 Uniform continuity, equivalent ways in terms of sequences, every continuous function on a compact set is uniformly continuous. Here is one more example of continuous functions which you may find is useful. A function, you may come across these things in some other courses. So, let me introduce that. A function f from a domain D in Rn to R is said to be Lipschitz. Mathematician who first gave this definition, so goes by his name, is said to be Lipschitz if for every x, y belonging to the domain D, if I look at the distance between the images, that is less than or equal to some alpha times x minus y. So, we have got the distance between images of any two points is bounded by the distance between x and y for some constant alpha. So, sometimes one says it is Lipschitz of order alpha or does not matter actually, some power sometimes, but this is a simplest kind of Lipschitz function you can think of. So, claim f Lipschitz f uniformly continuous, is that ok? Because it directly gives a measurement between distance between f x, f y and x and y. So, if x and y are closed, obviously f x and f y are closed obviously or if you like, if a sequence x and y n goes to 0, then f of x n minus f of y n will go to 0. Either way you can look at it and say this is a class of functions which is uniformly continuous. There are many other useful results because the domain is compact. So, let us say f is from D contained in R n to R n. So, let f is 1 1, D is compact. Domain is compact and the function is given 1 1 function. If the function is 1 1, then on the range there is a inverse function available. Every 1 1 onto function has a inverse from the range to the domain. So, that function is normally called f inverse from the range f of D to D. So, what is the inverse function? x point x goes to y. So, inverse function is you bring y back to x from the range back to the domain. So, you can think of as composition of these to be identity map. f composite f inverse is identity map on the set D. The claim is continuous if f is continuous. So, what we are saying is if the domain of a 1 1 function is continuous, then the inverse function automatically is a continuous function. So, let us prove that. So, what is to be shown? So, let y n converge to y in D in f of D. I want to show inverse function is continuous. So, what is the domain of the inverse function? That is f of D, the range. So, take a sequence there y n converging to y, so that the image converges to the image to show f inverse of y n converges to f inverse of y. So, that is to be shown. So, what is given to us? The domain is compact and the function is continuous. So, from the range, somehow we have to come to the domain because something about the domain only is given and how can I come back? Because y n and y are in the domain, they should be the images of something in the, they are in the range. So, there should be image of something in the domain. So, let us write that. So, let x n, x belong to D such that f of x n is equal to y n and f of x is equal to y. So, x is the pre-image of y and x n is the pre-image of y n. So, now x n and y n are x n is in the domain, x is in the domain. So, now, since x n belongs to D and now where comes our compactness? D compact implies there is a subsequence, a subsequence of x n. See, how compactness always gives you nice properties. x n such that x n k converges. x n k will convert somewhere. x n is only a subsequence. Converges somewhere, let us call it as say x 0. In f continuous, we will imply what? f of x n k converges to f of x 0. But what is f of x n k? That is y n k, because x n is the pre-image of y n. Where does y n converge? We are given y n is convergent to y. So, that implies what? y n k, so that converges to y, because y n converges to y. So, what we should have? On the other end, it should converge to f of x 0. And this y is equal to f of x. Let us also write, actually this is y. So, what we want to show? We want to show f of x is equal to y. So, what we have gotten? So, implies f of x 0 is equal to y. This converges this, this also converges to y. So, we have gotten this. Can I say x is equal to x 0? Can we say x is equal to x 0? I know f of x is equal to f of x 0. f of x is equal to x 0, but function is 1 1. We are not used that fact anyway. Note f of x 0 is equal to f of x, because x n k converges to x 0. Continuity implies f of x n k will converge to f of x 0. But this also converges to f of x, y is equal to f of x. What is y? y is equal to f of x. So, these two are equal. So, implies x is equal to x 0, because f is 1 1. f is a 1 1 function. So, x must be equal to x 0. So, what we are saying? That y is equal to f of x. That is what we want to show. So, hence f inverse is continuous. So, it is a continuous function. So, that is one consequence of saying it is… I think I should make a remark. You should try to remark that d compact cannot be removed. So, what does that mean? In this theorem, we had the condition that d is compact. That condition, if the domain is not compact, even if the function is 1 1, inverse need not be continuous. So, try to construct an example. Otherwise, in the tutorial