 So, the last thing which we did was a theorem that any product of connected spaces is connected. Okay, the proof. So that's a nice proof. Product topology, it works as a product. For the box topology, so example. Our omega. With, so our omega with the product topology is connected. Now our, that we have to prove still. R is connected, okay. That we will do today anyway. Our omega with the box or uniform topology. Uncle uniform topology is not connected. The reels itself are connected. That we will prove, okay. So with the box, with the product topology, this is connected, okay. But with the box or uniform, it's not connected. Proving not connected is easy. Now we have to find one separation, okay. Separate, we have to find the separation. What is the separation? We take separation, our omega. So we take the separation is, so one is all bounded sequences. So our omega are real sequences, no? Real sequences. So we take all bounded sequences. And then we take all unbounded sequences, okay. That's a separation, all unbounded. This is a separation. Well, box or uniform, okay. Product is connected. It's a separation. So why is it, well, of course it's disjoint, okay. It's disjoint. A sequence is either bound or unbounded. So we get everything. So what we have to prove is that it's not empty, no? So what we have to prove is open, okay. These are open. That's a crucial point for separation. Open, both are open, okay. So let, but that's trivial, no? Take any bounded sequence, okay. Let Rn, n in n. So this is an R omega. We are in R omega. B, a bounded sequence. Take any bounded sequence. Then in the box topology, we have a neighborhood. Then, so what is the neighborhood? In the box topology, we have arbitrary products of open sets, okay. Open times open times open arbitrary. So what we do is we take Rn minus epsilon, Rn plus epsilon. Well, epsilon, I can, one, for example. Rn minus one, Rn plus one, n in n. It's a neighborhood of Rn in the box topology. That's open, all open intervals, okay. It's a neighborhood of Rn in the box topology, obviously. Which is, which consists of bounded sequences, okay. Because Rn is bounded, okay. And now, you can go one ahead or down, okay. So it's a neighbor of Rn in the box topology, which consists of bounded sequences. So let me call this B here. This means that B is open, okay. So this means B is open. Well, but the same works for unbounded sequence. This is an unbounded sequence, Rn, no. Then clearly, this neighborhood of this point consists of unbounded sequences, no. Because in some way, with Rn, you go, it's not bounded. And this interval around it, it's, okay. Similarly, there's an excellence. R omega, the unbounded sequences are open, okay. And the neighborhood in the box topology consists of bounded sequences. So B is open. Similarly, unbounded sequences are open. Instead of R omega minus B is open, okay. But this is easy, no. Proving not connected, you find some separation, and that's it, okay. This is easy. Proving connected, that's interesting. And in the uniform, what is the uniform? You take it as a neighborhood. This is a neighborhood in the box topology, no. In the uniform, we have open boards, okay. So we take B uniform, row bar, Rn, n and n, with radius 1. For the uniform topology, instead of this neighborhood, you take this neighborhood, okay. Take this as a neighborhood in the uniform topology. And the proof is the same then. So in the box topology, this is the neighborhood. In the uniform topology, this is the neighborhood. And if Rn is bounded, this consists of bounded sequences. And if Rn is unbounded sequences, this consists of unbounded sequences. Is that clear? That should be clear, okay. I mean, what is a bound? If you have a bound C here for Rn, then C plus 1 is a bound for, okay. And here's the same, by the way, okay. Uncle C plus 1 is okay. So that's it. But we have to prove that the reals, the interesting thing is that in the product of a relay is connected. Now that's much more important, of course. And we have to prove that. So definition, that's the definition. In order set L, X, X is linear continuum. We prove it a little bit more general. Linear continuum is the following hold. It's the following two conditions hold. So the first one, linear continuum. The first one is L, X has the least upper bound property. The least upper bound property has the reals. What does it say? So least upper bound property says, what? Exactly. Any non-empty, let's say, non-empty bounded, not sequence, any subset. Any non-empty bounded subset of X has a least upper bound, okay. The smallest, least upper bound. That's the least upper bound property. Least upper bound property. Any non-empty bounded subset of X has the least upper bound, smallest upper bound. That's the first condition. It's important. The other one is easier. And this says that we have enough points, okay. Because this is called contino, okay. So this means if we have two points, then we always find the third one. So if X is smaller than Y, there exists a third point in X such that X is smaller than Z. For all X, Y, okay. Recall that this means smaller. This means smaller. Smaller equal, we have to write this way, okay. This means really smaller, okay. I mean, this means subset or equal, no. This is not exactly, okay. This means subset, but maybe equal. But this