 So, we have discussed the contraction mapping principle in the last class. Let us recall what it says once again that is if x is a complete metric space and f is a contraction map on x, then f has a unique fixed point. Now, this principle as I said has applications in several areas in particular differential equations. One of the standard theorem about the existence of differential equations what is called Picard's theorem uses this contraction mapping principle. One can also ask how does one come across this contraction mappings and one of the very standard ways of constructing this contraction maps is by using what you have learned in the undergraduate calculus as mean value theorem that is if you take suppose f is a function which has derivatives and suppose you take say 2 points x and y and suppose f has derivatives in the interval containing x and y, then you know that f y minus f x this can be written as some f prime at z multiplied by let us say y minus x assuming that x is less than y where z lies between x and y that is some point which lies between x and y. Now, suppose f is such that suppose f is differentiable suppose f is differentiable let us say on some interval a b and suppose for all points in a b suppose we know that absolute value of the derivative is less than or equal to some number alpha which is less than one then we can show using this that that map f is a contraction map that is and suppose we know this that is f prime at x is less than or equal to alpha and this alpha is strictly less than 1 and suppose this happens for every x in a b then using that what we can say is that mod f y minus f x then we can say that mod f y minus f x less than or equal to alpha times mod y minus x this will happen for every for every x y in a b and that means which is same as saying that f is a contraction map on this close interval a b f is a contraction map on this close interval a b and we have seen that a close real line as a metric space is complete and a close subset of a metric space is again again complete. So, any closed interval is an example of a complete metric space. So, if you take a map which satisfies this property on any close interval then that should have a that should have a fixed point unique fixed point another thing that we discussed about this remember all this started with some certain special types of uniformly continuous functions another type that we discuss was what we called isometry. Let us again recall what was an isometry suppose we take say two metric spaces x. So, let us say x d and y rho let us recall that we had said that f from x to y is called isometry if it preserves the distance isometry as the word means the map preserves the distances that is what should happen is that if you take set any two points x and y here and look at the distance between f x and f y f x and f y then that should be same as distance between x and y. So, that is for every x y in x in other words isometry is a distance preserving map and what we have already observed is that isometry will always be one work if f is an isometry it will always be one work it may or may not be on to and if it is on to then we say that x and y are isometric to each other if there exist an on to isometry from x to y then we said the two metric spaces x and y are isometric to each other and we also seen that if the two metric spaces are isometric to each other then as metric spaces they are essentially the same we can disregard the difference between the two any two isometric space only thing that differs is that the names or labels of the elements change otherwise the distance between the two elements are exactly same as whatever happens in x the same thing should happen in y and whatever property x holds the same property will be there for y also. So, in view of this what is usually done is that see let us let us now take one case now suppose f is an isometry suppose f is an isometry suppose f from x to y is an isometry it may or may not be on to but it that is it may or may not be on to y but this it will always be on to f of x f of x will be a subset or subspace of y. So, f from x to f x will always be on to and so these two spaces x and f x this is from x to f x this is a on to isometry. So, we need so we can disregard the difference between x and f x and f x is contained in y f x is contained in y. So, this kind of thing we can say that we express this by saying that x is isometrically embedded in y. So, the thing is called isometric embedding this space f x as I said that we did not regard that x and f x are different we can regard x and f x are the same. So, in by means of this isometry we can regard x as a subspace of y we can regard x as a subspace of y. Now, let us say that y is complete suppose y is complete x may or may not be complete suppose x is any arbitrary space and y is complete then regarding this f x as a subset of y we can always say that we can look at f x closure f x closure f x closure will be complete if y is complete f x closure will be complete. So, even if x is not a complete metric space f x closure is a complete metric space and it contains x as a dense subspace it contains x as a dense subspace. So, such as metric space is called a completion of x it is called completion of x what is meant by completion again let me repeat it is a it is a complete metric space which contains the given metric space as a dense subspace that is called completion of a metric space. Now, obvious question is whether such a completion exists given any or can every metric space be completed and is such a completion unique then the answer is yes we can always find a complete metric space which contains the given metric space as a dense subspace and then the completion is unique in certain sense it is unique what is what we say is that is unique up to isometric that is if you take any two completions of the given metric space then