 Probably, before proving it, because we are proving too many things, let me give you some consequences of this and then come back to a proof of this. So, let us look at, because this theorem is useful in producing many examples of connected sets. Suppose, A alpha is any family of connected sets, any collection of such that A alpha intersection A beta is not empty for alpha not equal to beta. Any two of them intersect at least once somewhere is a collection of sets. For example, let us look at R2. Look at two parallel lines, one parallel line, other parallel line. They do not intersect. So, we will not be considering that kind of example here. So, to look at any two lines which are not parallel, they will intersect. Now, intuitively a line in R2 looks like a connected set in R2. Intuitively, I am just saying, a line in R2 looks like a connected set. Why it is connected? Why is a line a connected set in R2? It looks like I cannot break it into two parts, which are disjoint kind of, which are not, which are separated. But, let us look at, a line is the image of the x axis. So, let us, I think, I am going in a different direction, which I have not thought of. But, let me, it is a nice thing to look at. So, look at, let us look at lines through the origin only for the time being. What is a line through the origin? What is the equation of that line? y equal to m of x. So, a line, so l m, a line with a slope m, I can write it as points x comma m of x, x belonging to R. That is one way of writing a line through the origin as a subset in R2. Now, look at the map x going to x m x. So, this is a function from where to where? This is a function from real line to R2. Is it continuous? Is this function continuous? If x n converges to x, then the first component of this converges, m of x will converge, m of x n will converge to m of x. So, it is continuous, because x n converging to x implies x n comma m x n will converge to x comma. This is a continuous function and we proved already. Is it okay? There is a continuous function. How do we approve? This function is continuous. In the domain, if x n converges to x, the image should converge. That is what we are saying here. Precisely. That is what it says. In R2, that sequence will converge. A sequence in R2 will converge if and only if each component converges. So, it is continuous and we proved that if a set is connected, its image is also a connected set. Continuity preserves connectedness. So, real line is connected. So, its image, the line through the origin also is connected. So, every line through the origin is connected. So, if you look at all lines in the plane passing through origin. So, this is a family of connected sets in the plane such that any two of them intersect. The intersected origin is a common point for all of them. So, this is one example of this kind of a situation. A alpha is a collection of sets in Rn. Here is an example. It was R2. So, that any two of them intersect. The claim is if each A alpha is connected, then the union also is connected. So, that is the point. So, here let me rewrite again. If A alpha is such that A alpha intersection A beta is non-empty, I do not have to write for alpha not equal to beta because alpha equal to beta is non-empty anyway. And each A alpha connected, then union A alpha is also. So, arbitrary union of connected sets need not be connected in general. For example, in the real line take one interval 0 to 1, other interval 2 to 3. Both are connected, but their union is one interval here, one interval there that is not connected. But if they intersect, so if any two of them intersect, then the union also becomes a. So, the kind of connected you can, from one you can go to another through that connecting point, intersection point. You can think of that way also and that is what we are going to do. So, proof, if I want to prove this is connected, let us take a function f and union A alpha, alpha belonging to y into 0, 1. I am going to use that theorem that we are not yet proved, but I am just giving an illustration of it. How useful that will be? And continuous. I should show this is constant. Claim, if then f is constant. So, how do I prove? For any two points, the value should be same. That is what we want to say. If they both belong to A alpha, then we are through anyway, because A alpha is connected. So, on each A alpha, f is going to be connected. And each A alpha, because A alpha is connected, f restricted to A alpha cannot take two values, either 0 or 1, it will be constant by that theorem. So, let us take one element in A alpha, other element in A beta. So, let x belong to A alpha and y belong to A beta, alpha not equal to beta. I want to show fx is equal to f of y. Any two points should take the same value, then it is a constant function. But what is f of x is equal to? From A alpha, I want to go to A beta and I am provided a root that there is a point of intersection between the two. A alpha intersects. So, let us go there and then come to the other point. It is equal to fz if z belong to A alpha intersection A beta and that I know is there. At least there is one point z in the intersection and that is equal to f of y, because A beta is connected. f of x is equal to A z, because z belongs to both A alpha and A beta. So, f of x must be equal to A z, because x and z both belong to A alpha. A alpha is connected and z and y belong to A beta. So, they should be equal. A alpha and A beta are connected. So, we have used the fact that both are connected and intersection is non-empty. So, that implies f is constant. Is it clear what we are saying? Here is A alpha, here is A beta. Just for the sake of illustration, A is A beta, here is z, here is x, here is y, f is a function either taking 0 or 1 on x. f of x has to be equal to f of z, because A alpha is connected