 So, last time we were trying to prove a theorem that is subset S of real line is connected if and only if it is an interval. So, we had already proved one way. So, namely if S is connected then S is an interval. So, let us prove the converse part of it. So, conversely, so let S be an interval and suppose S is not connected. So, that implies if S is not connected it must be having a separation. So, let us say S is equal to A union B, a separation of S. That means S is written as a union of two sets where A and B both are separated. So, let us choose an element A belonging to A and an element B belonging to B. So, here is converse separation. So, S is equal to A union B. When A and B are of course, non-empty sets and A is separated from B and B is separated from A. So, let us choose any element A in A and B in B and look at the element which is the midpoint of the two. So, then the midpoint should belong to, it is between A and B. So, let us write that it is a point. We know that this is in between A less than less than B implies we have got S is an interval. So, anything in between must be a part of an element in the interval. So, implies A plus B by 2 belong to S. Now, if it belongs to S, S is A union B. So, it will either belong to A or to B. So, two possibilities. So, either A plus B by 2 belongs to A or A plus B by 2 belongs to B. So, if this is the case, if this is the case, so then consider A1. Let us call this point as A1, A plus B by 2 and B1 equal to B. So, if it belongs to A, then I call it as A1 and the right point B, that point B, we call it as B1. So, what does it give me? Then A1, B1 is such that A1 belongs to A and B1 belongs to B. So, what I have done is taken two points A and B, look at the midpoint. Midpoint must be a part of the interval. So, it must belong to either A or B. If it belongs to A, call it A1 and call B as B1. So, what is the second possibility? If this is the case, then A1, so then this belongs to B. So, define A1 equal to A. So, let NB1 equal to A plus B by 2. Again A1, B1 is such that, so what is happening? It is such that again A1 belongs to A and B1 belongs to B. So, either case I have gotten is interval A1, B1 such that the left end point belongs to A and the right end point belongs to B, whichever the case may be. Another property is that A1, B1 is half of AB, the interval AB, because one of them is a midpoint. So, either it is A to A plus B by 2 or A plus B by 2 to B1. So, it is only half of it. So, length of this interval A1, B1 is half of the length of the original one. So, let us continue this process. So, what does it mean continue? I have got A1, B1 and now I will take the midpoint of A1, B1. Both A1 belongs to S, A2, A B1 also belongs to S. So, midpoint must belong to S again. So, whatever we were doing for AB, I am doing same thing for A1, B1. So, continue to get A2, B2 such that A2 belongs to A, B2 belongs to B and the length of A2, B2 is equal to half of length of A1, B1, which was equal to 1 by 4th length of AB. So, I have got a way of generating a smaller interval from AB. A was in A, B was in B. I am able to generate a interval of half the length with the same property. Left end point is in A, right end point is in B, whatever the case may be. So, continuing this process, what I will get? I will get a sequence of intervals A and Bn, which is nested, right. Each one is in the previous one. They are close bounded intervals and the length is going to, what is happening to the length of them? It is becoming smaller and smaller, it is going to 0. So, let us write. So, this will give us a nested sequence A and Bn of close bounded intervals such that An belongs to A, Bn belongs to B for every n and Bn minus An goes to 0 as n goes to infinity. So, I have manufactured a nested sequence of intervals. Now, we approach nested interval property of the real line that says there must be a point in the intersection of all of them. So, imply the nested interval property, intersection of A and Bn, n equal to 1 to infinity must be equal to some point, let us call it C, right. So, obviously, A less than or equal to C less than or equal to B, right. Now, A and B are both in S. So, C must belong to implying C belongs to S, right, because A and B both belong, S is the interval, so it must belong. But S is equal to union of, it is a separation A union B. So, either it will belong to A or it will belong to B, right. So, in case C belongs to A, then what happens? What is C? Basically, it is a limit of Bn, right. So, close to C, there will be points of B, right. So, I cannot find a neighborhood of C, which will be disjoint from B. And this is a point in A, but A and B are separated, so that is not possible, right. In case B, Bn converges to C, implies A and B are not separated. Is that clear? Because every neighborhood of C must have some Bn, after some stage onward, actually all, right, because Bn is going to converge, right, nested interval property. Bn's are right-hand points must be converging, okay. It is a nested sequence, it is a decreasing sequence, okay. That means there is no neighborhood of the point C. C is in A, which is disjoint from B, because one Bn at least will come in. So, A and B are not separated, not possible, not true. So, another possibility is C belongs to B, same thing. Now, you go to the left-hand points. An converges to C, implies A and B are not separated, so not true. So, in other case, we get a contradiction. So, our assumption, right, so what was our assumption? Our assumption was that S can be written as A, U and B. A separation of S is possible, right. We have shown it is not possible, that means S is a connected set, right. So, implies, so let us write, hence S is connected. So, we have shown that we have characterized all connected subsets of the real line. A subset of the real line is connected if and only if it is an interval, right. A similar question one can ask for Rn, okay. One can prove some theorems. We will do it a bit later. We need more techniques, right, more properties of some other concepts. So, we will later on