means smaller and not equal, okay. So there's a certain, so this is a linear continuum. Example, many examples are, that's the prototype, okay. The real numbers, that's a linear continuum. What about Q? No, the first condition doesn't work, okay. You have all numbers which are smaller than square root of two, okay. All rationals, which are smaller. This has no least upper bound in the rationals, okay, in the rationals. So no, that's exactly the reason why we need R instead of Q, okay. Q is not complete, okay. And that, we don't have the second property now. The first one is okay now, okay. But the second one not, right. So no, sorry, no, yes. That's the main example. We will see another example before I will prove something. Okay, so this is the least upper bound property in the theorem. If X is a linear continuum, in the order topology of course, okay. You have the order topology, then L is connected, X is connected. The reals are connected, for example, many examples, okay. Then X is connected and every interval or way in X is connected. And also every interval. So interval means open, closed, half open, doesn't matter, okay. Any interval, open, half open, closed, okay. Any interval and ray. So ray also open or closed, okay. This goes to infinity on one side, okay. And also every interval and ray in X is connected. That's the theorem. No, you prove it is connected, no? Okay, so let me prove that. So proof. So let A and B be disjoint non-empty sets. Let A and B, A, B be disjoint non-empty open subsets. Well, linear continuum and also every, and also every interval and ray Y in X, sorry, okay. Open subsets of Y. Maybe Y is X, okay, maybe. So we have all of these objects here, okay. Either everything on interval or ray, okay. That's what we have. That's called Y. X is everything. Maybe disjoint. So we have disjoint non-empty open, okay. Disjoint non-empty open. If the union is everything, there's a separation, okay. So we will prove that the union is contained, of course, in Y, but it's different. Now I have to write different, okay. It's not equal to Y. So there's no separation means, okay. No separation, so it's connected, okay. This implies Y has no separation and this means, of course, Y is connected. So this is what we prove, okay. So here are the conditions which we have. A, B are disjoint non-empty open, okay. So let's choose. They're non-empty. So we have to choose the point A and A, B and B. They are not empty. And suppose that A is smaller than B. Well, it doesn't matter. Otherwise we give other names, okay. So we can assume that A is smaller than B. Then this implies then that the interval A, B is contained. We are in Y, okay. So it's contained in Y. Y. What is Y? Y is an interval or a ray or everything, okay. If you have two points, you have everything between these two points, okay. Because this is an interval or a ray or everything, okay. So if you have two points, you get anything between them, right. They call convex subsets, okay. These are the convex subsets. Exactly the convex subsets, okay. So this is trivial. And now we reduce to this interval, okay. So let A0 be A intersection AB and B0 be B intersection AB. So we consider this interval in Y. And we intersect. So what this means is that A0 and B0 are disjoint still. They are open in AB now, okay. Which is the sub here. They are open in AB in this interval now. And they are disjoint. And we will prove that the union of these two is not the whole interval, okay. Then the union of these two is not the whole space. Okay. So this one. And let's see. This is the important definition of the proof. The least upper bound of A0. So we are in the interval now, okay. So the least upper bound of A0. This is least upper bound, okay. Least upper bound. Supremum, okay, if you want. So this is smaller equal to AB, to B, of course, okay. We are in the interval AB. And this is C. Anyway, this is an element of AB. C, this is important definition, the least upper bound. So this is, and this is the point which will be missing, okay. So let's prove that there are two cases. First case, suppose C is in B0, okay. So C might be in A0 and B0, okay. We will prove it's not in A0, it's not in B0. So that means this is a missing point, okay. A0 union B0 is not the whole interval, okay. This point C is missing. And this is not the separation. So first case, suppose that C is in B0. Well, start with B0. So I make a picture now. Here's this picture. Okay, so here's, this is X. This is the real numbers, no. But it's an order set, a linear continuum, no. So this is X. And then we have AB. This is very useful to make pictures, no, of the situation, what we have. So it's easier to see, no, it's a geometric, the geometry. And now we have a point C, the least upper bound. C is in B0, that's the case, okay. So A is in A0, no, of course. A is in A0, right. B is in B0, okay. So C is not equal to A. So C is somewhere here, okay. So it's bigger than A. C is not in A0. C is in B0. It might be equal to this, okay. That's why I don't care about that. So here we have smaller or equal, okay. It might be equal, but here it's smaller because C is in B0. Now B0 is open, okay, in AB. Well, we have the order topology. So what does it mean? The order topology is related by open intervals as a basis, okay. So we intersect with open intervals. With