 పారినిలుమ కింమరెని మాధ్ల్ల్లు. ప్లు చకినో ప్నోతి ఱుని చాంతలర్లాంధూ సంతి ఏరు మి కినియేత్ కేయరింద్ . తార్ను న్కిన్ల్ చాల్లానారందిక్లెదు పుక్నుందిందిరివుల్్టిన్లు నారికార్లినికి లికోపిమిignerally we cant compare house darkness and regularity. however if you mix up these two then something quite surprising comes out. there is a complete hierarchy, you can say, once there is such high hierarchy we will have to use numbers to indicate them. Zheng Philosophy crust ? A writes I want to warn I respect It is totally unthink it I do not know how they got to that mess like very good book like Simmons book has a different connotation altogether so my connotation is different from that so you have to be careful about that so I would prefer that the terminology I am for always more logical than the other part so I just want to warn it itself so we shall call a fresh space a T1 space and a Hofstor space a T2 space a space which is T1 and regular will be called T3 space and a space which is T1 and normal will be called T4 space so this is what I meant by mixing up T1 and regular will have a name named T3 and T1 and normal will have named T4 T1 and completely normal will be called as T5 now why these numbers these numbers have been chosen with some results already in mind namely a Tn where n is bigger than m will always implies Tm T5 implies T4 implies T3 implies T2 implies T1 so that is the hierarchy that looks like a beautiful way of putting it and easy to remember thing also okay so that is the theorem first theorem here okay so I will come back to these two things a little later so I will come back to that one so first let me go through this theorem I greater than or equal to j we have Ti implies Tj so Ty implies T4 is obvious T5 being completely normal plus T1 T4 is normal plus T1 okay so completely normal implies already normal so that is obvious though normality itself does not imply regularity if you put T1 on both sides there is an implication why because as soon as T1 is there okay singleton sets become closed therefore if you have a closed as a singleton point and a closed set disjoint from that as if you are having two disjoint closed subset therefore normality you can assume that there are open subsets around that which are dangerous so T1 assures that singleton points are closed that is why this works alright now T4 implies another one I will come to that one later on so now T3 similarly implies T2 why because T3 is two closed set and a single point but single point is closed and when you take two distinct points they are both closed subsets so you can apply regularity to get two open subsets around that so that implies closed offness so T1 assures helps to derive regularity from regularity closed offness only under T1 they will be equivalent that is what we have seen already otherwise it is not true and already we have seen T2 implies T1 okay now I go back to this other numbers here one number is so I will go back to that one so we are introducing little more few more numbers here the yeah let me come back here a T1 space which is completely regular remember there was a regularity and complete regularity also we have introduced okay so T1 space which is completely regular will be called as T3 and half space unfortunately there is no integer between 4 and 3 so you have to get 3 3 and half the whole idea here is that T4 implies T3 and half and T3 and half automatically implies T3 because complete regularity implies regularity add T1 on both sides you get T3 half implies T3 but what is it true is that T4 implies T3 and half because of your zones characterization remember that T3 under this complete regularity was actually an adapted version of reasons characterization right so that is the whole idea so this T3 and half has another name it is called tick off space okay so there was no integer to accommodate it so people cooked up this T3 and half name for it that is all but there is another thing one can do a weaker version of T1 okay so let us define that one we have not done that one yet so there is no regularity normality anything it is weaker than T1 space what is it at the political space is called T0 space so this time we are jumping or T half and so on T0 space if for every pair XY of points in X okay okay when I pair I mean X and Y distinct there exist an open set you containing X and not containing Y or an open set containing Y and not containing X I repeat given two distinct points you know you may have an open set around the first one not containing the second one or it may happen that there is an open set containing the second one but not the worst one just means that both of them can also occur I am just saying or I am not saying either or okay I am not saying only this or only that no the point is both of them can also occur so that do not have to I do not have to tell that but I want to make that one clear okay in the definition above we have used the word or not either or so it may happen that both are true as in the case of T1-ness in the case of T1-ness if you have two distinct points there is a neighborhood about one which does not contain the other now I do not say which one okay therefore it is applicable to both the points right so that is why a T1-ness automatically implies T0 but T0 may not imply T1 okay so this T0 space is just looks like a cooked up notion for I am high that is my