show, for example, that in Rn, what could be a generalization of an interval? What would be a ball, right? If you take an interval, the generalization in Rn is a ball. So, you would expect that every ball is connected, at least that much should be true, right. So, that can be shown. Actually, there is a concept called path-wise connectedness. Some geometry comes in. If any two points can be joined by a curve in a subset of Rn, then that subset is called path-connected. And one shows every path-connected set is also connected. But every connected need not be path-connected. So, this kind of things happen. So, slightly more complex, slightly more involved to study connected subsets of Rn or even in the plane. So, we will do some examples and some theorems when we come to continuity and such properties of Rn, okay, functions. So, for the time being, we are doing it only on the real line. So, till now what we have done is, we have looked at subsets of the real line and special properties of those subsets, right. We started with real line as a complete ordered field. And then after having done that, we looked at what are called intervals. They are some basic open sets called intervals. And then we looked at sets. We looked at sequences, whether they converge or not, when do they converge. And then we looked at sets which have the property that whenever a sequence of elements converge to some point, whether that point is inside the set or not. If all the limits are inside, that was called a closed set, right. Then we studied some properties of closed sets. If a set is not closed, you can make, put it inside a set which is smallest closed subset of real line called the closure. And then we defined open sets as those sets, those complements are closed. The set is open if and only if it is, complement is closed. So, open sets, properties and so on. And then we looked at properties of sets which are called compact sets. So, a compact set is something more than a closed set. Namely, a sequence of elements of the set may not converge, but at least there is a subsequence which converges inside the set. So, these are called compact sets. And then we characterize compact sets. Namely, we said every compact set, the set is compact if and only if it is closed and bounded, right. And then we looked at a characterization of compact sets called Heineborel property that every open cover has got a finite sub cover. And that we proved it only for the real line. True for Rn and probably we will see if time permits, we will do it later on, okay. All these are concepts which are also valid in general spaces called metric spaces. So, we will probably introduce that also a bit later, okay. For the time being, let us concentrate on real line and properties of real line, right. So, we have looked at real line as such sets in real line and special subsets. Of course, we looked at lastly the connected subsets, okay. Now, here is something very obvious, probably I should state that which is good. So, true for Rn also, S contained in Rn is connected, implies, okay. Let us write not connected means what? S is equal to A union, where A and B are separated, right. Now, that separation means what? Every point of A has at least a neighborhood, right, which is completely inside A. It does not intersect B, because it is separated. So, every point of A must be open. Every point of A must be an open. It must be an interior point and hence A must be an open set. Is that okay? Every point of A, it is separated from B. So, there must be an open neighborhood which does not intersect with B, right. So, there is a neighborhood of which is completely inside A. It does not intersect B. Is that okay? Oh, it does not intersect B, that does not say it should be inside. It does not say it is inside. So, let me revise this statement. So, what I am saying is not correct. I have to go to subset actually. That is not a good idea. What I want to say is something that needs subspace topology. But let me say a milder version of it. Let us take a set U contained in, so, forget this one, okay. So, U is a subset which is non-empty and contained in Rn. Let us take. So, let U be a non-empty subset of Rn. Can U be an proper subset? It is a subset of Rn. It is not empty. So, it has at least one element. It is not whole of Rn. Then, can U be both open and closed? Suppose it is. Suppose U is both open and closed. Then, what is real line equal to? It is U union U compliment. If true, then Rn is equal to U union R minus U. My claim is this is a separation of R, Rn. Why it is a separation of Rn? Because U is open. Take any point in U, then there must be a ball inside U because U is open. Every point of U will be inside a ball which is completely inside U. So, U is separated from U compliment and same if it is U is both open and closed, then U compliment also is open. So, every point of U compliment also is separated from U. Is that clear to everybody? Definition of open as says every point is an interior point. So, if I give you a point of U, it must be inside a ball which is completely inside U. So, it will not intersect U compliment. That means U is separated from not only it is disjoint, it is actually separated from U compliment and now let us go to the other way around. Take a point in U compliment. U compliment is also open because U is both open and closed. So, U is closed. So, U compliment is open. So, every point of U compliment will be inside a ball which is completely inside U compliment. So, U compliment also is separated from U for the same reason. That means real Rn is not connected because of what is separation. So, conclusion is in Rn there is no set non-empty proper subset which is both open and closed as a consequence of connectedness. So, let us write not possible implies Rn is not connected, not true because it is interval. So, hence there is no proper non-empty subset which is both open and closed. So, Rn connected implies this kind of a property. You cannot have subsets which are both open and closed of course non-empty and proper. Right.