open intervals, okay. And then this is, so what does it mean? We have something which is open. So we find, on this side, I don't care. But on this side, we find an open interval here. But on this side, we are maybe here. So we have to stop. But I'm interested on this side, okay. So there exists an open interval, half open interval, which is D, probably D, yes. So here's the point D. So there exists D smaller than C, such that the interval, DC. And now an open interval. But here I close at C, okay. Maybe I can go to the other side. I don't care about this other side, okay. Because maybe it's equal to B. Then it's finished here, okay. Then I may go here to the other side. We are in AB. So B is open in AB. Exists D such that DC is contained in B0. B0 is open. It's contained in B0. We have to order topology. Open intervals. Here we have finished at C. DC is in, but this is strange, no. We are looking for contradiction. Where's the contradiction? So what is C? C is the least upper bound of A0, okay. But all these points are in B0. And the other one. That means that D is an upper bound of A0. So this implies D is an upper bound of A0. Because the points here in between these are all in the other set. In B0, okay. There's no point. So, and there's a contradiction, no. Because it's not, so C was not the least upper bound. D is smaller, okay. So this implies contradiction. Contradiction C is not. So this implies that C is not the least upper bound of what? Of A0. And this is a contradiction. So I'll use sometimes this sign. I don't know what it is. So we have contradiction. So this case is not possible, okay. So second case. It's clear. You see, okay, the situation. Second case, C, suppose that C is, well, it might be in what is the other, A0. I make again this kind of picture now, okay. I like these pictures because you see the situation, okay. So what? So I make the picture here. This is again X. This is the same picture as this. This is AB, AB. And now C is in A0. So it means that C is somewhere here. It's smaller than B because B is in B0. But here I don't know on this side. I don't care too much, okay. Again, this might be equal to A. However, C is smaller than B. But now the same. A0 is open. Like here, B0 was open, okay. A0 is open. C is in A0. So now I care about this side, okay. So D. So there exists D which is smaller, bigger than C. Now we are on the other side. Such that the interval C, D, here close, half open interval, okay, is contained in A0. So here we are in A0. That looks also, see what is the least upper bound of A0, you know. Here we are in A0. So that doesn't look so good. However, we need the second condition, okay. Because it might be empty. So we find the point here, okay. By the second condition, we find this D is okay. But maybe D is in, I don't know where D is. D might be in B0, no? This D might be in B0. I want to find the point in A0 here, okay. D is not the right point. So here I need the second condition. By the second condition, for basis, no, for linear continuum, there exists, there is. Now I need another point here between, which is called E, okay. There exists E between C and B. So this means that E is where, is in the interval of course C, this is D, sorry. C and D, no, C and D, yes. So it's here, and this is contained in A0. So this means E is in A0. E is in A0, okay. But it's bigger than C. C was the least upper bound, okay. So contradiction, again contradiction. This finishes the proof. So this means then that there's no separation. So A, B, so to conclude the proof, then I can write. So all this implies then that Y has no separation. This is the end of the proof. Corollary, the main case of course, corollary. R is connected. And each interval and ray in R is connected. And also this is the main application of course, and also each interval and ray in R is connected. This is an important example from analysis, no. This is the important theorem, which is on the screen here, but with the proof. So I give another example. This is a nice example, and now the strange example, the remark. Here's an example of a linear continuum. The ordered square, it's a linear continuum. The ordered square. So it's written ordered square. This is I cross I with order topology, no? This order topology. With the order topology. What order topology? With the, how did we say the dictionary order topology, no? Lexicographic or dictionary. It's a dictionary order topology. This is the ordered square. One of the important examples in the book, okay? The ordered square. It's a linear continuum. Proof, the second condition is trivial, okay? I mean the second condition, so I write in the second condition is trivial. Well, let me make a picture now. Here you have the ordered square, and now you have any two points. Any two points, you have to find a point between. Of course, if you have these two points, you find here a point in between, okay? If they don't have the same coordinates, the two points, this one and this one, okay? Then you find many points in between, okay? So you find p, so the first condition is trivial, okay? If the first coordinate is equal, if the first coordinates are equal, you find something between the second coordinates, okay? Because it's the interval. And if the second coordinate, the first coordinates are different, you find something in