 Hello and welcome to the fourth lecture in this course. So, in the previous lecture we saw what a basis for a topology is and we are going to begin by proving a small lemma ok. So, let us begin with this lemma let tau be a topology on x let b be a basis for tau then union over all elements of b is equal to x. So, let us prove this it is clear the union over all w and b this contain x yeah as b is a subset of tau which is a powers subset of the power set of x and. So, w belongs to b implies w is a subset of x and we are just taking the union. So, we only need to prove that the reverse inclusion. So, let x in x be an element taking u equal to x and using the defining property of the basis and the element I am sorry and the point x in x we get that there is note that u equal to x this is in tau therefore, we can use the defining property of the basis there is a w in b we let us denote a w x such that x is in the set which is in b and obviously, w x is contained in x which is equal to u right. So, thus we get that this union contains x for all for all x in x thus x is contained in the this is because one of these w's is w sub x ok. So, this completes the proof of the lemma. So, now we will prove this very important proposition which tells us when a subset of the power set can be used to generate a topology. So, what it means to generate a topology we will explain just now. So, proposition let x be a set and let b contained in the power set of x be a collection which satisfies the following two properties. So, the first is the union of all w in b w should be equal to x. So, note that we have not given x at a topology right now. So, b is just a subset of the power set of x and the second condition is that given w 1 and w 2 in b and any x in w 1 intersection w 2 there is w in b such that this point x is in w and w is contained in w 1 intersection w 2. So, note that w 1 intersection w 2 may not be in b, but there is another w which is in b and which is contained in w 1 intersection w 2 and contains x. So, we have this collection b which satisfies these two conditions and we define a topology. So, define a collection tau which is a subset of p of x as follows tau is equal to those subsets u of x such that x in u there exists a w in b there exists a w in b such that x belongs to w and w is contained in u. So, then tau is a topology b is a basis for tau. So, let us prove this proposition in order to prove this proposition we have to check that tau satisfies the three conditions which define a topology. So, let us check these one by one. So, the empty set is in tau this is vacuously true since there is no point x in the empty set and. So, therefore, there is nothing to check the full set x is in tau. So, why is this as this union w belongs to b w is equal to x and. So, given any x in x there is it is in one of these w's such that x is in w and obviously, w is a subset of x right. So, thus we have proved that both the empty set and x are in tau. So, this proves the first condition let us look at the second condition here we want to say that finite intersections of elements of tau are in tau. So, suppose u 1, u 2 up to u n are in tau then we need to show that the intersection u i is in tau. So, for this we choose. So, choose x in this intersection u i and. So, then for each i as u i is in tau and x is in u i there is w i in b such that x is in w i and w i is contained in u i. So, we claim that property 2. So, this property 2 implies the following which we called 2 prime if x is in w 1 is in the intersection of all the w i's for finitely many w 1 up to w n where each w i is in b then there is w in b such that x belongs to w and w is contained in the intersection of i equal to 1 to n w i. So, note that this property 2 property 2 is same as 2 prime when n is equal to 2. However, we are saying that using an easy induction argument which is left as an exercise. So, the proof that 2 implies 2 prime is left as an exercise the hint is use induction ok. So, then using 2 prime we get that there is w which contains x and w is contained in intersection i equal to 1 to n w i which in turn is contained in the intersection i equal to 1 to n u i. So, thus this intersection u i is in tau as it satisfies the property defining tau. And finally, we have to check the third condition. So, given a set i and subsets u i of x such that u i is in tau we need to show that the union is in tau. So, once again we do the same. So, let x be an element in the union then x is in u j for some j in i and since u j is in tau this implies there exist some w such that x belongs to w and w is contained in u j which in turn is going to be contained in this union. So, thus so, this shows that the union satisfies the defining property for tau is contained and so, is contained. So, this proves that this completes the proof that tau is a topology. So, next we have to show that we have to show that we also need to show b is a basis for tau which means that we need to show that that is for every u in tau and x in u we need to show there exist w in b