those are isometric to each other and since in view of our idea of not regarding isometric spaces as different. So, we can say that up to isometry the completion is unique now how to go about constructing this completion that problem I shall not discuss right now that that will postpone for something. Now, let us go to next property of a metric space that is next property that a metric space may or may not have or subsets of a metric space may or may not have that is a property that is called connected. So, here let us say we are given a metric space x d and we are looking at a subspace subset a of x and we want to say what is meant by saying that a is connected or a is not connected a is connected or a is not connected. Now, roughly speaking a connected set should mean that it is a single piece it should it is a single piece for example, suppose you have suppose the set is like this say some points here and some points here then obviously we will regard this as not connected or disconnected only thing is that this what we feel intuitively about connectedness we have to make little more precise by giving the correct definition in the context of any metric spaces. So, the way in which it is defined is as follows now it so turns out it is simpler to define what is meant by not connected. So, we shall define that first and then we shall come to connectedness. So, we say we said a is said to be disconnected disconnected means not connected it is said to be disconnected you can keep this picture in mind for defining what is meant by disconnected what should happen is that we should be able to find two open sets two open sets such that a is contained in those two union of the two open sets and that those two open sets are disjoint should be disjoint and intersection of each of those sets with a should be non-empty. So, let me again repeat a is said to be disconnected if there exists disjoint open sets open sets g 1 g 2 in x such that what are the requirements first thing is that let me write spell out completely such that a is contained in g 1 union g 2 second requirement is that g 1 and g 2 are disjoint we did not write again, but anyway let us emphasize that g 1 intersection g 2 is empty and a intersection g 1 and a intersection g 2 both must be non-empty. So, a intersection g 1 is non-empty and a intersection g 2 is also non-empty. For example, suppose we take a set consisting of these two regions in either in a complex plane on in r 2 then you can obviously find some open set g 1 containing this some open set g 2 containing this the disjoint sets so such a set is disconnected. So, and if a is once having defined what is meant by a disconnected set now we can define a connected set is the one which is not disconnected. So, so we will say that a is connected if it is not disconnected connected if it is not disconnected of course you may not find it in this way in all books in a book you may find what is first it is defined what is meant by connected set, but then what will be the definition that such two open sets do not exist that is what is suppose I wanted to without going through this procedure suppose I want to define what is meant by a connected set then what I would have to say is that a is connected if they do not exist two disjoints two open sets satisfying all these properties or what is the meaning of that if you take any two open sets at least one of the requirement is violated that is the meaning of connected all right. Now, what are the obvious examples of connected sets or disconnected sets disconnected set we have already seen as far as connected set is concerned suppose you just take a single point suppose a set consists of only one point then obviously this last thing cannot happen that is g 1 and g 2 disjoint and a intersection g 1 and a intersection g 2 is non empty that sort of thing cannot happen right. So, a set consisting of one point is always connected whatever be the metric space a singleton set is always connected right. So, it is that is that is a standard example so singleton set is connected in any metric space of course a set can have more than one point and still be connected, but we shall see those examples little later on the other hand suppose you consider a discrete metric space suppose x is a discrete metric space and of course containing if it contains only one point obviously disconnected, but suppose x is a discrete metric space containing more than one point is it clear that x must be disconnected because you can just take let us say one point and then other as the complement because in the discrete metric space you know that any every subset is open right. So, you just write x as a union of two open sets that is for example you can take say g 1 as just one point and g 2 as it is complement then that will satisfy all this property right. And of course as a special case of this when a is the whole of x when we when we will metric space x itself is disconnected instead of this x is contained in g 1 union g 2 it will be x equal to g 1 union because g 1 union g 2 will always be a subset of x. So, and if x is also contained in g 1 union g 2 it simply means that x is equal to g 1 union g 2 it means that if you can write x as a disjoint union of two non-empty open sets then x is a disconnected metric space otherwise it is connected if you cannot write x as a disjoint union of two non-empty open sets then x is a connected metric space. And it is fairly straight forward to prove that this connectedness is a property which does not depend on whether you regard a as a subset of x or a as a subset of itself because we know that a itself is a metric space. So, we can say one might think that there may be some difference between regarding a as a connected set as a matrix as a subset of a and as a subset of x. We have seen that that kind of difference exists in open sets close sets etcetera whether you regard