between, okay? You can write that if you want, but I will not, okay? The interesting is the first condition, the least upper bound property, okay? So let, well, this is bounded anyway, okay? Bound is not so important here, you know? Let a be any subset, non-empty subset. Let a be any non-empty subset. So bounded, of course bounded, but it's trivial. Subset i ordered square. And we have to find the least upper bound. And also this is okay. I will not tell you my picture how to find that least upper bound, okay? So here's the ordered square. And now we have a subset a. I don't know this, okay? A, oops. For a finite subset, of course, it's trivial. No, that's not very interesting. The infinite subsets are interesting, but I cannot make infinite subsets. So this is a, some infinite subset, okay? And we have to find the least upper bound. To find the least upper bound. So what do we do? We consider first coordinates first, okay? First coordinates. So consider let, so we have two projections, no? Pi one and pi two. On the first coordinate, okay? First I consider pi one of a. So this is a projection onto the first coordinate. Of i cross i, okay? Project. So here you have this, you have this, this, this, this. Some points here, no? Okay, on the first coordinate. And then you have the least upper bound. Because now we're in the interval, the real interval. Now in the real interval. So let, so consider, sorry. Consider pi one of a, okay? And let c be the least upper bound, okay? Let c be the least upper bound x of pi one of a. The projection of a in the interval i. In this interval here, okay? So here's c. Well, somewhere here, okay? This is c. And then there are two cases. Or let's distinguish two cases. So two cases. If, so here you have this line, okay? This c coordinate first. If there are no points of a on this line. This is the first case. If there are no points. So I have to define this. So if, what means a point is on this line. That means that if there, these points, what is it? First coordinate is c. And the second coordinate is something. C x in i ordered square. First coordinate is c. Pi one of c. These are all points in the order square with first coordinate. These are all points in the order square. First coordinate is c. So all points on this line, right? If this is empty. If, let me call, give also name of this. I have too many names. If b is empty. Then we are done. So what is then the least upper bound of a? C times zero. Then c times zero. So it's then this point here. Okay? But now we are in the hole, okay? Then c times zero is the least upper bound of a. It's clear, no? There's no points here. So you come up really close to this, okay? So c times zero is the best cross, no? If this is not empty. So that means if it's not empty. That means now we have points here, okay? Some points here, infinitely many maybe, okay? Yeah, now we take the least upper bound of these points. Let d be the least upper bound of what? Of pi two of b. One second. Pi two of b, no? Okay? That's what it is. B are all points. First coordinate c. And now we take the least upper bound second coordinates of these points. And then what is the least upper bound now? So here we find now this now. And then the least upper bound is c times d, no? Then c times d. C times d is the least upper bound of a. Okay? So I gave a complete proof here of the second case, okay? First coordinates, then we have two cases. And then in the second case we have considered second coordinates. But not of all points, but only of the points which are on this line, the second coordinates, of course, okay? That's it. That's the proof. So it has the least upper bound property. So this means corollary is a linear continuum. So corollary is connected. It's connected. Corollary. The ordered square, ordered square. What is the ordered square? That's an interval. Which interval? Zero times zero. One times one. It's connected. Since it is an interval. Sorry? No. Yeah. That's our space. We are not R2. This other topology, if you go to R2, it's a completely different topology. Not interesting. Not very interesting. The interesting is this. In R2 it's not an interval. Because there are many points over and up and down of this. It's not an interval in R2. But our space here is the ordered square. Not R2. The ordered square with the ordered topology. Okay? The other one is much easier and not interesting. The other. So theorem. Come back to this. Come back to this. The first theorem, which is, so, and it's a base of analysis. So this, how is it called? Intermediate values here. Intermediate values here. So that you know from analysis. And now we have a little bit more general version here. So what does it say? Let f from x be continuous. Let f from x to y be continuous. So this means these are two topological spaces, obviously. Where x is connected. X is connected. And in y we have to be able to compare elements. Okay? Larger, smaller, intermediate value. Okay? So we have to compare. And y is ordered with the ordered topology. Okay? And y is connected and y is ordered. So it's a little bit more general than y is ordered with the ordered topology. So that's then, well, then if f of a, if f of a is smaller than r, smaller than f of b for a, b, what is our space? X. Then if this happens, there is c in x, such that f of c is equal