 Welcome to module 33 of point set topology part 1. Last time I mentioned some property of real numbers namely if you remove a point from an arc, from an, from an arc then it will get disconnected, how does it prove that? Intermediate value theorem from real analysis that will give you automatically that r minus any point is not path connected. So intermediate value theorem is something which is written in the definition of or in the construction of or in the creation of real numbers. Let us take a closer look at it which is because it is very, very relevant to the concept that we are trying to develop here. So I am just recalling it here, I am not going to prove intermediate value theorem. Let f from j to r be any map where j is some open interval given x less than y and f x less than some point z less than f y, there exists a w inside j such that w is between x and y and f w equal to z, so that is the intermediate value theorem. As a consequence of intermediate value theorem you must know, you must recall that we get Rawls theorem in calculus about functions which are differentiable in the open interval and blah, blah, blah, right. Also remember that Rawls theorem is false if you replace the code mean by R n where n is greater than equal to 2. Simplest example I would like to recall is theta going to cos theta sin theta which is differentiable everywhere, okay, the derivative is never 0 but several two points may have same end points and so on. So this is an easy example where you do not get Rawls theorem. In any case there does not seem to be any meaningful way to generalize intermediate value theorem if you replace the code mean with any other than some order to colloidal space because you want to say intermediate value, what is the meaningful intermediate value given two values, what is the meaning of intermediate value that does not make sense unless there is an order. So you must better take some order to colloidal that is all, that is one way of getting things do not work and so on but you do not give up. So try to keep the code domain as R itself and try to look at what is happening to domain. Why do we need the domain to be an interval itself all the time that is not obvious. So here is an example where domain need not be an interval, okay. So let X be a any path connected space, we have defined what is a path connected space, right. Recall that a path connected space means any two points can be joined by a path. So take such a space X that f from X to R be any map, map means of course continuous function. Now given X and Y in an inside X that is nothing between saying that X is less than Y or Y is less than X and so on, okay. But Z belong to R so Z, Z is between f X and f Y, these are elements of R. So then I say that there exists a point W inside X such that f W is equal to Z. So this is at least part of the like intermediate value theorem. Intermediate value has been obtained though the point wherein it is attained that point may be we do not know you cannot compare it with X or Y because there is no comparison in X. This is quite satisfactory generalization of intermediate value theorem, okay. So how does it prove one line proof? You have two points X and Y inside X join them by a path omega from 0 1 to X such that omega 0 is X and omega 1 is Y. Now we apply intermediate value theorem to f composite omega, omega is continuous, f is continuous, the composite is continuous, this will start from 0 1 and ends in the R. So IVEP is applicable, okay. So you get a T between 0 and 1 such that f composite omega T will be between f X and f Y. This omega T is precisely the point some W here, okay. So put omega T equal to W then you have W is equal to that, 4. So now comes the next question, you have done something fine but I am not satisfied. Can we replace the path connectivity assumption on X, okay, assumption on this domain here with some weaker condition in the above theorem, okay. Finally, I do not want to get into this path is at all, is there some way of telling that why I am telling you that, why I am asking that question, you may ask why this question at all, I mean you can question the question. So there is another point of view, instead of answering that I will take note, look at the another way. Consider the union of two non-intersecting closed disks inside a metric space. If you do not want too much of that you can do it in R2 or R3 and so on, any R, but they must be non-intersecting closed disks, let us say, okay. Immediately your intuition tells you that there is no path from one to the other, okay. So that space, the union of two non-intersecting disjoint closed sets, closed disks is not path connected. Why? Have you proved it? Now you cannot use intermediate theorem, intermediate theorem needs your space to be what, you know, code mean to be R. If you are doing it inside Rn, there may be a way to convert it, like we did the conversion here, but I am taking arbitrary metric spaces, okay. But you still feel that this is true, right. What is true? There is no path from one point in the closed disk, this closed disk to the other point in the other closed disk, right. Now how to make this one rigorous? So this is the question, how do you prove this one, that is all, okay. So it seems that we have been forced to use one of the most important properties of the usual topology of the real numbers here, right. In the real, there are no gaps. So this no gaps here, it may be difficult to understand in the case of arbitrary metric spaces, but we seem to understand it in the case of real numbers. In fact, this filling of the gaps was the motivation of construction of real numbers. The gaps were there in the real, in the rational numbers or any algebraic numbers and so on, okay. But so we have been able to convince ourselves that the real number system has no gaps, right. So that property, we come