 So, let us continue with our discussion on connectedness. So, the main result we are going to prove today is the following theorem. So, let x and y be connected topological spaces then the product x cross y with the product topology is connected. So, let us prove this. So, let us assume x cross y is not connected. So, then there are open subsets non-empty open subsets A and B such that this product is the disjoint union of A and B. So, in the previous lecture. So, in the previous lecture we saw that the image of a connected space topological space under a continuous map is again connected. So, the precise result was f is from x to y x is connected then f of x with the subspace topology from y is connected. So, let us use this result. So, we look at the map from y 2. So, fix x naught in x. So, fix an x naught in x and consider the map from y 2 x cross y given by y maps to x naught comma y. So, to check that this map is continuous we just need to check that both the factors the projections to both factors are continuous. The projection to the first factor is just the constant map which sends everything to x naught and therefore, it is continuous. The projection to the second factor is the identity map which is continuous. So, this map is continuous and has image equal to the subset x naught cross y right. So, therefore, applying this previous result. So, thus x naught cross y with the subspace topology since y is connected is connected. So, now let us look at we intersect this equation with x naught cross y. So, we get x naught cross y is intersected with a. This is an open subset of x naught cross y in the subspace topology disjoint union intersected with b right. So, if x naught cross y intersected a is non-empty and x naught cross y intersected b is non-empty then we get a contradiction to the connectedness of x naught cross y. So, therefore, one of these has to be empty. So, this implies that x naught cross y is contained in a or in b is completely contained either in a or it is completely contained in b right. So, similarly so, this happens for all x naught in x right. So, given any x naught in x when we look at x naught cross y that subset is completely contained either in a or in b right. And so, similarly similarly arguing the same way for every y naught in y the subset x cross y naught is contained. So, now, we can get a contradiction as follows. So, choose any. So, since a is non-empty let x naught comma y naught be a point in a. So, suppose a point x naught comma y naught is here. So, then if we fix x naught. So, this is the line this is x naught right this is x naught cross y right. So, this implies that. So, since x naught comma y naught is is contained. So, as x naught cross y is contained in a or in b and it has a point and x naught comma y naught which is in a point in this x naught cross y is in a this implies that this entire line x naught cross y is completely contained inside right. Now we can take any point let us say y 1 over here and we can argue in the same way we can look at the line this line right. So, this is y 1 and this line is x cross y 1 this implies for any y in y x naught comma y is contained in a and since x cross y is completely contained in a or in b and it has is one point x naught comma y. So, y 1 let me just take this point y this is this point y. So, this point is contained in a and this entire line is contained either in a or b, but this line contains this point which is in a. So, which forces that this implies that x cross y is contained in a right and this happens for every y for every. So, this implies that x cross y which is equal to the union of x cross y is completely contained inside right, but this contradicts, but this shows that b is empty which is a contradiction. So, thus x cross y is connected right. So, as a corollary we see that R n is connected. So, inside R n. So, let us see some applications of the results that we have seen so far right. So, we just showed that this we just showed this corollary R n is connected and as an application right. So, there is no surjective continuous map f from the interval 0 1 to let us say the interval 0 1 disjoint union interval 3 and 4 right. Why is this? Because if there was such a surjective continuous map then it will force that the image is connected, but clearly since the map is. So, if such a map existed then since it is surjective this will imply that 0 1 disjoint union 3 4 is connected, but clearly that is not possible this is not possible right. Because if we let x equal to this disjoint union and we let this to be u and this to be v right. So, then both u and v are open in x right. So, x obviously has obviously has I mean we are giving we are giving x the subspace topology from R right as the subspace topology from R. So, in particular these two spaces cannot be homomorphic. Another simple observation is that if x and y are homomorphic topological spaces then x is connected if and only if y is connected right. So, this is easy because if f and y are homomorphic that means there is a continuous bijective continuous map from x to y right. So, therefore, the image of f is all of y and since x is connected. So, that means y is connected right and conversely if you are given that y is connected then we can take the inverse of f right g and yeah once again since the image of g is all of x it will mean that x is connected. Now, the next application we have in mind is to show that the spheres are connected, but for that we need the following lemma let t 1 and t 2 be connected subspaces topological space x. So, assume that t 1 intersection t 2 is non-empty then the union t 1 union t 2 is nonempty. So, let us prove this so, let us assume if possible. So, if possible let t 1 union t 2 be not be connected. So, let us say our t 1 is like this and t 2 is like this. So, that means, where we can write t 1 union t 2 is equal to t 1 union t 2. So, let us call this space y. So, since we are assuming that y is disconnected we can write we can write y is equal to there are two open subsets. So, every open subset looks like y intersection u where u is an open subset of x disjoint union y intersection v right where u and v open subsets. So, let us take a point in the intersection. So, let a be a point in the intersection which exists because we are assuming that the intersection is non-empty right. So, then a is either in y intersection v or in y intersection v. So, assume that a is in y intersection v. So, since now let us look at t 1 right. So, intersecting this equality with t 1 we get that t 1 intersected u disjoint union t 1 intersected v right. Now, t 1 is connected as t 1 is connected one of these has to be empty that is t 1 is completely contained in u or t 1 is completely contained in v right. So, since t 1 contains a and a is in u as a is in t 1 and a is in u. So, this implies that t 1 is completely contained inside u right. Similarly, as a is also in t 2 this implies and t 2 is also connected. So, repeating the same argument t 2 this implies that t 2 is also contained right, but then this implies that y which is equal to t 1 union t 2 is also contained in v right. So, this implies that this y intersection v is in v has to be empty right because y intersection u is equal to y. So, therefore, this has to be this force to be empty this is a contradiction. So, now, let us use this lemma to show that the spheres are connected. So, Corollary of the lemma S n is. So, to show this recall that. So, recall the map projection from a point and using this map we had shown that using this map we had proved this was left as an exercise actually using this map we can show that we can show. So, this map is phi from S n minus so let us make a picture. So, we are taking this sphere we remove this north pole 1 point and then we project to this plane. So, we have shown that this minus 2 R n is a bijective continuous map whose inverse is also continuous. So, since so let psi denote the inverse. So, as R n is connected. So, let us call this inverse let us call this psi 1 ok as R n is connected this implies psi 1 of R n which is equal to this sphere minus this north pole is connected now similar to projecting removing the north pole we can instead of north pole we can remove the south pole and project once again. So, let us call this map phi 1 yeah similar to this map to phi 1 we have a map phi minus 1 this from the sphere. So, we remove the point the south pole 2 R n right. So, geometrically what does this map do? So, we have the sphere we are removing this point given any point on the sphere. So, this point is p and this point is x we join p and x with a straight line and x is sent to the point of the point q right. So, this is R n this is S n minus ok. So, in the same way we proved phi 1 is continuous is bijective continuous is bijective continuous with continuous inverse we can prove that phi minus 1 is bijective continuous with continuous inverse. So, let psi minus 1 denote the inverse of phi minus 1 right. So, then psi minus 1 of R n is connected as R n is connected right. So, our S n we can now write it as phi 1 sorry psi 1 of R n union psi minus 1 of R n. So, psi 1 of R n is S n minus this north pole and psi minus 1 of R n is S n minus this south pole right. So, as both these intersect. So, let us call this T 1 as S n minus the north pole and S n minus the south pole have a minus non-empty intersection and these are connected the previous lemma implies that S n is connected of course, always n is greater than equal to 1 over here. So, that is a nice application. So, we will end this lecture here ok and in the next lecture we will continue with connectedness.