class, we will discuss it. What more one should say? I think enough of uniform continuity. Let me also come back to the concept of connected subsets of R n that I said we will do it later. So, I think this is the right stage to look at connected subsets in. So, recall. Let us just, because we did it some time back. So, let us recall a subset x contained in R n is called connected. I am just recalling if x does not have a separation. So, what are the meaning of saying x does not have a separation? That means x cannot be written as A union B, A B non-empty, A and B separated. If a set can be written as a disjoint union, it can be partitioned into two parts. Not only partition, if that is a separation, then we say the set is disconnected. If no such thing is possible, we say it is connected. The typical example was that of an interval. Interval is a connected subset of the real line, because whenever you try to cut it into two parts, one part and the other part will be disjoint. You can cut it into two parts, but they will not be separated. Separated means not every point has a neighborhood which does not intersect the other set. Keep in mind neighborhood in R n. We said this is also equivalent to saying that there does not exist any set u, which is not empty, which is proper subset of x, u both open and close. So, this is equivalent to saying that there does not exist any subset of x, which is both open and closed. Because if it is both open and closed, this set and its complement in x will give us a separation, if not, certainly a separation. But one thing I should make it slightly more clear, which may not be that clear. What is meaning of saying u is open in x? u is a subset of x. So, u open in x is same as saying it is v intersection u v open in R n. It is part of the part of an open set in R n. For example, if you take a set which is open interval saying this is an open set, I can say it is an open set. If I look it as a subset of R real line whole thing, but can I say a part of it say this is 0 to 1 and let us like it 0 to 1 by 3. Can I say 0 to 1 by 3 is open in 0 to 1? Yes, but supposing I close it here, I take the close interval. Now, can I say that 0 to 1 by 3 is an open subset of the close interval is open in 0 to 1. It is open in 0 1 close, but not open in the whole thing, because I can write this as an intersection of bigger open set, intersection 0 to 1. So, that kind of thing you should keep in mind. Saying something is open in a subset means it is intersection of an open set in the whole space with that set. Let me say why? Let us just illustrate this a bit more. I am saying that 0 to 1 by 3 is open in 0 to 1. What does that statement mean? Let us elaborate it more. If I want to go back to this and look at in terms of interior points, so for every x belonging to 0 to 1 by 3, what should I have? I should have a ball around this. So, there exists. So, if x belongs, then let us look at say minus 1 to 1 by 3. Intersection with 0 to 1 is a subset of this point x belongs to this. Not this one. Let me just write slightly more better. Let x belong to this. Then x belongs to say minus 1 to 1 by 3, obviously. So, here is 0, here is 1 by 3, here is minus 1. But I can still say x belongs to minus 1 to 1 by 3. Intersection, if I look at this intersection with 0 to 1, which is a subset of 0 to 1. See, what we are saying is whatever be the set in R or in Rn, what is the meaning of saying that you are looking at open balls in that set? That set may not have the whole ball as of Rn inside it, but it is an intersection of an open ball in Rn with that set. That is what we mean. So, you can think it as a subset or subspace you can think of. So, that is the difference between saying some A is a subset and A is a subspace of Rn as far as the distance is concerned. So, what are open sets in a subset? They are intersection of open sets in the bigger intersected with that set. So, in that sense you should think of both open and close. So, for example, let me just keep that discussion a bit more. For example, if I look at the interval 0, 1, I am saying is open in 0, 1. It is open in 0, 1. Is it okay for everybody? It is like whole space is always open in itself. Whole space is also closed in itself. So, this because this way of looking at things helps to later on to define many things in higher courses. So, let me see what I wanted to do was the following. So, here is a theorem which is very useful. Let X be a subset of Rn. The following are equivalent. I am going to describe connectedness in terms of functions 1. X is connected and second, if f is a function from X to 2 points at 0, 1 is continuous, then f is constant. f is a constant function. So, let us just understand the statement. Saying a set is connected is equivalent to saying that if we have got a function on X which is continuous and takes possibly only two values 0 and 1, that is not possible if X is connected. It should take only one of the values either 0 or 1. It should be a constant function. So, continuous functions on connected sets cannot take two different values. Only one value is possible. It should be a constant function.