back again and again we have to use that one. So because of the importance of this property, there is need to formulate it, this no gaps concept, you know, this concept, independent of the order of real numbers, is there a possibility as a natural consequence of such an effort, if this effort is successful, we will be able to use it quite useful and it becomes quite useful because it will be available in a larger context. So this is precisely the so-called notion of connectivity. So all these I am talking to motivate why do we need the unintuitive notion of connectivity as compared to very, very intuitive notion of path connectivity, which is very easy to understand, okay. So let us see how connectivity answers these two questions that we have raised right now. Let x be a topological space and y contain inside x a subspace. So I am making some definitions now. By a separation of y, we mean two non-empty disjoint subsets A and B such that the union is equal to y, both end we are closed or equivalently both are open in one. So all these things I expressed by just writing very cryptically y equal to A separate B, to read it as A separate B, so it is a, the line is like a, you know, cardboard partition. Remember what are the things, both of them must be non-empty, they must be disjoint, both of them closed or both of them open, there are two different ways of looking at them. The union must be wide. So these are the few things I mean which you have to remember all the time just by disjoint one, okay. If there is no such separation, the negation of this is very important. Even if one of these conditions is not satisfied, it is a negation. So you have to understand that way, okay. If there is no such separation of y, then we say y is connected. So connectivity which is a very nice concept finally is defined in a negative way here, okay. So but I have done it as, to try to put it as, as much as positive, there is no separation, that is the separation will give you y is disconnected. So if there is a separation, then it is called disconnected, okay. I would like to include the space of empty set also in the definition. It does not come all that easily because I am assuming and we are non-empty. So an empty set cannot be union of this one. But I would like to take empty set as connected by definition, okay. So this is, this is not forced by this definition but this is convention. What has this property of y to do with the space x? See I started with y inside x, okay. X is to pull out equal to space y subset, okay. See closed in y, that is the phrase that will tell you what is that relation. What do they mean by closed in y? In the subspace, I have to take the subspace to call it from x to y, okay. That is the clincher. Once you take the subspace topology, you can forget about x. Then you close subspace, close in y means close subspace of y, that's all, okay. So only to get the subspace topology, this x is there, that is the, okay. So what is, what has this property to do, y to do with the space x? The answer is we are taking the subspace topology on y from topology, that's all. The definition of connectivity of y is completely given in terms of the topology on y. Sometimes it is useful to directly write down the condition in terms of the topology on x. As done in some expositions, they give you a different definition. So that definition will become a consequence of this definition and vice versa. So let me give you that. What is that? Let us denote a bar, okay, a bar denote the closure of a subset a taken inside x, the closure is taken inside x for every subset a which may be subset of y, doesn't matter, okay. Instead of writing this one, I can write this set of equation, namely y is union of a, b, a and b are non-empty. Instead of saying a and b are closed and so on, all that I say is a bar intersection b is empty and a intersection b bar is empty, okay. In particular, you will see that a is contained in a bar, so a and b are disjoint, will come automatically, okay. So the only new thing is instead of saying that a and b are closed inside y, okay, the topology of y is not bothered about it. Everything is referred to now x. So this is another definition. You can verify very easily that this definition is the same thing as, I mean gives you equivalent to this definition 21 here. Let us carry on with this definition and do something. A space is connected if and only if, the only subsets of a, subsets of a, a, a, I am taking a space now instead of y deliberately, which are both open and closed in a are empty set and a itself, okay, a could be one and empty set. So these two sets are improper, proper subset means what? Non-empty and not equal to the whole space. So there is no proper subset of a, which is both open and closed. So that is the connectivity. This is very immediate because suppose you have such a thing, then immediately you can write, suppose we have say b is a subset of a, which is both open and closed and neither empty nor a. Then you can write a as b separate b complement over, right. Conversely also, if you use 21, then it is obvious that if a is closed, sorry a is closed and b is closed, both a, that is another thing, sorry. Why I want to say is that a, a is a proper set, a is non-empty, a is proper because b is non-empty, okay. If a is closed, then b is open and b is closed, a is open. Therefore, a is both open and closed, okay. So we have proved that one anyway. So here I have not written down. So that part at least you should do on your own. A set, which is both open and closed, will be called a open set. So this is a short form saying both open and closed, that's all. So I may, may or may not use this at all, but some authors use this one quite often. Thus a space is connected if and only if only clopens of sets in it are what? Empty set in the whole space, improper sets. So this redefining