 For real line, a set is compact if and only if every sequence has got a convergent subsequence converging in the set or every open cover has got a finite sub cover or saying that the set is closed and bounded. All three are equivalent ways of saying it is a compact set and what is the big use of saying something is compact is basically in general settings is the last property saying every covering has got a finite sub cover that becomes very important. Bringing things from arbitrary, any arbitrary covering to a finite, so many of the proofs you will find which are true for arbitrary can be brought down to finite by using high-neighboral property and I think one of the immediate uses you may find, I do not know whether the proof is given in the probability courses or not. If you look at the length on the real line, if you look at the length of an interval for interval which is bounded b minus a is the length if n points are a comma b. If it is unbounded, you can put the length to be equal to plus infinity. So, you get a function from intervals to 0 to infinity and proving that this function is additive is not difficult that if you take an interval which is cut into two pieces in the length of the bigger piece is equal to length of the sum of the smaller pieces. It is something like keep in mind the probability of the set a is equal to probability of b union probability of plus probability of c additive property. But supposing you cut up an interval into countable infinite number of intervals, can you think of cutting up an interval into countable many sub intervals, disjoint and still the union being same, disjoint but union being same. So, let me just give you something where these things are going to be used, I do not know. Say for example, if I look at this interval a to b, what is interval a, b? I want to cut it into countable number of pieces. I can cut it into half, this close this open. So, what I get is smaller piece which is close. I can cut it into that into half. So, I will get this piece and I can go on doing it. So, I will be cutting up the close interval a, b into countable many disjoint pieces, disjoint pieces. Any two pieces are disjoint and the length of the whole thing is equal to summation of length as all the individual pieces. So, this is going to be called something this. If you think it is 0 to 1, length as the probability of something happening, probability of 0 to 1 is 1, length is 1 and you have got subsets which are events. The length of each is the probability of that event. So, it essentially says if an event a is a disjoint union of events a n, then the probability of a is equal to summation probability of a n. This is a property going to be used in probability theory called countable additivity. The length function is 1 such on 0 to 1 and proving it is countable additivity 1 use is compactness. So, when it comes there, you may get back to it. So, we looked at sets which are close, which are open and now everything is motivated by sequences. Now, here is a property we want to discuss which depends more on the geometric aspect of real line. That means the real line is a continuity of points. It is a continuity of points. There is no break. For example, if I take an interval, I take at some point then I keep on moving on the right side, then I keep on moving till I come to the end of the interval in some sense. There is no break. It breaks and then again something happens. This is a continuity of points. So, intervals are characterized by the continuity of points and we defined that way that if x and y are points in the interval and z is a point in between, then that must always belong to the set. That is how we have defined the interval. Such sets are important. They are called connected subsets. What we want to look at is what are called definition. We want to define. It does not make much difference in the real line, but later on it makes a difference. Let us define. Let us take A is a subset of R and x belongs to R. We say x is separated from A if there exists. You can define in Rn, but let us keep it to R if there exists an open interval i such that x belongs to y. That means there is a neighborhood of the interval, but this does not intersect A. There is a neighborhood which does not intersect A. Then you say x is separated from A. We want to say a set A is separated from B. What should be the definition? Every point of A is separated from B. This is 1, 2, A and B subsets of R. We say A is separated from B if every A belonging to A is separated from A. Let us write A B contained in R such that S is equal to A union B and A B separated. If a set S can be written as it is cut into two pieces where A and B are separated, we say S is not connected. What is connected? S is connected if there does not exist any separation of A. If A union B is a separation, then they have to be disjoint. Is that obvious? Every point of A is separated from a point of B. If there is an intersection, then that point has a neighborhood which intersects both of them because x itself is in both of them. Take any neighborhood that will intersect both. If A intersection B is non-empty, then it cannot be separated. So, separation implies that A and B are separated, implies that they are disjoint already. It is something more than disjoint. Separated means every point of A