MAT3500/ Mandatory assignment 2013 Solutions

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1 MAT3500/ Mandatory assignment 2013 s Problem 1 Let X be a topological space, A and B be subsets of X. Recall the definition of the boundary Bd A of a set A. Prove that Bd (A B) (Bd A) (Bd B). Discuss how Bd (A B) relates to Bd A and Bd B. Justify your conclusion. Since A B = A B and int A int B int (A B) we have that Bd A Bd B = (A\int A) (B \int B) A B \int (A B) = Bd (A B). We have that Bd (A B) = Bd(X \ (A B)) = Bd ((X \ A) (X \ B)) Bd (X \ A) Bd (X \ B) = Bd A Bd B. Problem 2 Let X be a topological space that is T 1 but not Hausdorff. Prove that X is infinite. Is it possible to prove that X must be uncountable? Hint: Try to give meaning to the two-point compactification of N. If X is T 1 every finite set is closed, and if X in addition is finite, every subset of X will be open. Then X has the discrete topology and will be Hausdorff. The contrapositive formulation of this is that if X is T 1, but not Hausdorff, then X must be infinite. There is a countable space that is T 1, but not Hausdorff, for instance N with the co-finite topology, so we cannot prove that a space like this must be uncountable. This example is better than the one suggested in the hint. 1

2 Problem 3 Let X be a compact topological space, Y a space with a topology induced by the metric d and let Z be the set of continuous functions f : X Y. If f and g are in Z, we let d (f, g) = max{d(f(x), g(x)) x X}. Let X = Y = [0, 1] with the standard topology and metric, and let Z be as above with the topology induced from d. Show that Z is not compact. Discuss what we may learn from this example. We may either show that Z is not sequentially compact or show that Z is not totally bounded. If we can prove that there is a sequence {f n } n N of functions with mutual distance 1, we have achieved both. There are many ways to do this, so what we give is just one example. Let f n be constant 0 up to 1, then grow, along a stright line, to the value n+1 1 at 1 and then be constant 1 for the rest of [0, 1]. n The lesson to learn is that the class of compact metric spaces is not closed under exponentiation with the uniform metric on function spaces. Problem 4 Let X be a topological space, X the set of connectedness components of X. Let p : X X map an object to its component, and let X have the quotient topology. a) Show that X has an isolated point if and only if X has a connectedness component that is both closed and open. b) Show that the components of X will consist of single points, and that X will be a T 1 -space. a) If A is a connected component in X, then A is closed, A X and p 1 ({A}) = A. Since p is a quotient map, we have that {A} is closed in X for each component A and that {A} is open in X if and only if A is open in X. {A} open in X means that A is isolated, and A is open in X means that A is a component that is both closed and open. 2

3 b) We actually saw in a) that each singleton in X is closed, so X is a T 1 -space by definition. Now assume that Y is a connected subset of X. We will prove that p 1 (Y ) is connected in X, so p 1 (Y ) will be one component, and thus Y will be a singleton. If A, B is a separation of p 1 (Y ), we cannot have that A and B separates any of the components that are subsets of p 1 (Y ), so both A and B will be saturated with respect to p. It follows that p(a) and p(b) will be a separation of Y, contradicting the assumption. Problem 5 Let X and Y be nonempty topological spaces, Z the set of continuous functions from X to Y. There are several useful ways to define a topology on Z, one of them is the compact open topology. This topology is generated from the sub-basis consisting of all sets {f Z f(c) U} where C X is a compact and U Y is open. a) Show that the compact open topology is finer than the topology on Z induced from the product topology on Y. x X Explain why Z with the compact open topology will be Hausdorff whenever Y is Hausdorff. b) Prove that if C X is compact and D Z is compact in the compact open topology, then E = f(c) is compact in Y. a) Since a sub-basis element in the product topology will be of the form f D B x,u = {f f(x) U} 3

4 where x X and U is open in Y, and since {x} is a compact for each x X, we have that each sub-basis element in the product topology also is a sub-basis element of the compact open topology. Thus the compact open topology is finer than the product topology. We know that an arbitrary product of Hausdorff spaces is a Hausdorf space, so Z with the product topology is Hausdorff when Y is Hausdorff. Then all finer topologies are Hausdorff, and in particular, the compact open topology. b) Let U be a family of open sets that covers E. For each f D, f(c) E is compact, so there is a finite set U f U that covers f(c). Let U f = U f. Let B f = {g Z g(c) U f.} Then B f is open in the compact open topology, and {B f f D} will be an open covering of D. Since D is compact, there is a finite subcovering B f1,..., B fn of D. Then n will be a finite covering of E. This shows that E is compact. i=1 U fi Problem 6 Let X 0 = { 1, 0, 1} with the discrete topology. Let X = X0 N = n N X 0 with the product topology and let p : X [ 1, 1] be defined by p(f) = 2 n f(n). n=1 a) Show that p is surjective and continuous. b) Show that X is compact, and use this to show that p is a quotient map. 4

5 The elements of X 0 are known as negative binary digits and p is known as the negative binary digit representation of the reals in [ 1, 1]. The important property is that we for every continuous φ : [ 1, 1] [ 1, 1] have a continuous ˆφ : X X such that φ p = p ˆφ You are not supposed to prove this. We call ˆφ a lifting of φ to the space of digitalized representations. a) Each x [0, 1] can be written on a binary form 0.a 1 a 2 where each a i is either 0 or 1. If we let f(i) = a i we have that x = p(f). If x [ 1, 0) we write x on binary form 0.a 1 a 2 and let f(i) = a i. Then, again, we have that p(f) = x. Thus p is surjective. In order to prove that p is continuous, let f X, x = p(f) and let ɛ > 0 be given. Choose n 0 so large that 2 (n 0 1) < ɛ. For all g X such that g(1) = f(1),..., g(n 0 ) = f(n 0 ) we have that p(f) p(g) = 2 n f(n) n=1 2 n g(n) 2 2 n = 2 (n0 1). n>n 0 n=1 Since the set of such g is an open neighbourhood of f, this proves that p is continuous. b) We can always use Tychonoff s theorem to observe that X is compact, but let us give a more direct argument: - X is metrizable since X is a countable product of metrizable spaces. - It is sufficient to prove that X is sequentially compact. - Let {f n } n N be any sequence from X. - There is an a 1 X 0 such that is infinite. A 1 = {n N f(1) = a 1 } 5

6 - Then there is an a 2 X 0 such that is infinite A 2 = {n A 1 f n (2) = a 2 } - By recursion we may find a decreasing set of infinite sets A k and a k X 0 such that A k = {n A k 1 f n (k) = a k }. Then f(k) = a k will be a limit point of a convergent subsequence. Since X is a compact Hausdorff space and [ 1, 1] is a Hausdorff space. p will be a closed map, and thus a quotient map. An even easier argument for the compactness of X is as follows: - X is homeomorphic to the set of reals x in [0,1] that only use the digits 1,2 and 3 in the decimal expansion. - This set is a closed subset of a compact set. - Thus X must be compact. END SOLUTION 6