 Welcome to module 11 of point set topology course. We have seen, we have introduced the notion of equivalences among us metric spaces and topological spaces, isometry, similarity and homeomorphism. And you have also seen that isometry implies similarity, similarity implies homeomorphism. So today we shall discuss the topic of reversing implications here. Does homeomorphism imply similarity? Does similarity imply isometry? The guess is that answer is in the negative and we will see. So for producing negative answers we have to produce examples. So this way the concept of all these isometrics, similarity and homeomorphism may be a little more clearer to us. So that is the idea. How to see that two metric spaces are non-isometric, you know, on one side producing an isometry may be very difficult. But how to say it is non-isometric, the only general method is to find the so-called isometric invariants such that one of them possesses state, the other one does not possesses state. But what we want is similar ones which are not isometrics. Two space, two metric should be similar. So if your invariant is preserved under similarity also then it will not be no use. So the invariant should be fragile enough so that when you look at similarity it is not preserved, it is violated. So such things we have to hunt around. So having said that there are several ad hoc methods to implement the above idea depending upon what two spaces you have in front of you. This method may not work, that method may not work. Something which appeals to you may just try it out, it may work. So that is the kind of approach we have because you are not, to begin with you are not given the spaces. So here are two similar spaces, are they isometric? This is the question, you do not know. So that for that you have to understand both the spaces properly. This is one thing which happens here what happens does not happen. This is the way you have to go around. So that is why it is an ad hoc method that we have to quite often do that. In any case, obviously we are more concentrating on topology so we will not go deeper into this aspect. That will be taken up by differential geometries or more surely geometries, combinatories, people and so on. So we will have only few easy examples. So let us take for example, being boundedness which you have seen that is an isometric invariant. Suppose one metric space with the metric it is bounded. On the same set you may have another metric which is not bounded. Then immediately you can say that these two are not isometric but can they be similar also if they can be similar also that is the whole idea. So being bounded is an isometric invariant but it is also a similarity invariant. So this will not work. Here is an example, easy way to use quite usual concept also which is an isometric invariant but not a similarity invariant. So I am trying to cook up something like that. So this is the new concept that we have here which is very canonical and very geometrical in nature. Take a metric space take any subset can be empty also do not worry about that. The diameter of that subset A is defined as so I have two notations here delta A varying I have ignored the metric but if the metric has to mentioned there are two different ones then that has to be mentioned then I use this notation delta A all and D. This is nothing but look at all the numbers distance between x and y where x and y are arbitrary points of A then take the supremum. This supremum may be infinite I do not care. If A is empty this will be minus infinity also makes sense supremum of any set of real numbers if we include minus infinity it makes sense. If delta A or delta x is bounded namely finite okay. So its number is bounded so the real number is finite then we say x is bounded okay including delta x equal to minus infinity that will not happen because our metric spaces are usually non-empty. If you take A it will be empty delta A may be non-empty infinity empty set is bounded that is okay no problem. So delta x finite we say x is bounded in that case we also say that the metric itself is bounded okay because I have taken the entire x here okay. The metric D may not be bounded but a subset may be bounded like any finite set you know is bounded okay. So all these concepts are there in Rn also in R also so is it nothing new as such so you are just testing the concept by just looking at the definition supremum of dx y where x and y range over all points of A okay. So I am making this remark here delta empty is minus infinity and delta A is 0 if filled only if A is a single term. As soon as there are two points the distance between them we know is positive so supremum will be positive okay only if A is a single term then delta A is 0 okay. So all these things may not be very useful but it will make the definition clear all right. Now look at a isometry from x1 D1 to x2 D2 okay. Then look at all the this all the real number D1 xy where x and y range over x1 and on the other side D2 of AB where A and B range over x2. So these two are two sets of real numbers I want to say there is a bijection between them induced by this f comma f from x1 cross x1 to x2 cross x2 okay namely put A equal to fx and B equal to fy. That will give you for each point D1 xy f D2 of fx fy which is equal to even so the same number will come here. So if you can use the reverse f inverse and you go here. So what happens is this set