and z and y. So, z and y also and the same A beta, so that f continuous connected implies they should be same. So, that implies it is a constant function. So, as a consequence of this, you can think of this kind of a set, this kind of a set is connected in R2. Here it is in R2. This is what is called a broom, a very common object in the household, a broom. So, that is a connected set. Another illustration of the same, let us give me an… Now, what do you think of this kind of a thing? Do you think intuitively it is a connected set in R2? I am just looking at pictures in R2. It is a very common thing again in household. What is that? It is a comb. Is comb a connected set? We can think of each vertical line as a set collection A alpha and bottom is a line, call it B and B intersects each one of them and claim that it is also connected. So, let me write second A alpha B connected such that B intersection A alpha is non-empty for every alpha. B intersects each alpha somewhere, then that implies union of A alpha along with B. I should write union, then implies union of A alpha, alpha belonging to I, union B is connected. Again, the idea of the proof is straight forward. Given any two points, let me keep that picture. One point is here, other point is here. I want to say that value at this point x is same as the value at the point y for any continuous function. In this union, the value is same. It should be a constant function. Any function defined on this set is a constant. That is what we want to say. So, take in the picture it looks like this kind of a set. Each A alpha is the vertical one. B is the horizontal one. They at least intersect at one point. So, value here is same as the value here. Value at x is same as the value at this point. So, I call it x1. x1 belong to the horizontal line line. Now, that value is same as the value at y1, because that is a connected set. Then I can go up to y. The value should be same. Same idea basically. That is also connected. So, intuitively, if you want to write, you can write take any function from this set into 0, 1. For any two points x and y, if x and y both belong to same, then no problem at all. If you belong to different one, then why are these kind of a root? Because of non-input intersection, you can pick up points and go. So, that is the basic idea. So, this gives you lot of examples. Let me prove one more before giving the abstract proof of that theorem. Let us look at three. Let y be contained in Rn or x contained in Rn such that x contained in y contained in x closure and x connected. So, what I am saying? y is a set which is trapped between a set x and the closure of the same set x. If x is connected, this implies y is connected. If something is trapped between the connected set and its closure, a connected set need not be closed. For example, open interval 0 to 1 is an open set which is connected. It need not be closed. So, again, what will be a proof of this? Same consequence of the same theorem. If I take any function on y, then it should take only, it should be a constant function. So, f y to 0 1 continuous implies f constant. So, that is what we want to show. Then, y will be connected. If any function on y 2 0 1, 2 point sets 0 and 1 is continuous and we are able to prove it is constant, then y is connected by that theorem. So, let us take two points or let us take, either way, it is okay. Let us take a point y in y. So, let implies y belongs where? y belongs to? y is in an enclosure. In an enclosure implies what? There is a sequence xn belonging to x, xn converging to y. Is that okay? Because we just now said, sorry, we have this situation. x is inside y, is in x closure. Take a point y in y that belongs to x closure. So, it must be a limit of a sequence in x. So, that means what? f continuous implies what? f of xn must converge to f of y. f is a continuous function on y. So, it must converge. But what kind of sequence f of xn is? Where are xn's? They are in x and x is connected and f is a function taking values in 0 1. So, either it can take everywhere the value 0 or it can take only the value 1. It has to be a constant function. So, if f on x is the constant function 0 and each f of xn is equal to 0. So, a constant sequence f of y is 0. For every point y, f of y is 0. If is the other one, f of f restricted to x is constant function 1, then each one is equal to 1. Then the value on y is 1. So, claim f constant on x implies f constant on, because of this. Whatever value on x it takes, if it is 0, then each of f of xn is 0. So, that is f of y. So, the constant is same whatever it was on x. Here is an interesting thing about this. Let us look at this example. Look at what is the equation of the y axis. That is, x component is 0 comma y, y belonging to R. So, that is the y axis. Let me write union of this with, intuitively, what I am trying to do is, look at the function sine 1, sine of 1 over x. What is the graph of sine 1 over x? Have you come across? It is like sine x. So, it is wavy. But 1 over x, that means near 0, it is getting cramped. It is minus 1 to 1. So, it is minus 1 to 1. So, that is 1. That is minus 1. And the graph looks like it goes like this. So, I am looking at the graph of sine 1 over x, union with the part minus 1 to 1 of y axis. That is connected, intuitively clear. The line is connected. That part is connected. What about the graph of sine 1 over x? It is a curve kind of thing. Let us intuitively assume that it also is connected. Or you can think it as a continuous image of 1 over x. That is a continuous function except at the point 0, sine 1 over x, if you do not look at it. So, let me write graph of sine. I am just trying to give you some interesting examples. What do you think is the closure of this set? Call this as x. What do you think is the closure of this set? The