 Welcome to module 15 of point sector biology part 1. Last time we did something about the closure of a set and close sets. So similar thing we will try to do for interior of a set and boundary of a set and such things. So most of them are similar in nature but you have to be little more careful that solve. So similar to what we have done for closures but you have to be little more careful that solve. So start with topological space again a and b are any two subsets of x. Then I have these seven conditions, seven properties here listed. The first one is interior of empty is empty it is similar to closure of empty set is empty. And interior of x is the wall of x. Again closure of x was x a similar product. So there is some similarity but do not get carried away. You have to be very careful. Interior of a is contained in a. In the case of closure it was a was contained in the closure of a. So this is also easy because by the very definition what is the definition of interior? It is the union of all open sets contained in a. Each open set is contained in a that union will be also contained in a. So that is a interior. A contained in b implies a interior is contained in b interior. This is similar to the closure property by the way. So how do we prove that? A interior is all those open sets union all those open sets contained in a. But then they will be contained in b also. So they will be contained in b interior. Because b interior is all those open sets contained in b you have to take and then take the union. So this is also obvious. Finally a is open if and only if a is a closure. So this is similar to a is closed if and only if a is a closure. Here a is open if and only if a is interior. a is equal to a interior. So why if a is equal to a interior? a interior is always open set because it is union of open sets. So a will be open. If a is open this is the largest open set already. All other open sets are already there. So I have to take this also. That will be interior. So a will be equal to a interior. b interior is the largest open set contained in b. So this is somewhat you know complementary to similar but complementary to closure. Closure was what? The closure of a set is the smallest closure set containing that set containing that b. So it is exactly you know like a Jim Morgan law exactly the open steps. b interior is the largest open set contained in b. I could have put a here but deliberately I have put it here you will see. So b interior is open set we know. Suppose we have another open set contained in b but then b interior is the union of all such things. So it is contained in b. So b interior is the largest open set. So these are all just restatements of whatever we have done they are not very difficult. Slowly you have to be careful. Interior of a intersection b is interior of a intersection interior of b. Here again interior of anything is an open set. And it is contained in the a intersection b here. Therefore it is contained in a. Therefore it is contained in the interior of a. But the same argument will show our interior b also. So it is contained in the intersection. So this is contained in the this side LHS contained in RHS is obvious. Now how to show this is contained here. Look at this one. Interior of a is an open set. Interior of b is an open set. The intersection is an open set. Interior of a is contained in a. So this intersection is also contained inside a. Interior of b is contained inside b. So it is intersection is contained inside b. But it is an open set. So it is contained in interior of a intersection b. So this last one here. Interior of a union b contains the interior of a union b. So there is no equality assertion here. It is only one way and that is obvious because interior is an interior of a is a subset of a. So it is subset of a union b. Similarly this subset of a union b. So this whole thing is a subset of a union b. But this is an open set being the union of two open sets. Therefore it is contained in a. So one way is clear. If a and b are disjoint open sets or disjoint closed sets. Disjointness is common. Both of them are open or both of them are closed. That is what we have to assume. Then equality holds. So let us see this one. So all these things I have written down here. Starting with the definition, what is the definition of this one? That a interior of a is the union of all open sets contained in a. This is your definition. So you can take different definition. Then your proofs may be somewhat slightly different. That is all. I have taken this definition. So I have gone through 1, 2, 3, 4, 5, 6 and so on up to here. Interior of a intersection, interior of b. Seventh one, one part we saw. First part directly from 3. Now I want to show the second part here. Namely suppose a intersection b is empty. If both a and b are open, then the union is also open. Therefore interior of a union b, a union b is already open. So interior of you know a and b. But a is interior of a, b is interior of b because a and b are open. So equality is obvious. So perhaps you have not used the intersection as empty here at all. Now suppose both a and b are closed. This time you may have to use that. Suppose both a and b are closed subsets and I assume the intersection b is empty. Let u be an open set contained in the union. If I have to show that interior of this one is equal to interior of a and interior of b, I must show that this open set first of all is contained in the interior of a union interior of b, separately. So first of all you can write this u as union of two subsets here. One is u intersection complement of b and u intersection complement of a. This is possible only because a intersection b is empty. a intersection b is empty would mean that the complements of a and complement of b will be the whole space. Therefore every set is contained in the whole space. So it is u intersection complement of b, u intersection