a as a subset of a itself that will always open, but you if you take a as a subset of x it may or may not be open, but in case of connectedness that is not nothing a is connected or not connected that does not depend on whether you regard a as a subset of x or any or a itself or any set lying between a and x that can be proved, but we shall not go into proof of that right now. Now, let us come back to the again very most familiar metric space namely the real line what are the connected sets in the in the real line. And there we have the most satisfactory theorem or most satisfactory answer intervals are the connected sets in the real line and those are the only connected sets. So, let us say let us write that as a theorem let us say that let a be a subset of r by the way what about empty set for example, can such a thing happen if a is empty it is not possible right. So, an empty set is connected set that is virtually or trivially it is a connected set. So, let a be a subset of r then what a theorem says that a is connected if and only if a is an interval. So, this gives a complete description of connected sets in the real line a is connected if and only a is an interval. Let me recall what is we have seen that how an interval is defined we have seen that interval means when we set a set is an interval that if you take two points x and y if x and y belong to a then all the points lying between x and y must also belong to a that is the definition of an interval. Let us first prove it this way suppose a is connected and we want to prove that a is an interval suppose a is connected now if a is not an interval suppose a is not an interval we should get a contradiction right suppose a is not an interval if a is not an interval what should happen yeah they should exist three points let us say x y and z let us say x less than z less than y such that x and y belong to a, but z does not belong to a. So, let us say then there exists let us say then there exists x y z in r there exists x y z in r such that x less than z less than y x y belong to a and z does not belong to a. Now, what we can do is that we can construct two open sets g 1 and g 2 satisfying all those properties listed there. So, let us say so for example consider g 1 is equal to consider g 1 is equal to let us say minus infinity to z open interval minus infinity to z and g 2 is equal to open interval z to infinity or g 1 and g 2 open sets there open interval so when open sets are they non-empty obviously I mean they are infinite sets is this true that a is contained in g 1 union g 2 in fact g 1 union g 2 contains the whole of r except this point z g 1 union g 2 is the whole of r except this point z and that point z is not in a. So, a is contained in g 1 union g 2 and what about a intersection g 1 a intersection g 1 contains x a intersection g 1 contains. So, x belongs to a intersection g 1 so that is non-empty and a intersection g 2 that contains y. So, y belongs to a intersection g 2 so that is non-empty. So, we have found sets g 1 and g 2 satisfying all these properties here. So, that means a is not connected a disconnected and that is a contradiction we had started with assumption that a is connected. So, every connected subset of real line is an interval let us now see how we can prove it the other way. So, we shall we shall discuss this proof again tomorrow. So, for the time being we shall assume that whenever a is an interval a we a is also connected. So, whatever is the missing parts we shall we shall discuss this again in tomorrow's lecture. Let us proceed further. Next what we want to say is that what happens to the images of the connected sets under the continuous functions. So, let us look at this. So, what are the theorems suppose x d and y rho are metric spaces f from x to y is connected. Then suppose you take a connected set in a then its image in y is also connected. So, that is what we want to prove. So, if a is contained in x is connected then f of a is also connected. Basically the whole proof depends of course is a important theorem about connectedness. The whole proof depends basically on the fact that inverse images of open sets are open under continuous functions which is what we have seen sometime earlier. So, suppose f of a is not connected then we shall we shall find they should exist open set g 1 g 2 satisfying all those properties which we have written and then consider their inverse images. They will do the same thing with respect to a and that will show that a is not connected and that will be a contradiction. So, let us go through that proof. So, suppose f of a is not connected f of a is not connected then there exists open sets g 1 g 2 such that whatever is required g 1 intersection g 2 is empty f of a is contained in g 1 union g 2 then f of a intersection g 1 is non-empty and f of a intersection g 2 is also non-empty. What we do after this is that of course these open set g 1 and g 2 those are in y open set g 1 and g 2 in y because f of a is subset of y. So, what we shall do now is that consider the inverse images of g 1 and g 2 under f and then just rewrite these equations in terms of those. So, then first look at f inverse g 1 is f inverse g 1 non-empty why f inverse g 1 is non-empty because this shows that f a intersection g 1 is non-empty. So, there should exist some x such that f x is in g 1 that because suppose that f x f a is a set of all f x of the form x is in a its intersection with g 1 is non-empty that means there exists some x in a such that f x intersection g 1 is non-empty. So, f inverse g 1 is non-empty. So, f inverse g 1 is first of all f inverse g 1 and f inverse g 2 are open sets in x why they are open because g 1 and g 2 are open in y and f is continuous and we have seen that inverse image of an open set under a continuous function is open. Now, look at this equation f of a is contained in g 1 union g 2 f of a is contained