this theorem, you know, reformulating this theorem, that's all. In the real analysis course, you may have learned that every interval in R is connected. I will come back to this one now. With this definition, you may have come across with this definition is what I am telling you. If you haven't, we will do it here, don't worry. Indeed, the following theorem tells you that connectivity of intervals, existence of greatest lower bound, existence of least upper bound, okay. Greatest lower bound, least upper bound, I will go hand in hand. And the intermediate value property, these are all equivalent to each other. Therefore, the notion of connectivity that we have introduced with pure topological now, you see, is, you know, it could have been used instead of JLB and LUB in the construction of real numbers. At least, we will see that these things are equivalent. Right? So I am redefining this one here, intermediate value property. A subset A of R is said to have intermediate value property. If the following is true, given any continuous function F from A to R and points from, points between X, X less than Y, okay. X and Y are points of where I am just putting some order that were X less than Y or Y less than X is also allowed. That's all. And T is a set, T lies between Fx and Fy. There must exist A belonging to such that X is less than, X is less than S less than Y and F of S is T. So whatever the property of, in a statement of intermediate theorem, I have made it to a property. The theorem says that if A is an interval, then this is true. Right? So we are going to prove that one and whatever. So that is the part of the game here now. So let I contain ZR. So I have used this notation I for a real interval. Here, I am not making the, I suppose I is interval, okay. Then the following conditions are equivalent. I is an interval, I is connected, I has IVP. So I think I want to prove these things are equivalent. That is the whole idea here. So in particular, it will prove intermediate value property for intervals. So consider, now we are bringing the connectivity in between that. Consider the first special case where I is closed interval AP. Okay. One implies two means what? If it is an interval, it is connected is what I have to show. So take a special case now, I is AP. Suppose I is, I has a separation. That means, suppose it is not connected, then I must get contradiction that it is an interval. Everybody knows the definition of interval by that. Okay. So suppose there is a separation. With A belonging to A, B belonging to B, this is my choice because once it is separation, both A and B are non-empty. You choose them. Let C be the least upper bound for A. Now I am using the least upper bound property here, okay, for the real numbers. The whole thing A, B are subsets of the closed interval. They are bounded. So take the least upper bound. Bounded subsets all have a least upper bound. It follows that this C must be between A and B. So A is less than or C less than B. And hence C is in the interval because I have assumed I is an interval, so C is inside A. Okay. So you see A, B itself, so there is no no no violation. Since A is closed, it follows that C is inside A. So this is another property of least upper bound I am using. Least upper bound are limit points of of the corresponding set. So in particular, C is less than B. Why? See C is inside A, it cannot be right. They are disjoint. That's all. Yeah, you are right. So C is less than B. Okay. First I said C is less than or could be. Since A is an also an open subset of I. You see we have assumed that it's separation. So both of them are open and closed. Okay. So A is an open subset of I. Okay. It follows that there exists X and Y such that A is less than or X less than or equal to C less than Y less than B. See C is a point of A now. Okay. And A is open. And of course C is a less strictly less than B. Therefore, I can always push a Y here. Okay. So C is less than Y and that Y is also less than B. Okay. Right. This elementary property of real numbers. But this side, I don't know, C may be equal to A also. So I don't know that. So I may have to take X, X, X, X and Y to put A. I have to put equality. Okay. But what is the conclusion such that the whole XY is contained inside A? Why? Because A is open in in the closed interval AB. A is open in the closed interval AB means for every point there is a neighborhood of this nature inside A. That neighborhood can be chosen so that this Y is smaller than B is extra thing. That's all. Okay. In any case, if you have chosen it contained inside A, automatically it will be less than B. That is no problem. But this contradicts the fact that C is an upper bound for A because there is Y also now bigger than C. So in other words, what I have used here is if you have an open set of real numbers bounded, the upper bound or even the lower bound okay cannot be a point of A. So that's all I have used here. All right. So what we have proved, we have proved that every interval, so only closed interval is connected. Now let us prove the full thing. Now consider the general case wherein I is any interval with I having a separation. We may assume that there are points A less than B belong to I such that A is in A, B is in B. Okay. By interchanging A and B, if you want. There are one point here, one point here. Y A this should be less than that one. I don't know. But you interchanged that's all. Then you can assume this one. We now take the restricted separation. Look at this closed interval AB. Okay. Once A is inside A, B is inside B, both of them are inside I. The entire interval AB is inside I by I is interval. That is what is assumed. Right. So entire interval AB is contained inside I. That is the definition of interval. So look at that. A intersection AB, B intersection AB, they will form a separation for this AB now. Only thing you must see that both of them are non-empty. Other things will be automatic. They will be disjoint. Union will be the whole thing and both of them are closed. Why this is non-empty? A is there here. Why this non-empty B is there here? Fine. So we are now back into the first case. So the proof is over. Now let us go to 2 in price 3. So what is 2? 