to r. If you have two values here, everything between these two values is also a value. Okay? These are values, no? f of a, f of b, this is between two values. So it's also there. Now that's everything, every element, every element between two values is also a value. It's also, value means image, no? That's what it says, okay? Very short, but we have to be able to compare elements, of course, in y. Okay? Smaller, equal, larger. In this form, this is a trivial theory. It's very important, but it's trivial, okay? The proof is trivial, as you see now. Trivial proof, so proof, proof, two lines, two line proof. So suppose that r is not in the image of f, okay? Suppose that, by contradiction, that r is not a value. That is, r is not in the image of f, no? r is not in the image. Suppose r is not a value. Then, x, we get the separation of x immediately. What is the separation of x? We take f minus one of, so what is not a value? r is not a value, okay? So I take minus infinity. This is a ray, no? This is a ray. It's convenient to write in this way, okay? Minus infinity, r. And then we have f minus one of r infinity, plus infinity, no? Then x equals, this is a separation of x. Is a separation of x. Why? Well, here we have some point, which point? We have, a is here, no? So here we have a, f of, where is f of a, okay? It's smaller than r, no? f of a is smaller than r. Here we have b, right? So they are not empty, these two, no? Not empty, not empty. They are open because f is continuous. Premature of open sets. They are disjoint because obviously these two are disjoint. The only point then is that the union is all everything. Why? Because r is not the value, so we don't miss r here, okay? If r would be a value, we have a problem, okay? But r is not a value. There's no point in x going to the r, okay? So there's no problem. We have the union. So this is a separation, what I wrote before. So this is a whole proof. Two lines, okay? Very trivial theory. Very important theorem. So large part of analysis depends on this theorem, okay? Together with the maximum value theorem, no? Which we also, it's equally trivial, but we will see that. So the whole analysis, one of the first things, which is all supplied this theorem, no? And still it seems to be trivial. What is, why, so you don't get something from trivial things, no? In general, I mean that's not possible. So if you want to apply this, you have to prove, of course, that something is connected first, okay? For example, the real, real interval. So the work is in this part, okay? You have to prove first, this is connected. My space, okay? A real interval or whatever, okay? And then you can apply this, okay? So the work is improving something is connected. Not in the proof of this, okay, theorem. The proof of this important theorem is trivial, okay? The only thing which we apply is what is the definition of connected, no? It's just a definition of connected. No separation. So it follows immediately from the definition, okay? But it's still, it's an important theorem, of course. Okay, so as we said, as I said, what is easy is if you prove something is not connected, okay? As we saw today. Our omega with a box of what is not connected. You just give some separation. You see some separation, that's for you, no? But that's not useful, no? To prove that something is connected, you have to work, okay? And this is useful for the applications. That's a philosophy, okay? Okay. So now, we have a second definition of a version of connectivity. Path connected. So path in the space X, path in the space X. In the topological space X, of course. It's a continuous map. It's always continuous. F from, and now we have this interval of zero one, okay? Two X. Well, that's all. So this interval is I, as you know, okay? Not equal. This is I, okay? That's why we don't use iPhone index, as I said, no. Index is J, because I is the interval. It's a continuous map. Initial point from F of zero. This is the initial point, or starting point. Initial point. It's just a way to talk. Initial point. Two F of one. And this is called the final point, endpoint. And so we have a picture, which is not very. So here we have, a picture is always this. So we have something F, which goes from initial point to the endpoint. End point, okay? End point. But passes are not so easy in general, okay? They're very complicated passes. Continuous maps. So this is a pass. A pass is a continuous map. A space X is pass, so this is the first one. You know the important condition. A space X, a topological space. A space X is pass connected. Pass connected. If for every two points, A B in X is pass connected. If for every two points, A B in X, there is a pass from A to B. There's a pass in X from A to B. So this is A and this is, for any two points, you find the pass, okay? That's also a very natural notion, okay? Well, of course, connectivity is more basic notion, you know? That's clear, because you don't need anything. Now here you need a real interval, okay? And so it's, the other one is more basic, okay? But this is also very natural, of course. And so we want to compare. So the first observation is the propositional theorem or whatever. Proposition X connected. X pass connected implies X connected. So pass connected space is connected. Proof is, suppose we have a separation of X. We