 So, last time we studied the implication science the other way around namely homomorphism does not imply similarity, similarity does not imply this was our topic, but we covered only one of them namely similarity does not imply isometry. So, today let us cover the other one namely topological equivalence does not imply similarity homomorphism does not imply similarity. Before that I want to give you another example here of isometry this time. Remember last time we proved that L1 and L2 are not isometrics and similarly L2 and L infinity are not isometrics, but this may come as a very surprise when I noticed it I was really very happy to have noticed such a thing namely L1 and L infinity are isometric all that you have to do is inside R2 all that you have to do is rotate by 45 degree if you do not like that you can say okay that is more complicated okay you write xy going to x plus y comma x minus y which is a linear isomorphism okay this is a linear isomorphism all right what I want to say that it is an isometric namely you take here L1 norm and on this side take L2 norm if you take the same norm this is not an isometric okay take L1 norm here and L2 norm here what is the L1 norm mod x plus mod y right if both x and y are positive what do you get maximum of that will be modulus of x plus y okay so there is a L2 norm if one of them is positive one of them is negative then this will be maximum if both of them are positive negative then again this will be maximum and so on so maximum is equal to the sum mod x plus mod y always right so this is an isometric all right the smallest norm L1 the largest norm whatever L infinity they are isometric on R2 but can you do it over complex number the definition is the same okay think about it can you do it with R3 R4 R5 and so on think about it I don't want to make any more comments I want to go to this topic now namely topological equivalence does not imply similarity so come here start with a metric on a on a some non-empty set x okay now define another metric which I am denoting by capital D DXY is equal to D little DXY divided by 1 plus little DXY DXY is non-negative so 1 plus DXY is never 0 therefore I can divide out by this they mark this number so the definition makes sense okay the point is you have to verify why this is a metric once again condition D1 is obvious because this is always non-negative and if it is 0 means the numerator must be 0 that means x equal to y so that is fine and D2 is symmetry if you change x and y here this formula remains the same because D is symmetric it will D is symmetric so 1 and 2 D1 and D2 are easy D3 may be a bit difficult okay so for this look at this one some it is this function is something like R divided by 1 plus R so look at that function this is a it is a familiar function to you okay this function is a homeomorphism we have studied earlier from 0 infinity to 0 1 in fact I could have taken minus 1 to minus infinity plus infinity and then put a mod here to get 0 1 if I don't put a mod minus 1 to plus 1 it would have come I am interested in the positive part therefore I am taking 0 infinity to 0 1 okay so this is not only arbitrary homeomorphism it is monotonically increasing function so that you have to check okay that will be used here now the only thing this this map property of this map what is important is monotonically increasing function okay use this to prove that this capital D is a what is a metric namely satisfies a triangle inequality okay the same thing I am doing another one here D prime of x y equal to minimum of D x y and 1 I should not take maximum then it will not be metric okay if D x y is 0 then minimum will be also 0 D prime x y will be also 0 and converse if D prime x y is 0 this must the minimum which what is to be 0 then x equal to y so D 1 is obvious symmetry is obvious here again so once again verifying the triangle inequality you have to work a little bit okay for this one you will have to do some adopt method namely look at all the points for which x y is less than equal to 1 that is one case then this one will not be there at all because I am all the time taking minimum of x y minimum of D x y and 1 so it will always D x y D x y D y z D x z suppose all of them are less than equal to 1 then the formula for D prime is the same thing as formula for D x y so triangle inequality satisfied so first case is obvious if one of them is bigger namely D x y is positive bigger than 1 then you see what happens okay irrespective of the third point D x z then you what happens look at what happens to this one then you so argument you have to break it okay as soon as D x y is bigger than 1 the definition D prime will be equal to 1 because this is bigger than 1 okay so what you will get is something like 1 is less than about 1 plus something 1 is less than equal to 1 plus 1 something so you verify that the importance of this example sees that whatever metric D is both D and D prime are bounded by 1 you see this never goes beyond 1 okay it won't even reach it 1 okay it is bounded by 1 see so this also by definition it is minimum of 0 1 so it is bounded by 1 so this does not depend upon D little D at all capital D and D prime are always bounded by 1 okay yeah so to each metric you can associate another metric in this method there are many many methods actually but this has some charm so let us see what is that so it is bounded by 1 now if you start with D as an unbounded unbounded metric then clearly D is neither similar to D capital D nor to D prime because if something is unbounded then similarity will preserve that so that will be also unbounded right therefore these two are not similar never D is not similar to D little D D prime is also I can use one of them to produce perhaps a counter example now these are not similar okay my aim was to get a metric which will give the same topology right so the claim is the topology underlying D and tau D is equals to tau D capital equal to tau D prime capital so they are all same topology so they are homeomorphism underlying topology is the same okay remember here