 So, welcome to the first lecture of this course on point set topology. So, lecture 1. So, in this lecture we shall see the definition of a topological space and some very basic examples. So, before we define a topological space. So, recall the definition of a power set. So, let X be any set, then the power set of X denoted P of X is the set consisting of all subsets of X. So, in particular the empty set which we denote by the symbol phi and the full set X are both elements of P of X. So, having recall this definition. So, let us define what a topological space is of a topology. So, let X be a set, let tau be a subset of the power set of X which satisfies the following three conditions. So, the first condition is the empty set the full set X are both in tau. So, recall that tau is a collection of subsets of X and what we need is that tau contains the. So, the empty set and X are both elements of the power set and what we need is that tau contains both these elements. The second condition we need is suppose u 1, u 2 up to u n are finitely many subsets of X and all these are in tau and all these are elements of tau. Then this intersection i equal to 1 to n this finite intersection which is a subset of X should also be an element of tau. So, over here let us emphasize we emphasize that in this condition there are only finitely many. So, this is the second condition and the third condition is let i be any set possibly infinite. Suppose for each i suppose for each i in i we are given a subset u i of X such that u i is an element of tau. Then the subset of X which is the union you know all these i is u i is this subset of X should be and once again let us emphasize that we emphasize that in this condition the collection u i may be infinite. So, suppose we have a collection tau like this a collection a subset of P of X which satisfies the above three conditions is called a topology on X. Then we shall write let X comma tau be a topological space. So, by this we mean that X is a set and tau is a topology on X. So, this is the definition of a topological space and now we will see some examples of topological spaces. So, the first example so, this is called the trivial topology. So, in all these examples let X be any set. So, here let tau be the subset of power set of X consisting of just two elements. So, namely tau is equal to it just contains the empty set and the entire set X. So, to check that tau defines a topology on X we have to check three conditions. So, let us check these. So, first condition is the first condition was that it should contain the empty set and X. So, clearly tau contains the empty set and the set X. So, the first condition is satisfied. So, the second condition for being a topology was that if we take finitely many elements in tau if u 1, q 2 up to u n are in tau then the intersection is in tau. So, in this case an element of tau is either empty set or the full set X. So, first consider the case where any one of the u i's is the empty set then the intersection u i is the empty set and so is in tau. So, next we consider the case. So, if this does not happen that means that all the u i's are X. So, if this the only other possibility is that all the u i's are equal to X and in this case also intersection i equal to 1 to n u i is again equal to X and so is in tau. So, therefore the second condition is also satisfied. So, finally, let us check the third condition. So, the third condition says that let i be a set and assume that for each i we are given an element u i in tau. So, it is in u i is an element of tau which means it is a subset of X and moreover since tau is contains of just 2 elements. So, then each u i is either empty or X and if all the u i are empty sets. So, then this arbitrary union is equal to empty and is in tau otherwise there exists some j such that u j is equal to X which implies that when we take the union this is going to become equal to X which is in tau. So, therefore the third condition is also satisfied. So, this implies that therefore, so this shows that tau which is phi comma X satisfies all 3 conditions defining conditions to be a topology on X. So, thus tau defines a topology on X and this is called the trivial topology. So, this completes the first example. So, in the same way let us see another example this is called the discrete topology. So, here again X is any set and we take tau to be the entire power set of X. So, once again to check that tau defines a topology on X we need to check that tau satisfies the 3 defining conditions. So, let us check them one by one once again since the clearly tau contains the empty set and X because tau is all of the power set and the power set contains phi and X. So, this condition is satisfied the second condition is that if we take or let u 1 u 2 up to u n be finitely many subsets of X which are in tau then their intersection u i should be in tau, but this condition is also satisfied because the intersection as the intersection is a subset of X it is the power set which is equal to tau. So, therefore, the second condition is also satisfied and similarly the third condition is satisfied. So, let i be any set and suppose we are given for each i a subset u i in X u i of X a subset u i of X which is in tau. So, then we need to show that the union is in tau, but again exactly as in the previous point this is clear as the union is a subset of X and so is a member of the power set of X which is equal to tau. So, thus the third condition is also satisfied. So, therefore, tau defines a topology on X which is called the discrete topology. So, the first two examples were fairly simple. So, let us take the third example this is a little bit more interesting. So, this is called the finite complement topology. So, here once again X is any set and let tau be those elements of the power set of X or in other words those subsets of X such that either u is empty or X minus u is a finite set ok. So, for this tau let us check that it satisfies the three defining conditions for a topology. So, the first recall that first condition we need to check was that phi and X are in tau right. So, clearly phi is in tau right and also since X minus X this is the empty set is finite of cardinality 0 we get that. So, this implies that X is in tau therefore, the first condition is satisfied. So, the second condition was the following. So, if u 1, u 2, u n are elements in tau then we need to show that intersection u i of these u i's this also in tau ok. So, let us first consider the case. So, let us first consider the case where one of the u i's is empty right. So, in this case. So, then the intersection is empty and so is in tau. So, if this does not happen. So, if so the other possibility is so the other possibility is that u i is non-empty for all i equal to 1 to n. So, in this case what happens is that. So, this will imply that for each i the set X minus u i is a finite. So, now let us consider the. So, let us consider X minus this finite intersection. Now, some simple set theory shows that this is equal to the union i equal to 1 to n of X minus u i and moreover. So, as each X minus u i is a finite set this implies that the finite union is also a finite set right. So, therefore we have proved that. So, thus X minus this intersection i equal to 1 to n u i is a finite set which implies that intersection i equal to 1 to n u i is in tau. So, this shows that tau satisfies the second condition for being on topology and finally, let us check the third condition. So, here we have to start with. So, i is any set let i be any set and suppose for each i in i we are given a subset u i which is in tau. So, what does this mean? So, we have to check that the union of all these this is in tau. So, once again we consider two cases. So, first consider the case when all the u i's are the empty set. So, in this case and clearly the union this is equal to m t which is in tau. So, if this does not happen that means that the other possibility is that there is 1 j for which at least 1 j for which u j is non-empty. So, this will imply that x minus u j is a finite set. So, now as before let us consider the set x minus union of these u i's and some simple set theory tells us that this is equal to intersection i in i x minus these u i's and this intersection is obviously contained in each of these pieces each of this subset. So, in particular this is going to be contained x minus u j right which is a finite set which is a finite set. So, this shows that x minus this union of i is a finite set which implies that union u i is in tau. So, therefore, tau satisfies the third condition also. So, thus we see that tau satisfies the. So, the tau we defined is here using this condition this collection tau it satisfies the three conditions which define the topology. So, this topology is called the. So, therefore, what we have seen so far is that given a topological space x any I am sorry given a set x any set x we saw three examples of topologies that we could put on it. So, we will end the first lecture here and in the next lecture we will see some specific examples which we will often encounter in this course.