Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Manifolds & Topology, MAT3009/3035 Test 1, Monday February 27 2012 Question 1: a) Give the definition of a homeomorphism. [2] b) Find an explicit homeomorphism between the following spaces (all with standard topology), or give an argument why there cannot be any: (i) (−1, 1); (ii) R; (iii) ((−1, 1) × {0}) ∪ ({0} × (0, 1)). [4] c) Let S = {z ∈ C : |z| = 1} be the unit circle and let f : [0, 1) → S be defined by f (x) = e2πix . Is f a homeomorphism? Justify your answer. 1 1 [4] Question 2: Let (X, O) be a topological space and A ⊂ X. a) Give the definition of a boundary point. [2] b) Assume now that X = (0, 1] with Euclidean topology. Take A = (0, 13 ) ∪ ( 23 , 1]. Identify ∂A. [2] c) Same question as b), but now when X has co-finite topology. [3] d) [for BSc] Is X with co-finite topology Hausdorff? Justify your answer. [4] e) [for MMath] Is X with co-finite topology separable? Justify your answer. [4] Question 3: a) Give the definition of a (coverings-)compact set. [2] b) Show that in a metric space, an open ball is not compact. [3] c) [for BSc] Prove that in a Hausdorff space, every singleton (i.e., a set consisting of one point) is closed. [4] d) [for MMath] Is the Sorgenfrey line R metrizable? Justify your answer. [4] Total [30] 1 Solutions Test 1 (Manifolds & Topology) Question 1: a) A homeomorphism is a continuous map with a continuous inverse. (In particular, a homeomorphism must be a bijection. x b) f : (−1, 1) → R, f (x) = 1−x 2 is a homeomorphism between the first two. [0, 1] is homeomorphic to neither : (−1, 1) nor R, because the first has a branch point which (when taken out) divides the set in three components. No such point exists is in (−1, 1) or R. c) f is a bijection, and continous but not a hoemomorphism, because the inverse is not continuous. Indeed, [0, 12 ) is open in [0, 1), but its image f ([0, 21 )), containing boundary point 1, is not open in S1 . Question 2: a) x is a boundary point of A if every neighbourhood of x intersects both A and Ac . b) ∂A = { 31 , 23 }. c) ∂A = (0, 1]. d) X is not Hausdorff. For example, if x = 13 and y = 23 have neighbourhoods Ux and Uy respectively, then #(X \Ux ) < ∞ and #(X \Uy ) < ∞. Therefore Ux ∩Uy ⊃ X \((X \Ux )∪(X \Uy )) 6= ∅, because X is an infinite set. e) X is separable. Every infinite set is dense in X, in particular sets that are countably infinite. Question 3: a) A set A in a topological space is compact if every open open of A has a finite subcover. b) n ) : n ∈ N} is an open cover without finite subcover, Let B(x; r) be an open ball, then {B(x; r n+1 because if I only take the first N such balls, then I cover B(x; r NN+1 ) but not the entire ball B(x; r). Alternatively, in a Hausdorff space (so also in a metric space), every compact set is closed. If r is small enough, B(x; r) is not closed as it won’t contain all its bounadary points. Therefore B(x; r) is cannot be compact. c) For every y 6= x, let Uy 3 x and Vy 3 y be disjoint open sets. Then ∪y6=x Vy is open and {x} = (∪y6=x Vy )c is closed. Alternatively, for every y 6= x, there is an open neighbourhood Vx disjoint from {x}, so y is interior to {x}c . That means, {x}c contains only interior points, so it is open, and therefore {x} is closed. d) The Sorgenfrey line is separable (Q is a countable dense subset), but not second countable (you need an uncountable basis of the topology to obtain all open sets of the form [x, ∞). Separable metric spaces are second countable, so the Sorgenfrey cannot have a metric. General remarks: • In definitions, get the quantifiers right and in the right order! • In theorems (and hence proofs), get the direction of the implication right! • Some linguistics: An open set U can have boundary points, only none of these boundary points belongs to A. So A doesn’t contain boundary points. A intersects B means A ∩ B 6= ∅. A is in B means A ⊂ B. 2 • To use a ball B(a; r) requires a metric, and hence doesn’t make sense in a topological space without metric. In a topological space, an open set containing a is a neighbourhood of a. If you have a metric, the ball B(a; r) is an example of a neighbourhood of a, but a has many other neighbourhoods as well. • Also the notion of bounded requires a metric. • If A is not closed, that does not mean it is open. For example, in the Euclidean line, [0, 1), Q and { n1 : n ∈ N} are examples of sets that are neither open nor closed. 3