the underlying set hasn't changed at all so perhaps we don't even need a arbitrary function the identity map itself will give you the two topologies are the same okay let us see so how to do how to see that consider the first case namely when I take this function okay r divided by 1 plus r dx y divided by 1 plus df the capital D let us first look at that okay now use the fact that again this phi r r divided by 1 plus r is a monotonically increasing function okay so it is a bijection is all that you have to observe phi of dx y by definition is capital DX y this is the what I have done if you replace r by dx y what you get is capital D okay therefore it follows that therefore dx y is less than r okay then phi of this will be less than phi of r because phi is monotonically increasing okay so phi of dx y is less than phi of r so phi of dx y is capital DX y right so capital DX y is less than equal to r and why is our size goes back okay okay this implies that every open ball in x d little d here is also an open ball in x capital D only thing is the diameter has changed to the radius has changed to from r to phi r okay and vice versa take anything here the diameter of the original thing will be phi inverse of that the radius of that one will be phi inverse of that okay so every ball here is a ball exactly is ball there the open balls are themselves the same same means what of course their diameter their radius has changed so the set of balls there that is the same therefore when you take the union whatever thing things you are taking corresponding things you can take on this side also so for each open set here there will be no open set there okay the set as tau d and tau d capital D are the same okay so the topology is the same all right now look at d prime for d prime you need a different argument altogether which one is easier that is left to you that is not one is easier or both of them are easier but before considering that let me make one of the points very clear to you about the topology tau d itself okay how is topology tau d defined given any metric d on x an element of tau d is described as a union of open balls in the metric d right now let r not get bigger than equal to 0 be any fixed number okay since every open ball of radius bigger than r not is also a union of open balls all of them of radius less than r not if you take a ball of radius 1 for example take any point inside that I can find a ball of radius something less than 1 and by all these all of them if I take I can write the the entire union as equal to the original ball similarly for any r bigger than r not okay I can take balls of radius smaller than r not even smaller smaller very smaller I have to take all very small also but they are all smaller than r not I do not want to take anything bigger than r not it is possible so that will be the it will give you all these balls will give you the original ball of radius r okay therefore it follows that every member of tau d is also a union of open balls all of them radius less than r not you don't have to take any bigger balls at all smaller things you can't control okay is that clear sir can you please repeat once let us say for example r not is 1 okay or let us say it is half maybe 1 is to be too good let us say it is half can you write the entire real line as a union of okay entire the whole of it as a union of intervals of length less than half yes can you take any open interval now not the whole of it okay can you write it also intervals of length less than half yes yeah so now question is can you take any open set not not intervals can you do that the same thing I don't want you to take any intervals of length than half so it is like a measurement you know your your scale is only six inch scale you are not taking one yard foot but you have to measure all the way from here to Delhi is it possible or not I am asking yes no that's precisely what it is here okay suppose your measurement is total measurement is something five but what you have is a ten foot roller can you measure that smaller one it may five it may four you don't know right but the other is always possible that is the only thing that is ignored here okay so take any r not okay any fixed number less than that I can always take and all the things which are less than that can be measured for all the boss now see your your r not maybe too large say it is 50 but your metric space itself is bounded by one then is there a contradiction I can take anything less than 50 means I can take less than one also right I can take half I can take one third I can so all those smaller numbers are always there there is no restriction there right it should be less than r all of them balls which are of radius less than r okay each ball which you take must be of radius less than r not never take any ball bigger than r not or equal to r not even can you write any ball as first of all union of such things is a question okay once you find once you find one ball contained inside another ball center is the same center whatever you wanted to choose you have chosen okay original ball center is different you take a point and that is the center now inside that ball you find a we have found a open ball inside that one okay after that you can take smaller smaller smaller so I want it to be smaller than r not so take the minimum of the two over okay yes yeah so so now you use that here all that you have to do so I have put d prime as minimum of dx y and one right suppose I take only balls of radius less than one then whether I take d prime or d it is the same thing look at a ball of radius less than one in the metric d when you measure the distance all of them will be less than one right right therefore d prime will be also equal to d the same thing because d prime is a minimum of dx y and one okay so you put this r not here equal to one here then what you get is any ball here okay whatever your small ball you have taken they are sufficient here also and vice versa so tau d will be equal to tau d prime in this part no ball will be of radius bigger than