opinion cooked up notion from freshness there is only one instance wherein some extra hypothesis there wherein T0-ness will imply T1-ness we will see that one okay there is only one instance of that one so anyway the numbering is completely justified because of our theorem now T5 implies T4 implies T3 and half implies T2 implies T1 implies T0 okay for this complete hierarchy alright whenever I is bigger than j T I will implies Tj okay so here is a another example now which which is a house door space but not a regular space okay the house door-ness does not imply regularity regularity does not imply house door-ness either but regularity plus T1 implies house door-ness okay house door-ness is same thing as T2 so here is an example which is house door but not regular again on the real line we take a collection tau of all subsets U which satisfies the following condition given X belonging to U there exists an open interval I such that X is inside I but instead of saying that I is contained inside U which will be the usual topology what we say I intersection Q is contained inside U much weaker condition okay if the whole of I is contained inside U well and good that will be usual topology but this is much weaker condition okay nevertheless this condition defines a topology on X namely R with this topology R will be called rationally extended topology okay that is the name obviously it is finer than the usual topology because usual topology satisfy this condition right the whole of I will be contained inside once it is finer than the usual topology it is house door anything finer than house door space is house door so half the part is over so what we want to prove is that it is not regular okay for seeing that it is not regular we take F equal to O Q complement set of all irrational number the set of rational number is an open subset here right because take a point in the rational number say any interval all the rational points in the interval are contained inside Q so Q itself is open therefore Q complement is a closed set okay now take X equal to 0 or any rational number for that matter just take X equal to 0 that is outside right so we must find what we must find U and V such that X belongs to U and F is contained inside V U intersection V is non-empty that is regularity but now we have to show that no matter what U and V are the moment they are open and contain X and F their intersection is non-empty is what I would show to conclude that the space is not house door we could have chosen any other point we could have chosen any other ring this is our choice so F is QC so we will try to do this one if it fails doesn't mean that it is regular because we have made a choice here okay so assume that U is an open subset containing a containing 0 and V is an open set containing all the irrational number that is what we have started so since it is an open subset U is open subset there will be an open interval such that this X is inside I remember X is just 0 you can or any rational number X belongs to I and I intersection Q is contained inside U okay so I is an open interval therefore you can take any S inside I intersection F what is F? F was set of all irrational numbers so it has lot of irrational numbers okay obviously this S will be different from X no problem then there must be another open interval J such that S is inside J J intersection Q inside V because V is an open subset containing F by our assumption okay look at these two intervals I and J okay they are common point S I and J are open intervals this is a common point S means they are intersecting so intersection of two open intervals if it is non-empty in general interval only right therefore since S in C I intersection J I intersection J is non-empty open interval but then this non-empty open interval intersection with Q is also non-empty now if you look at J intersection Q that is inside V but if you intersect I intersection Q that is inside U therefore this intersection is both inside U and V so U intersection V that is the thing that we wanted to prove that such open subset cannot be disjoint so that proves that the rational extended topology on R is half star but not regular okay okay can it be normal can it be normal no why because we have seen that T3 implies T4 the other way around T4 implies T3 normal plus T1 implies regular plus T1 so T4 implies T3 but we have seen that it is not regular but it is host also it is T1 okay so it follows that this cannot be normalized so let us say once you prove that it is not regular okay now we will take another example which will give you regularity does not imply normality this space is regular but not normal okay we have already seen such an example but we would like to do this one for reason that this again another modification of the real topology Euclidean topology so in exercise 4.48 we have indicated that the semi-interval topology product with itself is completely regular but not normal see what we have proved is regular and not normal but it is actually completely regular is what we have what we have indicated in the exercise there it was an exercise okay but now we will prove this one the other example this is not regular this is regular but not normal okay so what I do I take the upper half plane H all x y belong to R2 y positive okay the second coordinate is positive the upper half plane open upper