complement since for any omega in k i of a complement f i of omega is a subset of k i of a complement. And the reason for that is because k i of a complement is itself a union of elements of f i elements of f i that are not a subset of it ok. So, what this basically shows is essentially the property that I had talked about which is which is that k i of k i of a complement is equal to k i of a complement ok. Now, remember I have used here that this is this is not some arbitrary set right. So, if I put some set c here k i of c is not equal to c ok not necessarily, but k i of k i of a complement is equal to k i k i of a complement is this clear ok. Again just one more tongue twister for you if what if I put the complement inside do the is this still true yes or no it is true because when a player knows a set he knows that he knows when a player knows an event he knows that he knows the event the event here is just a complement that is it. Privileged actually this is this is this is true from the property we discussed last time ok. So, this is also true, but it is not but this is trivial as compared to this one which is which is more subtle ok alright. Now that we understand this the structure here no that is that gives us a chance to basically prove the theorem that we mentioned that I stated at the end of the last class. So, what I will do is I will just stated again here theorem ok I will just define the graph also once again. So, define that define the graph g, g is comprised of vertices from y and edges e omega, omega dash belongs to e if and only if there exists an player i such that omega dash belongs to f i of omega ok. And c of omega is the connected component of g containing containing omega ok and our theorem was this an event a is common knowledge in omega if and only if a contains c of omega alright ok alright. So, let us let us do the proof of this. So, first let us show that so, suppose a is common knowledge in omega ok and now suppose so, we will suppose that a is common knowledge in omega and prove we will show that c of omega is a subset of it ok alright. So, now omega is so, a is common knowledge in omega so, necessarily a obtains. So, omega is a subset of a sorry is an element of a ok. Now and omega is the also an element of c of omega by definition. Ok, omega is also part of the c of omega is the connected component containing omega. So, omega is in c of omega ok. Now, what we will do is we want to show that c of omega is a subset of a. So, what we will do is we will take an arbitrary element of omega let us call that omega star or omega dash. So, let omega dash belong to c of omega. So, this is an arbitrary element of c of omega and what we will show is that omega dash is in a ok. So, we will show that omega dash belongs to a ok alright. So, how are we going to show this? So, see omega dash is in c of omega omega is also in c of omega c of omega is the connected component containing omega omega dash is in that is in that connected component. What is the definition of the connected component? It means that between any two vertices in that connected component there is a path right. There is a path that connects these two vertices. So, this means that there exists a path let us call this path omega 0 omega 1 till omega omega r suppose. So, omega c till omega r omega r is itself omega dash and omega 0 is omega. So, there is a path like this starting from omega 0 equal to omega 2 omega 1 to omega 2 dot dot dot till omega r where omega r is actually omega dash in g which means that this what is what does it mean for there is a path that it just means that these are the the consecutive vertices here are connected by an edge ok. So, omega 0 is connected to omega 1 omega 1 is connected to omega 2 dot dot dot omega r minus 1 is connected to omega r ok. So, there is such a path in in g. Now, since there is such a path in g what does this mean? If you take any two consecutive vertices there is an edge which means take any two consecutive vertices there is at least one player who cannot distinguish between the two ok means for every consecutive every consecutive vertices that means take omega k and omega k plus 1 that is going to be a player who cannot distinguish between these two all right ok. So, actually because of this structure let us just prove this by induction prove by induction that omega k belongs to A for any. So, actually all we wanted to show was that omega dash belongs to A, but we will end up showing something stronger we will show that the entire path actually belongs to A ok. So, that comes for free throughout through the whole argument ok. So, we will show that omega k belongs to A for any k between 0 to r. So, obviously it is true for k equal to 0 because omega 0 we took as omega itself ok. So, true so base the induction the it is true for k equal to the induction hypothesis is true for k equal to 0 that is your base case ok true for k equal to 0. Now, assume it is true for some k assume it is true for some k now if it is true for some k now remember I said as I said omega k and omega k plus 1 is an edge which means that there is a there is a player who cannot distinguish between these two ok. So, assume it is true for some k which means omega k is in A and now we also know that there exists a i such that omega k plus 1 is in f i of omega k now omega k plus 1 is in f i of omega k ok. Now, remember what would be which means player i cannot distinguish between omega k and omega k plus 1 ok. Now, omega k plus which means that A is common knowledge also in omega k ok. So, let us go through this step by step see essentially what are we going to show we will show that A is common knowledge in A is also common knowledge in omega k plus 1 and therefore omega k plus 1 has to be in A ok. So, essentially so let us let us do this ok. So, I will just erase this and I will we will be able to show something even stronger. So, we will prove that we will prove the by induction that omega k belongs to A and the and that A is common knowledge in omega k ok. So, this is stronger actually, but in fact this is eventually going to be necessary because the entire you know connected component is going to be shown as a subset of A ok. So, so this is this will make our proof a lot easier. So, we will prove by induction that omega k belongs to A and also that omega k is common knowledge in A that A is common knowledge in omega k ok. So, this is trivially true for omega equal to for k equal to 0 true for k equal to 0. Why is it true for k equal to 0? Because we said that omega 0 is omega and A is assumed to be common knowledge in omega ok. So, this is true for k equal to 0. Now, assume it is true the induction hypothesis for some k ok. So, assume the induction hypothesis for some k ok. It is a k greater than 0. Now, there exists a player i in n such that omega k plus 1 belongs to F i of omega k ok. And now if you remember one of the things we had shown last time was this that if player i cannot distinguish if there is a player who cannot distinguish between two states and an event is common knowledge in one of the states then it is also common knowledge in the other states right. This is something we had shown if you you can you we showed this by taking we are taking a sequence of players which was once one length longer and then we drop the first guy and then we said we can make the we can apply the same the definition of common knowledge on the remaining sequence essentially that is how we showed this. So, now we now what we are in exactly that situation we have that there is a player who cannot distinguish between omega k and omega k plus 1. A is common knowledge in omega k ok. So, therefore, A has to be common knowledge in omega k plus 1 ok. So, which means that A is common knowledge in omega k plus 1. Now, A is common. So, this is give gives us that A is common knowledge in omega k plus 1 and once A is common knowledge in omega k plus 1 A actually occurs in omega k plus 1 which means A and omega k plus 1 has to therefore be a element of A ok. So, this basically establishes by induction that so, which means by induction omega dash is in A ok. Now, what did we start of seeing? We started of seeing that let omega dash be an arbitrary element of C of omega and we just showed that that arbitrary element is also in A which means that the all of C of omega is actually in A ok. So, I will just write that here which means that C of omega is a subset of A ok alright. So, this is shown that what have we shown we said that well suppose A is common knowledge in omega then C of omega is a subset of A that is the proof that is what we have shown. But the theorem asks us to show something stronger right it says that an event A is common knowledge in omega if and only if if and only if this holds.