one here there are so that is why you start with bigger balls here it does not matter okay but then first thing is you write them as union of balls of radius less than one after that they are the same thing as this one all right so the two topologies are the same okay now we shall now introduce an important metric concept related to convergence of sequences once again I say I am introducing but these are all in elementary analysis okay copied down definition is copied down except the modulus is replaced by d that is all go back and check your definitions you are same thing okay at x t be a metric space a sequence is said to be a Cauchy sequence if for every epsilon positive there exists a natural number such that if I take n bigger than that k m could be any arbitrary I have written like this x n and x n plus m so both of them are bigger than equal to k that is all okay you can write it as say x i x j i n j both bigger than equal to k distance between such points inside a sequence must be less than epsilon so that is the Cauchy sequence okay a metric space x d is said to be complete if every Cauchy sequence in it is convergent and the elementary analysis says that every convergent sequence is always Cauchy but a Cauchy sequence may not be convergent very easy examples take a Cauchy sequence take a sequence which is convergent to a point and remove that point in that space the Cauchy sequence will remain but the convergence point you have removed therefore it is no more of convergent okay that is the way you can produce Cauchy sequences in r minus any single point inside r you cannot because every Cauchy sequence in r is convergent so this you must have studied namely completeness of r so I am going to use that one here I am not going to prove that r is complete here right now okay so x t is a complete metric space every Cauchy sequence is convergent okay so I am assuming that you know that k which is either r or c in with with the usual metric namely our modulus metric there right this also is always convergent if x n is a sequence in so you can take k power k k copies of k k cross k cross k when is a sequence convergent or respectively Cauchy you find only if each coordinate x n i okay I am getting x n i coordinate functions so x n 1 x n 2 x n k there are n coordinate sequences here each of them must be Cauchy then x n will be Cauchy if x n is convergent each of them will be convergent and conversely alright so this much we have seen already actually so therefore each k power k is also complete metric space alright now comes the point namely see our our title is study of similarities and isometrics and so on right suppose you have a similarity x d 1 x 2 d 2 then a sequence x n in a x 1 is convergent or Cauchy if and only f of x n is convergent in x 2 or Cauchy so Cauchy implies f of x in the Cauchy convergent in f of x n is convergent in fact if x n converges to x then f x n will converge to f x okay so these things are elementary okay so you should remember it like this similarity preserves Cauchy's sequence and completeness you don't need the strong hypothesis namely isometry isometry will also preserve because isometry is a similarity okay so similarity preserves these things alright so next question is whether homium or visor will preserve this one right we will see that that is not the case so these things are metric dependent the notions are metric dependent okay so how to verify such a thing if and only one is one is convergence another is Cauchy there is if and only if there are four statements here right I mean four sub statements of one single statement you can break it into four different thing right visor no see there is if and only if part so one's one convergence implies affection conversion then affection converges into an x convergent I have to prove similarly Cauchy is Cauchy implies is Cauchy is Cauchy right but look at similarity if f is similarity f inverse is also similarity therefore if I prove one way here then the other way is also coming right therefore instead of four statements you have to prove only two statements but even those two statements are very much similar here all that you have to do is for every epsilon there exists some k blah blah exactly similar things you have to do therefore I will prove one of them say I will prove that x n is Cauchy implies f x n is Cauchy okay just to remind you what are these Cauchy sense on okay so instead of four statements I will prove one of them x n Cauchy implies f x n Cauchy so recall that that f is a symmetry f is a similarity implies the one half I am taking namely d 2 of f x f y is less than or equal to some positive constant c 2 times d 1 of x y this was the definition actually there is another inequality here d 1 of x y c 1 times d 1 of x y is less than or equal to this one is another part which I do not need now that can be used for f inverse okay you use this one okay now I am assuming that this x n converges right so given epsilon choose a k such that d 1 of x n x n plus 1 is less than instead of epsilon put epsilon by c 2 that the c 2 factor is there to c 2 so I am using a Cauchy sequence not converging suppose x n is a Cauchy sequence right Cauchy implies this one option then that is this is the condition you get what are n n is greater than or equal to k that is such a k and m is any number this will be true now put this x equal to x n and y equal to x n plus 1 here that c 2 c 2 cancels out what do you get d 2 of f x n f of y n plus m that is what I am going to put here f f of x n x n plus 1 instead of y y equal to x n plus 1 right is less than equal to the c 2 c 2 cancels is less than epsilon okay this whole thing is less than epsilon so other three things also you can write down write it down for your practice if I keep doing it you won't get a practice you have to do that now I will take this opportunity to complete one of the little more knowledge about one of our you know favorite example namely take x to be any set and take all functions from x to k so that was a vector space that was a ring and so on inside