half plane I am denoting L by L the real line R cross 0 y equal to 0 okay I am including that also in the along with upper half plane and that is my x so this is the closed upper half plane but I do not want to call it that way because this part I can call it as upper half plane but here I am going to change the topology so I am using a different notation X here so here is the topology coming now two sub basis are declared S1 is equal to set of all open balls okay around points inside capital H namely y coordinate is positive x coordinate anything B epsilon of x y it must be contained inside H therefore the radius must be less than y that is all 0 less than epsilon less than y so I am taking all the open balls completely contained inside the upper half plane okay these are standard open balls right the second one is slightly different that is where the Krusk of the matter lies they are open balls x comma y center the radius is equal to y not epsilon so y coordinate becomes the radius of that okay so it is touching the x axis right and one point what is that point x comma 0 so you include that also that point is not there here it is tangential so include that point also that is the that is the that is one of the elements in this set such that x y's are inside H okay start with a point in the upper half plane take the maximum open ball contained inside that that is the meaning of this B y x y okay you cannot take bigger than that then it will go below the x axis that is not allowed right so if you take maximum open ball this is what it is then put that point x comma 0 also in that so this is going to be one of the one of the sets inside this S2 take collection of all of them so that is your S2 now you put a union of these two call that as as a sub base for a topology on x any collection of subsets can be declared as sub base that we know okay so this is a base for topology on x whatever that topology is it has the property that by the way I have made a remark here namely this is actually a base so let us not bother about this this is sub base is enough for us note that this topology is finer than the Euclidean topology because you see on the on the upper half part this is actually Euclidean topology okay everything open in R2 is there and vice versa on the x axis you intersect this balls with the x axis what is it it is just a single point x comma 0 therefore each singleton point on the x axis becomes an open set therefore the induced topology on the x axis is discrete in any case it is finer than the usual topology alright therefore this entire topology is finer than the usual topology okay in particular it is so all right now we have to show that this is not what this is not normal so I want you to observe this namely if you take by x y union the singleton x 0 okay that is an open subset of of this topology it is in a second part of you know s2 part its closure is all those f a b belong to R such that x squared minus a squared actually yeah a square plus y squared minus b squared less than equal to y the full you know the full ball will come when a b raise over all R okay so the closure will be just the closure of all that is all that is what x y is already there but the closure will contain all the rest of the circle also regularity at points of edge follows easily by the euclideanness of of the upper half plane take a point in the upper half plane take an open subset you do not have to worry about weird open subset but you can take inside that you can take a regular open set to open ball right and verify the regularity it is already euclidean space so it is regular so there there is no problem the problem arises when you take x cross 0 inside L namely on the x axis okay for x comma 0 belong to L and u is an open set containing it okay there exists a y positive such that b y x y inside u so this is these are because these are basic open sets now we can take v equal to b y x y 1 union x 0 where the y coordinate the this y coordinate is less than 1 okay and then check that v bar less than y sorry and then check that v bar is contained inside u okay so even for points on that one you can verify that it is regular to see that x is not normal is our point here okay so for not normal we again take a equal to the entire q and b equal to r minus q okay similar to the earlier example okay then we show that there is no opens of sets containing a and b which are this year okay so that is what we want to show it is similar to r cross L but some somewhat easier maybe and say for each r in r choose y r positive such that b y r r comma y r y r positive r comma y r will be in the paraft plane right now I am taking a the full ball maximum ball of whatever possible radius that will be radius will be y r put the r also this is an open subset now so such an open subset will be inside g where g is either you or we accordingly where wherever you are r may be inside you or or inside that means rational or rational okay for both of them argument is same okay so you can change you can use such a thing that is always I think so that this this is contained either inside you or inside we according r inside a or b now for any inside n let us define f n to be all s inside b okay such that y s coordinate okay is bigger than 1 by n see for each each r I have a y r okay I have chosen so I look at all those s such that its y s is bigger