has a neighborhood which does not intersect that and every point of B has a neighborhood which does not. For example, look at the rationales as one set, set of irrationals as another set. Real line is union of these two. Real line is union of rationales and irrationals. But can you say rationales are separated from irrationals? No, because every neighborhood of a rational has to have an irrational inside it. So, it will intersect. So, rationales and irrationals, though they are disjoint, they are not separated. So, saying a separation is something more than saying they are disjoint. So, the theorem that we want to prove, time is too short. Theorem, we will prove it next time. That S contained in R is connected if and only if S is an interval. The only connected subsets of the real line are intervals. So, let us see how much we can prove. So, two parts. Suppose S is connected to show S is an interval. Keep in mind we had called empty set also as an interval. So, if S is empty set, is there anything to prove? Yes, because separation does not make sense. Separation of empty set into two parts does not make sense. A union B or you can say empty set is a connected set. You can forget about it. You can declare empty set as connected. Suppose it is a singleton. So, if S is equal to empty or singleton, is singleton point that is a subset of real line, is it connected? That means it should not have any separation, the only singleton point. So, it will be in one of the sets A or B only. So, there is no separation possible. So, it is connected. Finally, let us take two points. Let X, Y belong to S. What we want to show? S is connected to show it is an interval. We have got two points to show it is an interval. What should I show? Every point in between is also inside and let X less than Y, X less than Z and let Z belonging to R such that X less than Z less than Y to show Z belongs to S. So, let us take the picture. Here X, here Y and here Z. I want to show Z belongs to the set S and given S is connected. If Z does not belong to S, if Z does not belong to S, then I should, what I should show? If Z does not belong, then I should show S is not connected. I should produce a separation of S. So, let us write, suppose Z does not belong to S, then Z does not belong to S. That means, where is S? S is in Z complement, singleton. So, that means, then S is a part of R minus Z. That does not belong. So, it is part of the complement. So, what does the complement look like? It looks like two parts. One part is here, another part is on the other side. So, one is here, another is here. So, let us look at which is equal to minus infinity to Z, union Z to plus infinity. So, what is S equal to intersection of both sides? So, minus infinity to Z, intersection A, union minus infinity to Z, intersection B, because S is not containing Z. So, some part will be on the left side, some part will be on the right side. So, this is the part on the left side and this is the part on the intersection A. Where is A? There is no A, there is S. So, let me write properly some intersection S, union Z to infinity intersection S. So, intersection of S on the left side of Z, intersection of Z on the right side. So, that will be everything. Now, here is a set, call this set as A and call this set as P. So, S is written as a union of two sets A and B, they are disjoint. Is it a separation of A? Is it a separation of A, separation of S? Let us see. So, let us take a point on in A. So, if A belongs to A, where it will be? It will be somewhere here. So, I can take a interval because if this, it is not equal to Z. So, it is bigger than Z. So, there will be some interval I can take around A, which does not go beyond Z. So, for every point here, I have got a interval which does not intersect the right hand side. So, every point A in A will be having a neighborhood disjoint from P. Is the picture clear? Choose epsilon bigger than 0 such that A minus epsilon A plus epsilon intersection A, intersection B is empty. I am saying possible. Why it is possible? Because if you take any point A, it is going to be less than Z. So, I can always take a interval on that side so that it does not go on beyond Z. So, that means what? Every point A, that means A is separated from every point of A as a neighborhood which does not intersect. So, A is separated from B. Similarly, B is separated from A. Similarly, if I take on the right side, I can similarly do it. Is it clear to everybody? Similarly, means what? If I take a point A or B, then there is a neighborhood which does not intersect. It does not go on the left side. So, what I have got? If Z does not belong, I have got a separation of S, but S is connected. So, contradiction. So, implies S is equal to A union B is a separation of S not possible. Hence, S is an interval. So, we will prove the other way around tomorrow. Namely, what we have shown today is, if S is connected, it is an interval. We will show that if we take an interval, then it is connected. That means no interval can be cut up into two parts such that they are separated. The two parts are separated from each other. So, that will prove the only interval. That is why this is the reason in the real line connectedness. The important is not very interesting because only intervals are the points. We will see what it means in R n tomorrow.