have to prove X is connected, okay? Suppose, so it's a fake contradiction, but it's easy to write by contradiction, okay? So we prove it's not connected, then it's not pass connected, no? Suppose X equal A union B is a separation. So that means X is not connected. Suppose it's a separation. So a separation means, of course, the usual stuff, disjoint or non-empty, okay? Non-empty, so let A, B and A choose A and A and B and B. Fix two points, okay? Choose two points A and A and B and B. Let, so it's pass connected, okay? So let F be from I to X be a path in X from A to B. All of this situation, no? We have a path from A to B. Then the interval 0, 1 is equal to, now we have a continuous map, so we take pre-images, no? F minus 1 of A union F minus 1 of B. No, sorry. Yes, A and B, yes, that's what they call it, okay? So you take pre-images of these two, A and B. X is A union B, okay? And then we take pre-images. It's clear that we get everything, okay? A and B are disjoint, so these are disjoint. F is continuous, so they are open. The pre-images of open sets are open and non-empty because here we have 0, here and here we have 1. F of 0 is in A. So then this is a separation of 0, 1. It's a separation of 0, 1. However, we have proved 20 minutes before that this is connected. This is a contradiction since the real intervals are connected. So this is similar as the proof of the intermediate value in some sense, no? But this is more difficult because we have, the whole proof depends on the fact that the interval is connected, okay? That's an important factor. I is connected. This is an important factor of this proposition. The interval is connected and this implies that path-connected is connected. And now we have the other inclusion which is not true, okay? The other inclusion is not true. The other implication is not true, okay? So then we need an example of a space which is connected but not path-connected. That's what we want. We cannot go in this direction in general, okay? And this is not so easy. This is difficult, why? I'd like to explain why it's difficult, such an example. To prove that, so what we need is a space, so what we want is a space, okay? Yeah, that's what we want. So we want a space, we want to find, we want a space to see, to find a space which is connected but not path-connected. The other inclusion does not hold, okay? The other implication does not hold. And this is not easy. Why is it not easy? So we want a space which is connected, okay? As we said, proving that something is connected may be difficult also, okay? Proving that it's not connected is easy. For path-connected is exactly the opposite. So proving that something is path-connected is easy in general. Because you have to find a path, okay? For example, proving that something is path-connected is easy. Proving that something is not path-connected is difficult. Because then you have to prove there's no path, okay? So that means proving connected, we are on the difficult side. Proving not path-connected, we are on the difficult side on both, okay? That's why it's not so easy, this one, okay? It's easy to prove something is not connected, and it's easy to prove something is path-connected. Well, I'll give you an example for path-connected. So optimization, exercise, whatever. Our omega, it's trivial to prove our omega is path-connected, okay? Our omega is path-connected, that's absolutely trivial. Yes. Well, in the box of products, it's not even connected. If you don't say anything, product. Yes, product, product, product. It's path-connected. This is trivial. Why? So they're given two points. It's path-connected. If let, let me choose two points. Xn, n in n, and yn, n in n, b in r omega. Give, choose two points. Then we have to find the path, no? That means what? That means that xn, yn, are in the reals, right? Of course, coordinate, pair coordinate. But the reals are path-connected. That's also trivial. I mean, if you have two real points, you just have the segment. You have to re-parameterize, maybe. Going from 0 to 1, okay? But it's clearly r is connected, trivial, also trivial, no? Let fn be a path in the reals from xn to yn. Whatever, path, linear, or what? And then, of course, we have a path. Then f from i to r omega. So how do you find? f of t is equal, you take coordinate, pair coordinate, fn of t, no? n in n. It's a path from this to this, so called this x. Do we have any x here? No. So this is x, and this is y, sorry. Okay? Then f is a path in x is a path in r omega from x to y. In the product topology, no? In the product topology. Obviously, in the product topology. So this is important, no? Yeah, in the product topology. f is continuous in the product topology, right? When is something continuous in the product topology? Coordinate, pair coordinate, okay? In the product topology, since f is continuous in the product topology. For the box topology, we saw it's not even connected. Yeah, so this map cannot be continuous in no way in the box topology, okay? It's not component wise in the box topology. So note again, r omega in the box topology is not even connected. Also in the uniform, it's not even connected. So it cannot be path connected, okay? Path connected implies connected. So this is trivial, r omega in the, r omega in the product topology is path connected, no? So it's