as totally a totality on this set is equal to the totality of this set therefore supremum over this one is same thing as supremum over this one because the sets are the same but supremum on this one is delta x1 D1 and here it is delta x2 D2. So what we have proved is that under isometry the diameter is preserved okay diameter is an isometric variant. Now we can make this one we can use this one to show that similarity does not imply isometric not just boundedness but more fragile thing namely diameter that xD be a bounded metric now start with a bounded metric so that diameter is finite that is the definition right. So put delta of xD equal to m okay I am putting this positive because I do not want to take x as a non-empty an empty set that is all. Now define D prime of xy equal to twice D of xy so it is a definition of D prime again on x cross x okay for every xy belong to x. You can easily check that this is also a metric so I am producing another metric by just multiplying by 2 you could have taken any real number here non-zero that is all instead of 2. So this D prime is also a metric no rover look at the identity map that defines similarity now okay all that D prime of x twice D of xy is less than equal to D prime of xy is equal to xD xy so that is the similarity relation right and there is no further function it is identity function itself. So identity is a similarity between xD and xD prime but the diameter with respect to D prime will be exactly twice m okay for each number here in the definition of this one there will be a twice that number in the other one here in the other one right. So there are two sets like this so the supremum of this one will be twice the supremum of twice supremum of this one will be twice supremum of this one so that is what is D of delta of xD prime here is twice m okay I could have put any number r not equal to 1 not equal to 0 of course then also it would have worked. So this was an easy method you see easy method to see that only thing is I assumed bounded metric without which the this finite number doesn't make sense then even multiplying by infinity goes to infinity there is no contradiction. So the above cheap method to get non-isometric metric spaces which are similar to each other does not work in general namely when the metric is unbounded for instance you get a vector space v okay and take any norm on it for any positive real number r just like we did it for 2 here you could you can take r times the norm you can take 2 times the norm all right by and call it as norm prime. So you have two different metrics here d and d prime corresponding metrics okay then look at the vector space v to v okay here I put d here here I put d prime here the norm here is the original norm here you see is the norm prime which is r times that you are not now you take the function mu of x equal to x by r okay all that you need is r must be again as usual not 0 not equal to 1 then it won't be much of his choice here okay for here I should have taken r not equal to 0 that's all r equal to 1 it is the same they are the same so not equal to 0 not equal to 1 it's obvious okay you can define like this there is no problem but if you want something more like an isometry when you come here you have to put that one okay so this is an isometry okay you have to check that its norm of x is suppose it's 1 here in the second in the in the second norm the norm prime the same vector would be r times that but I am taking x by r so r and r cancels out so it will be normal so I am just giving an example but that is not needed just norm of x here will become x by norm x is could be equal to that one because because it is an r here in the denominator r here in the numerator that's okay so multiplication by some number may not produce the necessity thing in the in the case of infinite and vector spaces okay the same thing you just take the subset V any finite subset may do but take something nice namely all points such that norm of x equal to 1 that's a unit sphere okay and look at the two matrix on this set I am taking the first norm here okay so this set now you take the two different matrix and restrict it to this a okay when you restrict a metric to a subset that's another metric space that's what we have seen so then that metric space is ad ad prime will not be isometric isometric why because the diameter of of the first one is one but the same thing the diameter of the other one will be r times that okay because you are multiplying by r okay clearly if you take dx y equal to in fact dx y equal to 1 divided by r times t prime of x y for every x y inside a therefore the two matrix are similar by the very definition you can write r times dx y equal to d prime of x y or the other way around because they are given by the norm okay only when you go to the whole V it doesn't work but this works for all nice things namely bounded things now let us come to again our favorite examples namely r n and c n so k power n okay start with this elementary inequality which says that if you take mod x 1 mod x 2 mod x n okay take the maximum take the power p it is less than be equal to x 1 p x n p all added together okay why because there are n of them if maximum is one of them say x i one of them that will be already there here because all the n of them are there and the rest of them are all non negative so if you add something non negative it will be still bigger than