complement of a take the union. This is true for all subsets now because b complement and a complement is the whole space, whole set. It is just a set addicting right now. So for that I have to use a intersection b is empty. Now how do I use this one? u is now written as union of two two sets here. As union of two open sets u is open, b is closed, p complement is open, intersection is open. a complement is open, intersection is open. So I have actually this one gives you that u is the union of two open sets respectively. This one is inside a now because b complement is there. A intersection b is empty. So one of them must be inside a, the other one must be inside b. So that means that u intersection b complement is contained in interior of a and u intersection a complement is contained in interior of b. Therefore, u is in the interior of a, union interior of b. This proves that one way namely interior of a union b is contained in interior of a, union interior of b. The other way we have already seen. So similar thing we will do for now the boundary operator. Take any subset of a topological space, boundary of the empty set is empty. Boundary of the whole space is also empty, not the whole space x. So this is the difference now. So recall what is the boundary? Boundary of a set is the closure of a minus interior of a. Throw away all the interior points from the closure of a. Keep that, that is boundary of a. So closure is the whole space. If a is x, then x bar is x, interior is also x. So x minus x is empty set. So that is what the first thing says here. Second one says a point x is a boundary point of a if and only if every neighborhood u of x intersects both a and a complement. Take a neighborhood of a point. If that neighborhood, I have to show it intersects both of them. If it does not intersect the complement means what? That neighborhood is completely contained inside a, which means that that point is an interior point. But I have thrown other interior points from boundary. A point is a boundary only if it is the complement of all of your interior points. No interior point is a boundary point. Similarly argument is similar here. If it does not, suppose it does not intersect a, then it will be contained in the complement of a. If it does not intersect a, then it is contained in complement of a. If it is contained in the complement of a, then it is in the interior of the complement. Therefore it is not a boundary point at all because every point should intersect the set a. So this argument is similar here. Similarly, now interchanging a and a bar it seems a boundary of a equal boundary of a complement also. So this statement is symmetric. So this is also boundary of a is always equal to boundary of the complement. The fourth statement is boundary of a is a closer subset. That is also equally easy because what is boundary of set? It is a bar minus a interior which is the same thing as writing a bar intersection with a interior complement. A interior is open. So its complement is closed. Then you are taking intersection with a bar. That is also close. So intersection of two close sets is closed. That is what we want to show. So each boundary a is a closer set. Finally, boundary of a union b is contained in boundary of a, union boundary of b. This thing I will leave to you as exercise because I have done so many things. You have to do something on your own to get the feeling what is going on. So you have to start with I will just tell you how to do it. Take a point here you have to show it is either here or here. What you have to do? Take a point here which is not here then it must be here. This is what you have to do. This kind of argument also we have used already. So write the full detail as an exercise. The second part I will do it here. If a intersection b is empty and both a and b are open or both are closed it is similar to our interior part then the equality holds. I can leave this also an exercise but I will do this one so that you are not feeling that I am cheating. So we shall leave the first part as an exercise. To do the second part I have to assume that a and b are disjoint. Once they are disjoint the first thing you have to notice is this one. A intersection b is empty implies a intersection interior b is also empty because interior b is smaller subset. Similarly interior b intersecting b is also empty. So this is also because this is smaller subset than a. So this is just set theory because interiors are contained in the original set. But now this implies we have seen in the closure part that the closure of a intersection interior b is empty because interior b is an open set. Whenever an open set does not intersect a set it does not intersect a closure. That is what we have seen. So from here to here you get this also empty. From here to here you get intersection interior of a intersection closure of b is empty because interior of a is open. So this is all that I can say right from the assumption that a intersection b is empty. Now let us start. The first thing is interior of the union contains interior of a and interior of b. This we have seen. So boundary of a union b which is by definition the closure of the union b minus the interior right. Now closure of the union finite union only two sets is closure of a union closure of b. And this one is this is smaller interior this one is larger than this one. So I have thrown away larger set here I am throwing a smaller set. So I have a containment relation here not equality. Interior of throwing this one to throw interior of a union interior of b which is smaller set. So this is contained in here though this is contained here because we are taking the complementation here this is the minus. Once you have here look at interior of a this is a subset of closure of a. And I have just shown that interior of a does not intersect closure of b at all. So when I am subtracting this one from