that take the set of all bounded functions that was denoted by b on b we had a metric namely d of f g is equal to supremum of modulus of f x minus g x since each f and g are assumed to be bounded their difference will be also bounded so this supremum makes sense and we have verified that this is a metric okay so look at this matrix this what I want to prove that is that this metric is a complete metric space every Cauchy sequence here converges so this you might not have seen so pay attention and see that it is not at all difficult it uses the completeness of k and the metric the supremum metric okay convergence with the supremum metric we can we can talk about this as if it is a uniform convergence okay so that is the that is the key here okay so start with a Cauchy sequence f n inside b these are functions which are bounded that is the meaning and a sequence is Cauchy it follows that fix each x belonging to x just one one x then look at the sequence f n of x this will be Cauchy why because when f n is Cauchy inside b it is distance between this distance between f and g the supremum is less than something right f n of x f n of g for all x supremum is less than so each of them will be also less than that epsilon that is all and the supremum is bounded the each point is also bounded that is what it is here so each of them is a Cauchy sequence okay so where are the Cauchy sequences they are inside k because this f n are k valued functions right therefore they are convergent for each x you get a convergent sequence let us denote the limit by f n so this way I have already cooked up a function okay remember a Cauchy sequence if it is convergent any convergent sequence there is a only unique limit point okay so I can call that as f n that is happening inside k okay so first we claim that this f itself is bounded see we do not know whether this function is inside b for that I have to show that it is bounded therefore it is inside b okay how do you show that it is bounded look at any epsilon less than 1 0 less than epsilon less than 1 according to Cauchy because of Cauchy there will be some k here okay or n here actually a k equal to n I have put no problem because I have to put bigger than that and that is all there is an n outside d of f n comma f n plus 1 is less than epsilon by 2 for every m bigger than 1 for every integer bigger than equal to 1 so instead of epsilon I made epsilon by 2 now and I have cleverly chosen epsilon less than 1 any epsilon may do perhaps okay does not matter okay you have got this thing right now let m be such that the f n is fixed now select one n this f n is bounded by m supremum of f n is m so f n of x for all x modulus is less than equal to m okay now what happens if m is bigger than equal to 1 I can use this one distance between f n plus 1 of x and f n of x is less than epsilon by 2 right so you have bounded set and then these points are away from the bounded set by another epsilon by 2 right by triangle inequality f n plus 1 of x okay is less than equal to f n of x plus epsilon by 2 the distance between distance is less than epsilon by 2 is what see here it is modulus the usual metric on mod k okay so this is less than m plus 1 see would have been any epsilon fixed m plus 50,000 that does not matter all that I want it is bounded okay so this is true for all m bigger than n so this is a sequence which is bounded by this and it converges what happens when you take the limit of this as m tends to infinity the same thing as limit of f and we converge this so f x is less than equal to m plus 1 so I have shown that whatever x is okay see this this m was chosen independent of x right for all x this is true therefore this is true for all x so f is bounded all right now we want to show that this f is the limit of the sequence f n okay so look at this look at this equation which we have already got we can keep using this one again and again you do not have to write one more okay phi f of n plus m 1 f minus f of n plus m 2 of x just an epsilon by 2 actually so go by triangle inequality this will be less than epsilon now for every m 1 and m 2 bigger than n because they are at a distance from f n of x at epsilon by 2 okay for that reason I have chosen this epsilon by 2 there so this difference by triangle inequality less than epsilon now take the limit as m 2 tends to 0 m 2 tends to infinity keep m 1 as it is okay m 1 is some number here m 2 is some other number they are all some positive integers that is all take the limit as m 2 tends to infinity this becomes f x so f of n plus m 1 x is less than minus f x less than epsilon here is strictly less than epsilon I have to put epsilon less than recruit epsilon okay so this happens for every m 1 okay for every m 1 inside x so this is the same thing as saying that the supremum of this is less than recruit epsilon what is a supremum to put a distance between f f f n plus m and f okay so this is precisely the meaning of that this sequence converges to f okay so you do not need the first n turns here you see when a sequence converges if the sequence converges after n terms okay you truncate that one that is enough this is same thing as the whole sequence converges to f so we have proved that the this space b is a Banach space a complete normal space is a Banach space okay that is the name for that celebrating name for that one so I want to do something more about this one I will do later on but right now we will stop here and just look at a sequence of of exercises here for you okay so one of the exercise here is take a polynomial in r x okay of degree 3 real coefficients and one variable evaluate at each x x going to p x that gives you a map from r to r you have to show that it is a homeomorphism from r to r what is the homeomorphism we have a criterion that it must be continuous it must be on tow and it is monotonically increasing strictly monotonically increasing that is enough okay that is the thing okay so nothing is here is difficult if you think properly a little bit and use whatever little knowledge you have on journey okay thank you