than 1 by n so that is my definition of f n then for everything some y r is positive is what it has said so entire of b will be union of f n after once it is positive it will be bigger than some 1 by n so b is union of f n now by bears category theorem 1 r okay with usual topology if you remember what is b b set of all irrational numbers it follows that interior of f n bars cannot be empty for all of them right we have shown that the entire even we have actually shown that even irrational numbers cannot be even as countable union of norad and sets because then you can add another set of countable number containing r the whole of r will be written like that so that is why one of the f n must have property that interior of f n bar is non-empty for fix such an integer now okay so by just by looking at these definitions of this one and so far we have not used anything other than that that a and b view and we are open okay nothing else but now one of these has interior of f n bar is non-empty okay choose an open interval i contained inside f n bar interior in the usual topology these are all subsets of r now so there will be an open interval container and a rational number now are we down to i if epsilon is chosen appropriately we claim that for any irrational number inside see this s belong to r minus epsilon r plus epsilon r is contained inside this interval now for every epsilon is for every s inside this r minus epsilon r plus epsilon which is obviously I have chosen so that is contained inside i and which contains f n bar but I am not saying that for all epsilon is true appropriately chosen epsilon what we will have is b y r r y r intersection corresponding s is b y s s y s will be non-empty that is a contradiction because these things are supposed to be contained inside to disjoint open sets okay this once is r is rational number and s is irrational number okay so that will be that will imply that you and we are you intersection v is non-empty okay so this is our claim how to choose epsilon is the point so that this will happen so she has a picture what is happening here whatever r and s whatever they are there is some y r okay so this whole open bar along with this point is inside u or inside v that is how we have got it now I want that s and r are chosen such that they are intersecting here okay all the time chosen r has been already chosen s should be chosen close to r in such a way such that the corresponding ball will intersect this one this one because s y s are only chosen I have no control over that but I can choose s itself closer and closer to r so how close I should choose is indicated in this picture okay so now I just work out well standard mathematics here so 31 is true that means the intersection is non-empty r minus s square plus y r minus y s square this is the distance from distance from whatever you see the including distance is less than y r plus y s square so this is what I am saying r minus s square to go back to this picture r minus s square this distance okay minus y r or r y I have written r y y r it should be the y r minus y s y coordinate distance square so that is the distance between these two this distance total distance must be less than the length of this one plus the radius of this plus radius of that means y r plus y s so there is notation difference here y r this is a y r and y s that is all okay so so that is the first condition y r plus y s they are a radius sum total if I take the square root of this this is the square root of that so as I have taken the square of this one this distance will be less okay this is same thing as now simplify r minus s square is less than 4 times y r into y s you take this one on this side y r square y s square we cancel out okay twice y r s twice y r s we add up this will be 4 times y r y r the same thing as now taking a square root modulus of r minus s should be less than twice square root of y r y s okay therefore choose now epsilon to be less than twice the square root of y r by n okay suppose you choose this y r by n it follows that now what is this n remember this n was fixed with that the interval is contained inside the interior of f n closure so that n appears here it follows that r minus s is less than epsilon if you if you if you have this one is epsilon is less than twice y r by n that will be less than twice y r y s square root I want this one I want the last thing if I have satisfied this r minus s less than this one then the intersections will be non-empty the 31 will be true so now I choose this epsilon to be less than this one and r minus s less than epsilon satisfies okay so this is because y s is bigger than 1 by n so that y s part disappears here you see some condition should not be depending on s so that I am choosing s so this is purely in terms of r now okay so that will be automatically less than this because y s part is less than 1 by n so this completely proves that x tau is not normal so here is some exercise maybe you can have your own exercise also but I would not like you to I would not like to encourage you to go on studying just counter examples nevertheless if you are interested in there is a very good book written on this one long long back okay I have given the reference in right in the beginning so I will indicate it to you later so you can read that book okay so thank you we will meet next time okay