2 says I is connected. Okay. 3 says it has intermediate value property. Okay. So from this one, one implies 3 will come automatically. Every interval has intermediate value property. That is a theorem that you know in analysis. But here we are going to prove it. So first I have to connectivity implies the intermediate value property. Okay. So let f from I to R be a continuous function. Take x less than y belonging to I, fx less than t less than f y. These are the hypotheses for intermediate value theorem. Okay. Then you have to find out some element here between x and y such that that element goes to this t here. So that is what you have to do. To prove the, to prove intermediate value property that is what you have to do. So once you have given these things are given and these are given to you, put a equal to I intersection minus infinity, you take f inverse of that. All the points will less than t take inverse dimension. All points which go less than t are inside, inside is a. All those which are bigger than are inside b. Okay. Then a and b are non-empty because x is there in one another and y is there. Open disjoint subsets. Okay. Suppose t does not belong to f y. See we want to show that there is some z between them whatever wherever inside I such that f of that z is t. That is the same thing as this t belongs to f of I. Right. If t does not belong to f of I that is the only point which is missing from here. Then it follows that these two sets will cover I. Right. Right. Therefore I, this a b becomes separation. I equal to a separately. And that is a contradiction because we have assumed I is connected set here. Be careful here. I have never used that I is an interval. The last part I am using that I is a connected set. Otherwise that is a contradiction. Okay. Now theme plus one is a very straightforward thing. If I has an intermediate value property, okay. Then we consider the map I from inclusion map here. I to R. What is the meaning of I is an interval? Given any two points x, y, x and y everything between them must be there. Right. The inclusion map has a intermediate value property means what? Because it is continuous I must have intermediate value property. The between x and y point must be there that is all. So this three implies one is more of a methodology when it comes. Okay. Apply the apply the property three intermediate property property to the inclusion map. Okay. In particular it follows that R itself is connected because R is an interval. Notice that we have used crucially the fact that every bounded above set in R has a least upper bound. That is the only thing which you have used which is equivalent to saying that every bounded below set is has a least greatest lower bound. We have also used the fact that a least upper bound of a set A is in the closure of A which is true in any ordered topology. It has nothing to with real numbers. It is true with ordered topology. Indeed the following generalization gives a two way general result on the ordered topology that we have discussed in example 1.39 in the first chapter. Okay. So you see I wanted to show you that three things here are all say connectivity of intervals, existence and IVB. So first I took only this one. Now I am attacking this GLB, ULB itself. Okay. So here is a theorem. To understand this one I have to go to arbitrary totally ordered sets with certain property. I do not want to impose myself that I am working R. That will be like a vicious circle. Okay. So start with a totally ordered set and give the ordered topology on it. It is ordered complete if axis connected. Order complete means what? Every bounded above set has a least upper bound. That is what you have to go. Okay. And every bounded below set has greatest lower bound. That is the mean order complete. If it is connected I want to say that it is order complete. The converse you have to assume one more condition. Conversely suppose axis order complete and satisfy the property that given any two elements x less than y, there exists a third element between them. Okay. This is like what? This is similar to our equivalent to the Archimedean property of real numbers. Okay. Then x is connected. So connectivity is equivalent to this order completeness. Only thing you have to assume that totally ordered set which has this Archimedean property. See here there are no integers. So you have to reformulate the Archimedean property in some way. This is the way it has been done here. So all these goes back to cantor. This is not my invention or anything. So I am doing this because many standard books do not have these things. That is all. Okay. Suppose x is connected. Let A be any non-empty subset of x which is bounded above. Say there is a B in it. So set A is less than or equal to B for all A in A. Okay. That is the meaning of let us say upper bound. Now take B to be all x in it. So set A is less than or equal to x for every A in A. Okay. This B is non-empty because this little B is there. You start with a thing which is bounded above. Now you take all the bounds above upper bounds here. A is less than or equal to for every A inside A, then x will be there in that set. Okay. That is B is the set of all upper bounds of A. Okay. Clearly B is in B and so okay. And B is bounded below. Why? A is non-empty. Elements of A are all such that they are all less than or equal to elements of B. Each element here. So it is bounded below. Let C equal to the set of all x in A such that x is less than or equal to y for every y in B. What I am doing here