the union I am actually subtracting it only from closure of a. So it is closure of a minus interior of a. Similarly when you are throwing interior of b it is only thrown away from here no point of this one comes there. So this union is closure union minus the union of these two is equal to this minus this this minus this follows because of this relation all right. But this is boundary of a and that is boundary of a and b. So what you have what is boundary of a and b is contained in the boundary of a union boundary okay. Now now I put one more hypothesis what is the hypothesis is that a in a and b are both open or both closed that is the extra hypothesis here and we are both open or both closed. In that case we have proved in the previous theorem that what we have proved in the last part of previous theorem that interior of a and b is actually equal to interior of a union of b right. Remember that here is a statement here okay then equality holds right. So once equality holds go back to these steps there are four steps here the second step was only containment why because here there was no equality this was contained. If there is an equality here I can put equality then everything else. What is so one equality comes extra comes instead of this one provided both a and b are closed or both are open. So this containment becomes equality here and they are all other things are equal to all right. So now I will sum up something namely open sets where the foundations for our definition of topology right. The topology was axiomatized on depending upon open sets. There is a parallel we have already observed because De Morgan law is there and then you can you can just convert all these three statements t au and what fi au is arbitrary union fi is finite intersection. When you take the complements of this by De Morgan law the first statement remains as it is empty set and the whole space become whole space and empty set that is all. The second one arbitrary union for open sets becomes what arbitrary intersections for open sets for closed sets and then finite union for open sets for closed sets sorry finite intersection for open sets becomes finite union for closed sets. So you could have defined the same way with three axioms for closed sets right but that would be cheating what Krootowski did was even more entertaining. At the time when Burbaki adopted a modern a modified version of the axioms for topology okay that is the definition we have been taking okay this result was only the correct definition available what I am talking about the Krootowski's result. Yet for some reason which I do not know what it is they rejected Krootowski's approach and adopted a modification of Hofstorff's definition. If you directly take Hofstorff's definition you do not get all topological spaces as we consider because Hofstorff's are put extra conditions which will come to what are called Hofstorff's spaces. So they dropped out that one but instead of dropping out the whole Hofstorff just dropped out that condition retended one okay. So let us take a look at this Krootowski's approach here. Maybe it is of some use okay gives you some more insight that's all but we will not change our definition our definition of topology it depends upon open set zone all right let us do that let X be any set and see from PX to PX be a operator operator is just a self map okay don't worry about usually this word operator is used by function analysis okay. So PX to PX be an operator on the power set of X PX is power set it's a function from power set to power set suppose this operator satisfies the following properties these properties are called Krootowski's closure axioms okay the first thing says that the let us let me read that C as closure itself now okay just for just for a sake that is how Krootowski read it but we have different meaning for closure right now but let me just for this C of V instead of C of V I say closure of V is V remember this was a property we have proved okay so here it becomes an axiom in terms of for Krootowski there is no topology here now there is the operator a set is there and look at the all its subsets on that there is an operator so the first thing is closure of V is V whatever you assigned to empty set it's empty set okay whatever you assign to A in general contains A is contained inside C for every A C of A union B is C A union C B if I read the closure what is it closure of A union B is closure of A closure of B we have proved it this is a theorem there here it is an axiom it is true for every A B C of C A C A this also we have proved right closure of a closure is the closure itself so this is an axiom again here and that's it the four axioms he has selected he can create the entire topology that is the claim of Krootowski okay so how do you do that put tau C instead of tau take tau C C corresponds to this closure axiom all those U inside P X okay that means what all subsets of X such that the complement of this one C of the complement is equal to see itself to put all of them the complements of that one they form a topology on X and this topology has something to do with C what is that the the usual definition of closure A bar is equal to C of A for all A so this C becomes the closure that is why it is a closure object okay so when I read it first time I was really thrilled by this one so let us verify this one it is very straightforward not at all difficult so let us write F C instead of tau C F C as all those A such that A equal to C A so what is tau C remember it is a complement of these things that is tau C tau C is a topology you have to verify just now I have told you by demarcant law instead of tau instead of T I have to verify T prime but that is T T prime instead of AU AU I have to I AU prime is nothing but this is now AI that means arbitrary intersection and instead of finite intersection I have to do finite union because under demarcant