connected. This is another proof, it's connected. But this is a fake because the proof that something which is path connected is connected requires that the interval zero, one is connected. So here we have to prove something, okay? It's not possible, I mean as I said, proving that something is path connected is trivial. Proving that something is connected is not trivial in general, okay? But path connected implies connected. But this step from path connected to connected requires some work. The main thing is that the interval is connected. That's what you have to prove for that, okay? You have to prove the interval is connected. Then it's easy that path connected implies connected, okay? But for the, okay. So then we have this example. We want to find the space which is connected but not path connected. And we are on the wrong side for both, okay? But half of the work we did already. So, for example, well, here we are again. The other square is connected. That we proved and it's not trivial because we need, it's a linear continuum and it's an interval in the linear continuum. Intervals in the linear continuum are connected, okay? So we use our theorem. So it's not immediate, this one, okay? It's connected already proved and this is not called path connected but not path connected. That's an example. Our first example. We'll give two examples. This is the first one. The second one is closer to analysis. So it's also an interesting example. This is a very strange example. The other square. It's not so. So let's prove, it's not connected. Let's prove that I ordered the square. It's not path connected. Okay, so how to prove that? Suppose it is path connected. So suppose there is a path from 0, 0 to 1, 1, okay? We have these two points here. This is 0 cross 0 and this is 1 cross 1. And if it's path connected, there must be. So suppose there's a path in the ordered square. I ordered square. That's 0, not 0, no? From a path, sorry, f, as usual, f from i to the ordered square. There's a path f from the interval to the ordered square from 0 times 0 to 1 times 1. So we have some path here which goes from this point to this point, okay? The topology is very strange, no? It's not the standard topology. It's the ordered topology, okay? The ordered topology. Suppose you have a path which goes, so this is f, some path here, f, which goes from this point to this point, right? And we want the contradiction, of course. So this is nice proof. Intermediate value. This implies, so the intermediate, what? So intermediate value theory. So i is connected. The interval is connected. Now this is connected. So then the intermediate value theory. Yes, exactly. Intermediate value theory says that these are two values, no? This is f of 0 and this is f of 1. These are values, okay? These are the small, so everything else is intermediate, okay? So intermediate value theorem says f is subjective. Everything is a value. One second, one second. f is subjective, okay? So what do you say? Of i? No, they have the same cardinality. The same cardinality. Yes, right. They have the same cardinality. The real r and r2, the reals and reals times reals have the same cardinality. The same cardinality. Yeah, yeah. They have, yes, the one has to prove that. But that's, our n, all for all n have the same cardinality. They are all like the continuum, okay? Yeah, seems strange. Well, we don't want to prove that here now. It's interesting to think about the proof, no? Why have the same cardinality? Well, more or less an idea. It's a real number, what is it, no? A real number in the interval is 0 times infinite decimal expenditure, no? 0 times 1, 4, 1, 5, 9, 4, 3, and so on. Goes on forever and ever, no? Right? That's the real number, no? In the interval, 0, 1. And now what you do, you give, associate to this real number, two real numbers. You take all even coordinates and all odd. First all odd, the first one, the third one, the fifth, all odd ones, and then all even ones. That gives you two numbers, no? Between what? Yeah. And then if you have two real numbers, no? You can mix them together, no? From one, then from the next, and then the next, one, the next, no? And this more or less is the bijection. You have to be a little careful because the real number is not unique, no? You have this 9, 9, 9, 9, 9. One has to be careful with this, no? That is not completely unique. But more or less you see from this that it's not so different. R and R cross R, okay? You just divide. In odd coordinates, even coordinates, okay? You can divide. On the other hand, you can put together again, no? And these are more or less bijections. That's interesting also, okay? So F is subjective. That is okay, no? So now, what we need? We need a contradiction, right? We want a contradiction. We want to... So this path, F is subjective. It's not like this here, no? It goes everywhere, okay? If it exists, it's subjective. So we get everything. By the way, there are continuous maps. If you think now of I cross I, standard, not the order square. I cross I from analysis, okay? There are subjective maps, paths from I to I cross I. They exist. This is also strange. But there are subjective maps from I. So observation. There are subjective maps. Well, we almost said before there are bijections even, but subjective continuous maps, okay? That's a different