in g 1 union g 2. So, that means that a is contained in f inverse of g 1 union g 2. So, this means a is contained in f inverse of g 1 union g 2 and we can always say that f inverse of g 1 union g 2 that is same as in fact it is same as f inverse g 1 union f inverse g 2, but we can suddenly say it is contained in f inverse g 1 union f inverse g 2 actually this is equal. So, a is contained in f inverse g 1 union f inverse g 2 both these are open sets. Then what about a intersection f inverse g 1 that is where we come a look at this point again f a intersection g 1 is non empty it means there exist some x such that f x belongs to g 1 there is a some x such that f x belongs to g 1. So, that x should be in a as well as in f inverse g 1 see f a let me repeat f a intersection g 1 is non empty means there exist some x in a such that f x is in g 1. So, that x is in a and also in f inverse g 1. So, which is same as saying that a intersection f inverse g 1 is non empty a intersection f inverse g 1 is non empty and in a similar way we can say that a intersection f inverse g 2 is also non empty a intersection if inverse g 2 is also non empty, but that is a contradiction that shows that a is not connected. So, this means a is not connected now as a special case of this if this second space y is real line if the second space y is real line then it means that if you take a connected set in x its image must be a connected set in R and which we have just now proved that connected sets in R must be an intervals. So, that means image of a connected set under a continuous function will always be an interval image of a connected set under a continuous real valued function will always be an interval and this theorem of course which follows immediately from here it is special and the previous theorem is well known as an intermediate value theorem. So, let us say that let x t be a metric space f from x to R be continuous of course when I say nothing more about R it means it is with the usual metric if we do some other metric on R that will be specifically said this is something I mentioned earlier also. So, if a is a connected subset of x connected subset of x then f of a is an interval of course this follows immediately from this as such in a normal course we should have simply called this as a corollary of this because no new ideas are involved here if a is connected f of a is connected and since f of a is a subset of R and every connected subset of R is an interval. So, f of a is an interval why is it called intermediate value theorem because we can say that since f of a is an interval what we can say is that suppose there are any two points let us say x and y belong to f of a and x let us say and x less than y and suppose there exist some z x less than z less than y then this should imply that z also belongs to f of a z also belongs to f of a. So, in other words if f takes the value x and y x y belongs to f of a means what there exist some a such that f of a equal to x and y belongs to f of a means what there exist some b such that f of b is equal to y that is let us say that is f of a is equal to x that is there exist a b in a such that f of a is equal to x and f of b is equal to y that means the function f assumes the values x and y if it does that it should assume all the values in between those two values or all the values which are intermediate that is why it is called intermediate value theorem. So, if f assumes any two real numbers it assumes all the values which are in between those two values. So, for example in particular suppose at some point f of a is less than 0 and at some other point f of b is bigger than 0 then there should exist some point c such that f of c is equal to 0. So, this is something that is used fairly often again in numerical analysis in finding the first approximation of any roots of the equations. So, intermediate value theorem is again quite useful in basically deciding the starting points of many of the numerical algorithms and usually in applications this function f will go from some subset of r to r and that subset will usually be an interval that subset will be usual interval. So, image of an interval under a continuous function will always be an interval and this also answers for example certain kinds of functions cannot exist. For example, can there exist a continuous function from r to n let us say continuous and on to is that possible let us say is it possible f let us say f from r to n f from r to n continuous and on to is that possible because obviously the intermediate value property will be valid right n is not an interval n is not an interval. So, there cannot exist any continuous function going from r to r to n. So, such a thing is not possible all right now next we want to see how the operations on this connected sets take place. For example, we would like to know what about the intersections and unions of connected sets it is something similar to the intervals we have seen that intersection of any two intervals will again be an interval, but union need not be in case of union what we know that union will be an interval if the intersection is non empty. Something similar is true for the connected sets also that is if you take two connected sets and if their intersection is non empty then their union is again connected and this is true not only for two sets, but it is true for any family of sets. I will just state that theorem and then we shall see the proof tomorrow. So, suppose x d is a metric space and let us say that a i small i belonging to big i let us say i it is an indexing set is a family of connected subsets is a family of connected sets connected sets in x of f. Of course, such that the intersection is non empty such that intersection a i i belong to i is non empty then we want to say that then union a i i belonging to i is connected all we shall see the proof of this theorem in tomorrow's class.