graph of sine 1 over x and the interval 1 to minus 1 to 1 on the y axis. This is coming. It will be something. You cannot give a better geometric interpretation of this. But what I am saying is this is contained in. What is the closure of that? What will be the closure of that? That will be union of. So, what I want to say is this set is connected, graph of sine 1 over x and that they look like separate parts. It does not look like that they are connected. What I am saying is, if you look at the closure of that set, the graph of sine 1 over x is coming closer and closer to minus 1 to 1. So, union of that is a connected set. So, this is a connected set. I am able to visualize it. If not, I am just giving you some examples of connected sets in order to do not worry about whether this will be asked in the exam or not. Try to understand it. Intuitively, this is also a connected. So, very exotic kind of things you can produce by using that theorem alone. So, I think I have got some time to prove that. So, I want to prove this theorem here. X is connected and two points at 0, 1 continuous. It should be a constant function. So, let us assume one of them and prove the other. So, let us write once again the theorem. So, here is the theorem, here is the theorem. So, I want to prove it again. So, let X be connected f x to 0, 1 continuous. Claim f is constant. So, for the just sake, for the sake of visualization, it is good to, this is X, this is the two points at 0, 1. So, let me just draw a picture. This is 0, this is 1. Suppose not, not of this, it is not a constant function. That means, there are points where the value 0 is taken and there are also points where the value 1 is taken. We do not know what is where. So, look at then, look at the set A contained in X and what is A? That is f inverse of 0 and B in X, what is B? That is f inverse of, so collect all the points where the value 1 is taken, collect all the points where the value 0 is taken. So, intuitively, you will get this kind of a picture. So, this is kind of A and this is kind of, so this is A, this is B. So, every point in A goes to 0 and every point in B goes to 1. Now, look at the set A. So, X is equal to A union B. So, note A union B. Every point of X has to go either in 0 or 1. So, inverse image of both. Now, claim, this is a separation of, I have written X as a union of two sets A and B and A and B are separated from each other because I will look at any point of A. If I look at any point of A, I will look at, can you find a ball so that no point goes to the value 0? See, at every point, let me look at the picture here. At every point of A, I can have a ball. Even if you think this way, I will have a ball. This is a ball in A. Only the part in A is to be considered. So, that will not intersect with any point in B or if you like, here is another way of looking at it. F inverse of singleton, can you find a singleton is an open set or a closed set? Singleton set in Rn, A singleton point is open or closed. What is the complement of it? It is open. So, singleton is closed. Every singleton is closed. Inverse image of a closed set is closed, if you believe in continuity that definition, equivalent way of saying that. So, basically what we are saying is that F inverse of 0 and F inverse of 1 both are open as well as closed in A, in X. Both are closed and open in X or the two are separated from each other, whichever way it helps you to visualize. At every point of A, the value is 0 and between 0 and 1, there is a distance. So, I can take an open ball around 0. Take the inverse image that will not intersect anywhere in B. No point of that should go in B. You can also think it this way. If you like, take an open interval around it, which does not include 1. So, this is the neighborhood of 0. F is continuous. So, I should have a neighborhood of the inverse image. So, neighborhood of whatever point I am taking that will be in A. That should not intersect with 1, because that is going to be completely contained here. So, every point in A will have a neighborhood which does not intersect in B. So, A is separated from B. Is that convincing for you? Take a neighborhood of the point 0, which does not include, say for example, you can take 1 by 3 and minus 1 by 3 open interval. So, let me just, for example, here it is 0, minus 1 by 3 and 1 by 3. This is the neighborhood of 0. Let us take a point X belonging to A. Let us take a point X belonging to A. Then what is F of X? That is 0. So, X is a point, the value is 0. So, if I take this neighborhood of 0, what does continuity say? There should be a neighborhood of in the domain, which is mapped into inside it. So, there is a neighborhood of the point X, which is mapped into minus 1 to 1 by 3. No point of that, because that is inside this. So, no point of that neighborhood will go to 1. That means, that neighborhood does not intersect, this at B. So, that is another way of saying, this is a separation, similarly from B to A. So, every point of A is separated from B, every point of B is separated from A and that is a contradiction because X is connected. So, it is a separation, this is a contradiction because X is given to be connected and we are producing a separation if it is not a constant function. So, it must be a constant function. So, we have proved one way. If X is connected, then the image, then every function on X to 0, 1, taking two values at the most, it should only take one value if X is connected. Every continuous function on X into the two points at 0, 1 must be a constant function. The converse is also true, but I think we do not have time to prove the converse. So, we will do it next time. Namely, if X is such that it has properties that every function is a constant function 2 point, then it is a connected set. So, we will prove the converse next time.