point. Continuous maps now. There are subjective continuous maps. So this is a path, okay? Paths. We are talking about that. From I to I square, I cross I. But this is a standard topology, okay? This is a standard now. Subspace of R2, right? That's a standard from analysis, okay? Not the order square. This is a standard topology on I cross I. Subspace of R cross R. Standard also exists, no? There are subjective maps. Well, here we will prove something is wrong. But here they have, how are they called? Somebody has a certain name, how are they called? Some Italian name, no? They are called piano curves, okay? So they are known on this name more or less. Piano curves. He constructed in some way, nice way, such surjections, okay? Piano curves. So anyway, we want a contradiction, no? And we don't have the standard I cross I, but we have the order square, okay? One of our four favorite spaces, no? R omega, order square, RL, lower limit. And the fourth one, we still have to introduce. Which is maybe the... But now we have a contradiction, okay? So here's the contradiction. So consider this. Here's a point x, okay? So this is open. So let x be in the interval. So x cross 0, 1. What is this? This is the interval x cross 0, x cross 1, interval, no? That's an interval from this point to this point, right? That's an open interval. Where? In the order square now, okay? This is an open interval, so it's open. So open interval, so it's open. It is open. Well, it's an open interval, so it's open in the order square. Right? Okay. So f minus 1. Now it takes a preimage, f minus 1 of this. Of x cross 0, 1. It takes a preimage. What is this? It's open. Well, now we are in i, no? In the interval, all right? Preimage of open set is open. It's open. It's not empty. Non-empty. Why is it non-empty? Since it's subjective. Who said that? I heard something. Yes. Because it's subjective. That's important here, okay? Since f is subjective. So let me give this a name. This is, so I have to give a x. A we don't have. So A x. I don't know what is a good name. So this is x, no? A x. It's open and non-empty. Right? And now? Now we have all these A x. Somebody sees the contradiction already. For each x in x, we have A x, okay? So we take i, and we have x in x, and we take A x. We have this A x. For each x, we can take the preimage, and this gives us A x, okay? So these are all open, non-empty. All these A x are open, non-empty. Where? Open in, i. And if you take different x, if x is different from y, A x intersection A y is of course the joint, no? That's clear, because here you have this joint. So preimages are the joint, no? A x and a y are the joint. So where's the contradiction? We are in the interval now. Now we are in the real analysis, okay? We are in the interval. We get all these A x. They are open, non-empty. And they are all this joint. Where's the contradiction? We are in the standard topology, no? We are in the standard topology of the interval. And we have an open set. This contains many intervals, no? An open set, right? Open, that's the basis, no? An open set, non-empty, contains intervals. Each real interval contains what? Yes, contains a rationale. And now how many of these we have? Uncountably many, yeah? Because x is in the interval. So this is uncountable, no? I is uncountable. So how many rational numbers do we get in this way? Uncountably many, no? We have all these disjoint open sets. In each A x, in each we choose a rationale. So we have uncountably many rationals, okay? That's not possible. So in this way we would get uncountably many rationals, okay? Each, so let me finish. Each A x, so A x is what, this one, contains a rational number, contains a rational number. That's, okay, no? Each interval, okay? The rationals are dense in the reals, no? So varying x, we get uncountably many rationals, okay? So varying for something x in I, we get uncountably many rationals. And that's a big contradiction. Now the rationals are countable. The rationals are countable. This is uncountable, okay? But the rationals are countable, right? That's a nice, this is easy also. I mean it's not easy because we use, and I is connected again, we use intermediate value, you know? It's not immediate, the proof, but it's a nice proof. I mean, there's no path from these two, between these two points. If there would be a path, it would be subjective. And then we get all these uncountably many open sets in the interval. Non-empty open sets, all disjoint. And then we have too many rationals, okay? So in some sense it's, the idea is very nice and clear, okay? So this is a proof, so this finisher proof that there is no path, okay? So there's no path between 0, 0 and 1, 1. There's no path. So this is one example, okay? This finishes this example. So the other square is connected, but not path connected. It's of course a sort of strange example, no? It's a, tomorrow we'll give another example, which is closer to analysis, okay? This is a topology sine curve, and this is more standard for analysis, okay? So we start tomorrow with another example, which is closer to the standard analysis than this one, okay? But both examples are interesting. I suppose, well, the clock is not so, five minutes, but this here one minute, so, finish. Tomorrow you have exercises, right? Tomorrow, no? Not tomorrow, yes? Okay, that's it.