such that x is in w and w is in u. But this follows from the definition of u being in tau. So, this completes the proof with the proposition. So, later on we shall in the next lecture we shall use this proposition to generate topologies on various sets. So, this is a very useful proposition. Let me give an exercise over here. Notice that in this proposition we started with a set and a collection b which satisfy two properties and then we defined a topology tau on x using this collection b and moreover b turned out to be a basis for that topology tau. So, but we can ask ourselves what happens if we started with a topology and a basis for that topology. So, let x be a set and let before this. So, let us introduce some notation let x be a set b contained in power set b such that it satisfies hypothesis in the proposition. Then b defines a topology then we defined topology then we shall denote the topology we defined in the previous proposition tau sub b and note that b is a basis for tau sub b. So, with this notation here is an exercise let x be a set and let tau b a topology let b be a basis for tau. So, then by the lemma. So, by this lemma that we proved using this lemma using this lemma and the definition of a basis the definition of basis it follows that b satisfies the two hypothesis in the proposition. Thus b defines a topology on x which we denoted tau sub b. So, a priori this tau sub b may be different from tau and the exercises show that tau is equal to tau sub b. So, with this ok. So, now we will put this aside for a few minutes and let us define topologies on subsets of x given a topological space x. So, definition of subspace topology. So, from now on before we define subspace topology when we say let x be a topological space we shall mean that x is a set and it has a topology. So, with this let us define subspace topology. So, let x be a topological space and let us say with topology tau let y be a subset of x. Then using tau we can define a topology on y in a natural way. So, define subset tau sub y contained in the power set of y as follows is the collection of u intersection y where u belongs to tau. So, we simply take all open subsets in tau. So, recall that if I said u belongs to tau then we often then we had decided agreed to use a notation u is open in tau. So, we just take all open subsets in tau and intersect that with y and that gives a topology on. So, claim is that this gives a topology on y. So, this is being left as an exercise. Check that tau sub y satisfies the three conditions for being a topology and so defines a topology which we call the subspace topology. So, let us see some simple examples of subspace topology. So, first so, let x have the trivial topology. So, then recall that tau is just consists of the empty set and x and in this case clearly the subspace topology tau y also just consists of the empty set and y and so the subspace topology on y is the trivial topology. So, similarly so, let x have the discrete topology. So, recall that the discrete topology meant that we are going to take tau equal to power set of x. So, then clearly tau sub y is equal to the power set of y because given any v in y, v is a subset of y and so it is also a subset of x and so v belongs to tau this implies that v intersection y which is v belongs to tau sub y. Thus we have proved that tau sub y it contains all subsets y which implies that tau sub y is the discrete topology. Let us take a third example. Let us take r with the standard topology and let this is x and let y be the subspace z ok. So, the question is what is the subspace topology on y? So, we claim that the subspace topology on y is the discrete topology that it is equal to the discrete topology. So, to show this it is enough to show that if we take a point in y a singleton then this subset is open in the subspace topology because if we show this then this would imply that given any subset w in y since we can write w as the union over all n in w these subsets singleton n and since arbitrary unions of open sets are open this imply that. So, we are assuming that suppose we can show that each singleton n is open. So, each n is open and we take arbitrary unions. So, that is also open. So, this would imply that w is open. So, every open subset will be I am sorry every subset of y will be open and y. So, therefore it is enough to show that these singleton ends are open, but to see this. So, each n we can write as the open interval n minus half comma n plus half intersected with y right because. So, here we have the real line and here is point n. So, let us say this is n minus 1 this n plus 1. So, this interval this is open in the standard topology in R right. So, is open in the standard topology in R and u intersection y is exactly n right. So, therefore this singleton n is in tau y or equivalently it is open in the standard in the subspace topology and therefore and from what we have seen above all subsets of y open. So, we will end this discussion here and in the next lecture we will see a slightly more complicated example.