F 1 =. Setting F 1 = F i0 we have that. j=1 F i j

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1 Topology Exercise Sheet 5 Prof. Dr. Alessandro Sisto Due to 28 March Question 1: Let T be the following topology on the real line R: T ; for each finite set F R, we declare R F T. (a) Check that T is a topology and that (R, T ) is compact. For the fact that T is a topology: (i) and R are contained in T. (ii) If R F 1 and R F 2 are contained in T, then their intersection is R (F 1 F 2 ). Thus T is closed under finite intersections. (iii) If R F i is a family of sets in T, then (R F i ) = R F i. Since the intersection of (arbitrarily many) finite sets is finite, we have that T is closed under union. For the fact that (R, T ) is compact. Let O = {R F i } be an open cover. This means that (R F i ) = R, that is, F i =. Consider F 1, and let s = F 1 be the cardinality of F 1. Thus, there are at most s elements ( s ) F i1,..., F is such that j=1 F i j F 1 =. Setting F 1 = F i0 we have that O = {R F ij j = 0,..., s} is a finite subcover of O. (b) Let T std be the standard topology on R. Show that (R, T ) and (R, T std ) are not homeomorphic. If they where, then (R, T std ) would be compact. However, consider the following open cover: for each z Z, let O z = ( z 1, z ), and let O = {O z z Z}. Since every z Z is contained only in the set O z, if we remove any element from O we wouldn t have a cover anymore. Thus (R, T std ) is not compact and hence not homeomorphic to (R, T ). 1

2 Question 2: Show that [0, 1) [0, 1) is homeomorphic to [0, 1] [0, 1), but not to [0, 1] [0, 1]. Let A = [0, 1) [0, 1), B = [0, 1] [0, 1), and C = [0, 1] [0, 1]. We have that C is compact (product of compact sets), but A and B are not. Indeed, they are not closed sets and a set in R 2 is compact if and only if it is closed and bounded. Thus A and C are not homeomorphic. It is possible to write an explicit homeomorphism between A and B, but this would probably be just a very long formula that does not explain the ideas. For this reason, we will find homeomorphic spaces A = A and B = B and give an explicit formula only for A = B. Let A be the image of A under the homeomorphism between the square [0, 1] [0, 1] and the disk {(x, y) R 2 x 2 + y 2 1}, and similarly B (see Exercise sheet 3, Question 3.iii). (0, 1) A = A (0, 0) B = B (0, 0) If we look at A and B in polar coordinates, we have that the map f : A B defined as follows is a homeomorphism: (ρ, θ) if θ [ 3 π, 5π] ; ( 4 4 f(ρ, θ) = ρ, 5 4 π + 2 ( θ 5π)) = ( ρ, 2θ 5π) if θ [ 5 π, 7π] ; ( ρ, 3 π ( 1 3 π θ)) = ( ρ, 1θ + 3π) if θ [ 7 π, 3π] Intuitively, the map f drags the point in A to the point (1, 1) in B. 2

3 Question 3: Let X i, for i I, be a family of Hausdorff topological spaces. Show that X = i I X i is a Hausdorff space. Let x = {x i } i I, y = {y i } i I be two distinct point of X. In particular, there is a coordinate j such that x j y j. Thus, we can find disjoint open sets O j and U j of X j such that x j O j and y j U j. Let O = i I {j} X j O j and U = i I {j} X i U j. Then O and U are disjoint open sets of X such that x O and y U. Question 4: Write down an example of a topological space that is not Hausdorff. Note: recall that metric spaces are Hausdorff. If you have an example, check that is not a subspace of metric space (i.e. with respect to the induced topology). Let X be the set consisting of the points {a, b, c, d}. Let T be the following topology on X: T = {, {a, b}, {c, d}, X}. Every open set that contains a contains also b. Thus X is not Hausdorff. Question 5: [Hard exercise!] Let I = [0, 1] and consider the space I I (that is, I-many copies of I). Show that I I is compact but not sequentially compact. Hints: (i) You are allowed to use Tychonoff s theorem in the case of an uncountable product. (ii) You may want to think of the space I I as the space of functions f : I I. (iii) To check that a subsequence does not converge, it is enough to show that it does not converge on a coordinate. 3

4 Since I I is the product of compact sets, it is compact. We will show that is not sequentially compact. For each n N, let f n : I I be the function that associate to x the n-th binary digit of x, and consider the sequence {f n } n N. We claim that such a sequence does not have a converging subsequence. Suppose that there was a converging subsequence {f nj }. Then we can find a point x I such that the n 1 -th binary digit of x is a 1, the n 2 -th digit is a 0, the n 3 -th is a 1 and so on. That is, we can find a coordinate x I such that the sequence {f nj } restricted on x is 1,0,1,0,1,0,1,0,.... Thus {f nj } do not converge. Since the subsequence {f nj } is arbitrary, we get the result. Question 6: Let X be a first-countable topological space, x be a point of X and {O α } α=1 be a neighborhood basis for x. (i) For each n N, let U n = n α=1 O α, and let x n be any point in U n. Show that {x n } converges to x. Let V be an open set that contains x. Since {O α } is a neighborhood basis for x there exists α N such that O α V. In particular, for each m α we have that U m V. Hence there are at most m points in the sequence {x n } that are not contained in V. (ii) Let {y i } be a sequence such that for each n N and α N there is an i > n such that y i O α. Show that there exists a subsequence {y ij } of {y i } that converges to x. We define y i1 to be any point y s of the sequence that is contained in O 1, where O 1 is the first neighborhood of the basis {O α }. Now, suppose that the point y im was defined. We define y im as follows: let U m = m α=1 O α. Since U m is an open set that contains x, there is a neighborhood O αm of x that is contained in U m. By hypothesis, there is a point y s with s > i m 1 4

5 such that y s O αm. We choose y im = y s. Then the exercise is concluded by part (i). Question 7: Let (R, T ) be the real line equipped with the topology described in Question 1. Show that (R, T ) is not first countable. Choose a point of R, say 0, and suppose that there is a countable neighborhood basis {O α } for 0. We will construct an open set U such that every neighborhood {O α } is not contained in U. We know that every neighborhood {O α } has the form R F α, where F α is a finite set. Then F = α=1 F α is a countable union of finite sets, thus is countable. Since R is uncountable, there is a point x (indeed, uncountably many) that is not contained in F. Thus U = R {x} is an open set that does not contain any O α, for α N. 5