law intersection becomes union and so on so I have to verify these three axioms for F C okay then tau C will verify T AU and F U and that is why it is topology so first part will be over so let us verify these two the first T is nothing but T prime nothing but closure of empty set is empty and that is what is already there therefore empty set is there M C but X is already contained in C X so C X cannot be bigger than X because they are all subsets of X so equality holds here therefore X is in C X therefore X is in F C so tau prime is verified okay this T prime is verified sorry if A and B are in F C I have to C of A union B C A union C B therefore C A is A C B is B because they are in F C so C of A union B is A union so A union B is in F C so this verifies finite union I have to show arbitrary intersection take any family of A alpha inside F C inside F C means what C of A alpha equal to A alpha for every alpha then I have to show the same property for the equal section it is what I have to show okay no problem first we use another property we derive this is not a part of the axiom this we are proving as a consequence what is that if A is contained inside B then C A is contained inside C B why write B B is a larger set right as A union B minus A if you write this way then apply the operator C operator C B C of the union is C A union C of B minus A but C A is there so C A is contained inside C B that is all whatever happens to C B minus A don't care okay so A contained inside B implies C A is contained C B but the intersection is contained in every alpha A alpha therefore its closure here is contained in closure of each of them C of A alpha but C of A alpha is A alpha right so this closure is contained here but this equal to A alpha so if it is contained inside A alpha for every alpha it is contained in the intersection so closure of the intersection is contained in the intersection but intersection is contained in always closure because for intersection for any set it is contained in its closure so these two are there so there must be equality these two are same right so there is equality so this proves this proves the first part I will leave the second part namely A bar equal to C A just very nice you know just think about it it is not difficult at all in this topology you have to prove that A bar equal to C A for all A so that I will leave it as an exercise to you okay here are some exercises about pseudo metric just to tell you what is definition okay D1 is not there that is all a metric satisfies D1 D2 D3 remember D1 was the positive definiteness that is missing D2 is symmetry D3 is triangle inequality only these two are there all other things are working exactly same no problem okay you can put a tau d namely the topology associative to these things what are they they are unions of balls open balls open balls that is all these things are same same way you have to carry on okay so that becomes a topology you have to do that you show that D is a metric if and only if in this topology tau d okay every singleton is closed so that is the first exercise this part is easy you can put the topology and so on so that I am not bothered about first you have to show this observe this one this is also not difficult observe this one the second part is put an equivalence relation d of x y equal to 0 then x is related to y see if it is a metric d x y equal to 0 would have implied that x is equal to y since it is only a pseudo metric you do not know whether x is equal to y that is why you want to identify them so you put a equivalence relation you put a relation now you check that is an equivalence relation for that you know all that you have to do is symmetry d2 you have to use d3 you have to use it is very easy look at the equivalence classes that is my notation x hat here denote the equivalence classes on each equivalence class I am denoting by this bracket x bracket y and so on so I am defining a map from x hat cross x hat into 0 infinity by this formula d of x bracket y bracket equal d of x y I am just putting this formula but y this is well defined you have to check next you have to show that now this d hat here there is a hat here okay d hat becomes a metric on x hat okay so the associated to a pseudo metric space and a pseudo metric there is a metric space okay which is the quotient of this x because equivalence relations are there that is the that is the gist of this one verify this one there is all these things we have to verify so there is a similar exercise here about when you have union of increasing sequence of union of subsets x is countable increasing sequence of subsets xn minus 1 contain the xn and so on on each xn you have a pseudo metric dn okay actually sorry dns are metric pseudo metrics on x on the defined on the whole of it when you restrict it to xn it is a metric only when x y are inside xn dn of xy equal to 0 plus x equal to y if you go out of that into x that relation is not there that is all so on x they are all pseudo metrics restricted to xn they are all metrics of course that is a case then you do this funny thing this is typical way of summing up infinite things okay inside a when you have infinitely many real number only thing I have to assume that all the dns are bounded by one single number I have assumed bounded by one does not matter bounded by one single number it is enough defined delta xy equal to this summation of dn of xy by 2 power n it is bounded by m m divided by 2 power n summation that is convergent so this is also convergent so this will be finite number this has wonderful properties now this delta will become a metric on x okay and it has some wonderful properties all these things you have to verify namely a sequence in x is cosy with respect to this delta if and only if it is cosy with respect to each dn okay so the topology is very closely related that is what it means okay so let us meet next time thank you