now? I am looking at all the lower bounds of B. Take all of them. That is C. Okay. Then A is contained inside C because all elements of A are already lower bounds for B. But this set C may be very larger, much larger than A. Okay. So A is contained inside C. Now clearly x itself is union of B and C. So something either they must be inside B or they must be inside C. Okay. That is all I am claiming here. What is the meaning of that something is inside B? Inside B means what? Every element of A must be smaller than that element. Okay. If that is not true, then I want to say that this element what I have taken satisfies other property. Where all x belong to the axis of y for every y inside B. Okay. So now the claim is that intersection B intersection B C is empty. B intersection C is empty. Sorry. B intersection C is non-empty. Suppose we prove that B intersection is non-empty. Then you can take any S belong to B intersection C. Then S is the least upper bound for A. Elements of B intersection C are bounds for A. That is clear. Because they are also inside C, it has to be least upper bound. Among us all the upper bound must be least one. Okay. So how do you prove B intersection is non-empty? If B intersection is empty, we get both B and C are first of all open subsets. Okay. That we have not shown it. We will show that. Then it will show, it will follow that x is B separate C. So this will be separation of x. But we started with saying that x is connected. Okay. So we have to prove that it is non-empty. Sorry. It is empty. Then there will be a contradiction. Okay. So assuming that it is empty and then we are getting a contradiction. To get a contradiction, I have to show that both B and C are open. So that is what I have to do that now. So let x belong to C. Okay. Now x is not in B implied that there must be an A inside A contained inside C. Of course, such that this x is less than A. Okay. Because B is by definition all the upper bounds of elements of A. So at least one element of A must be bigger than x, then x will be less than. And then x is not inside B. So that is the meaning. But then we have all, so x belongs to the set namely all y inside x such that y is less than A. Right. So this set is much smaller than C because C set of all C by definition less than equal to y for all y inside B. Okay. So x belongs to this open set contained inside C. What I have done started with any element x. I have got an open set by definition this is an open set by the way inside the order topology. So it is contained inside C. Every point around every point inside C has a neighbor content inside C that means C is open. The same thing I want to prove it for B now. Suppose x is inside B. That means x is not inside C. Right. This implies that there is a B such that B is less than x. That is the definition of C after all. If it is not in C then there is some B which is less than x. But then look at all y inside x as B is less than y. That will contain this x and this is contained inside B. So this B is open. So this completes the proof that A has at least has a least upper bound. Okay. The proof of the converse is verbatim as in the proof of connectivity of an interval. One implies to how did you prove you have to exactly prove here. Instead of usual less than or equal to you replace it by this brick that is all this notation. Okay. So this is something which I have taken you somewhat deeper. So go through this one carefully because these things are purely logical here. That is how nothing goes. So you can try to write down, try to make a picture and so on. But when you make a picture you may be misled because then you may be already using properties of real numbers. So you make a picture but throw it away and see that everything comes from pure logic. So let me do a little more here. Let f from x to y be a continuous function. Then every path connected subset z of x, okay, x take a path connector subset here. The image of z is path connected. Okay. So now we are trying to prove something about properties of path connected. In particular if x is path connected f of x will be path connected here. In particular if y map f is surjective and continuous that's continuity or there is x is path connected will imply f of x which is y is path connected. So other than continuity I am not assuming anything but if you have subset here which is path connected the image is path connected is the correct thing here. So it is one line proof. Assume that z equal to x, okay, instead of no, instead of taking subset and so on you restrict the whole thing to z. So you can z if you itself x and f is surjective. What do you mean by that? You assume replace y by f z, z to f z take the function f itself then it is as if you have taken f is surjective, okay and assuming that x is path connected now, okay. Now, suppose a, b is a separation of y, y equal to a separate b. I must get a contradiction because I wanted to prove that y is path connected, y is connectivity. First I am looking at connectivity. Later on path connectivity will see that is obvious in here. So a and b are forming a separation for y. Then immediately it follows that f inverse of a separate f inverse of b the whole of x because what? Because now f is surjective. So f inverse of a union f inverse of b the whole of x they are disjoint because a and b are disjoint. They are closed because a and b are closed. They are non-empty because a and b are non-empty. Everything is from there it is pulled back it is under f inverse everything comes back. So this shows that x is if x is connected so is y, okay. If x is path connected so is y is more obvious because take two points inside y they are the image of some points here, right. Like a and b f a and f b are there inside y. Now a to b joined by a path f of that path f of gamma will join f a and f b. Maybe I should stop here now, okay.