so this is less than good this one okay but again if you take maximum of this okay repeat it n times n times that that will be definitely bigger than this one first you can put this p inside and then go outside also because the maximum first you take the maximum and then take the power p or first you take the power and then take the maximum they are the same so this is an elementary inequality take the pth root what do you get here you get the l infinity norm what do you get here you get any times the l infinity norm sorry n pth root of n times l infinity norm in between what do you get you get the l p norm okay therefore you get l infinity is less than or equal to l p less than or equal to pth root of n times l infinity okay so this shows that what does it show this shows that all l p's are similar to l infinity this is similarity relation the two norms are trapped this norm is trapped between two multiples of this norm so they are similar to each other okay so all of them are what are these l p's one less than one less than one infinity of course infinity is already there take anything other than that they are all similar to l infinity the similarity is a equivalence relation so all the l p norms on a finite dimensional vector space k n they are all equivalent to each other okay so very simple idea here namely this inequality of materials however we claim that l 1 and l 2 are non isometric this will be our second example of similarity not implying isometry so this is also important to know that l 1 and l 2 are not you know isometric to each other but they are similar okay so for this because they are environmental dimension so just we try to use boundedness it does not seem to work so we need some other ad hoc method here okay here is one such take the set of four points the coordinate points plus minus plus minus 1 0 and 0 plus minus 1 okay e 1 e 2 minus e 1 minus e 2 all the four are taken okay on the x y x a see these are all elements of r 2 now so I am working inside r 2 and just show that the l 1 norm is not isometric to l 2 norm okay what is the l 1 distance between any of any two of them what is the l 1 distance for l 1 distance you have to take the difference of the two vectors and take the modulus right the modulus will be always you know some of them have to take modulus this minus this modulus this minus and so on so it will be always equal to between any two points here okay so the sum of the distance whatever mod of that is equal to two if you draw a picture it will be like a diamond shape thing right here these two points here points here like that the distance will be two the l 1 distance okay suppose there is an isometry of these four points you look at the image inside again r 2 inside r 2 but with altometric inside altometric you must get four points which are at distance two from each other right that is what you want so such a thing is not possible okay so what I say is you have four points here which are distance equal to two from each other suppose here an isometry here then it follows that f a has four points and the l 2 distance between any two points of f a should be equal to two because it is an isometry the distance delivery so this is just false that is very easy from your school geometry if you have three distinct points which are equidistance that is already a what it is a equilateral triangle right now the fourth point which is equidistant from them is only one such namely the ortho center or in center all of them are same centers because it is an equilateral triangle and the center is at a distance much less than you can compute the distance I don't care it is much less than two the sides so you cannot have a fourth point which is at any distance equal to two from all the three points all right all right so you see this is just another method so there are ways of combining various things and so or distance you can extend this one one real aspect here is the same argument you can use to show that l 2 is non-isometric to l infinity also this time if you take the four points all four points you must take tell me one one one minus one minus one minus one and what is that minus one plus one take for those four points l infinity distance is always two from a from even the diagonally also it will be two okay if there is an isometric to l 2 again you get a contradiction all right so so there are such examples and so on maybe we will stop here when we get more examples we will we will have that one so here is a comment in functional analysis there is a not so difficult and popular theorem it says that on a finite dimensional vector space any two norms are similar in particular if you follow that all these LP norms on KN are similar okay but what we have done we have already proved it by by taking a little bit trouble but not too much of trouble by starting with this inequality this inequality and yeah these are all similar okay but this theorem says it will not produce all of them any metric any any norm any norm in a finite dimensional vector space okay it will be equivalent to the Euclidean norm for example so they are all okay so this is an important result so we shall prove it at an appropriate time in the course okay as an easy consequence of some typical result that we are going to develop we will not spend time to just to prove that one okay on the way we will prove that all right so let us stop today for this one thank you