Math 5801 General Topology and Knot Theory
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1 Lecture 13-9/21/2012 Math 5801 Ohio State University September 21, 2012 Course Info Reading for Monday, September 24 Review Chapter 2.19 HW 5 for Monday, September 24 Chapter 2.17: 3, 5, 9, 13 Chapter 2.18: 2, 5, 8a-b, 10 Midterm 1 Friday, September 28 Munkres Chapters ZFC proofs (I ll supply you with all of the axioms)
2 Definition 125 (Homeomorphism) Let X and Y be topological spaces. A function f : X Y is a homeomorphism if f : X Y is a bijection. f : X Y is a continuous. f 1 : Y X is continuous. Examples 126 (Homeomorphisms) The following are homeomorphisms: 1. f : R R where f (x) = x 3. f is continuous by Prop. 124(2) since f = idr idr idr. f is injective since f > 0 for x 0 (hence f increasing). f is surjective by IVT since limx f (x) = and limx f (x) = f 1 is continuous since f ((a, b)) = (a 3, b 3 ). 2. g : ( 1, 1) R where g(x) = x 1 x 2 Let h : ( 1, 1) R where h(x) = x. h = idr ( 1,1) is cont. by Prop. 119(4) (Restr. of domain) 1 h 2 is cont. by Prop. 124(1-2). And h(x) 0 for x ( 1, 1). Hence g = h 1 h2 is cont. by Prop. 124(3). g (x) = 1+x2 > 0 so g is injective. (1 x 2 ) 2 g surj. by IVT and limx 1 g(x) = and lim x 1 + g(x) = g 1 is continuous since f ((a, b)) = ( a 1 a 2, b 1 b 2 ).
3 Definition 127 (Product of functions) If f : A B and g : C D are functions then let f g : A B C D be the funtion See Chapter 2.18 Problem 10 (f g)(a, c) = (f (a), g(c)) Examples 128 (Continuity Proofs) The following are continuous: 1. f : R R R where f (x, y) = x y. f = π1 π2 is cont. by Prop. 121 (Proj. are cont.) and Prop. 124(2). 2. g : R R R where g(x) = (x, x) g = idr idr which is cont. by Example 116(3) and Prop Definition 129 (n-sphere) The n-sphere S n is the subspace of R n+1 where S n = {(x1, x2,, xn+1) R n+1 x1 2 + x xn+1 2 = 1} Example 130 (Continuous bijection homeomorphism) Let f : [0, 1) S 1 be the function f (t) = (cos 2πt, sin 2πt). Let g : R R R be the function g(t) = (cos 2πt, sin 2πt). g is cont. by Prop g [0,1) : [0, 1) R R is cont. by Prop. 119(4) (Restr. of domain) f is cont. by Prop. 119(5) (Restr. of codomain) Let U [0, 1) be [0, 1 2 ). U open in [0, 1) since U = [0, 1) ( 2, 1 2 ) f 1 not cont. since (0, 0) f (U) but every nbhd of (0, 0) in S 1 intersects S 1 f (U).
4 Definition 131 (Indexed family of sets) Let J be set and A be a set of sets. A family of sets indexed by J is a function f : J A. We write {Aα} instead of f and for α J we write Aα instead of f (α). Definition 132 (Cartesian Product) Let X be a set. A J-tuple of X is a function x : J X. Let {Aα} be a family of sets indexed by the set J. The cartesian product of the family {Aα} is Aα = {x x is a J-tuple of Aα s.t. x(α) Aα} Definition 133 (The product topology) Let {Aα} be a family of topological spaces indexed by the set J and let πα : Aα Aα be the αth projection function. The product topology on Aα has subbasis S = {πα 1 (Vα) α J and Vα open in Aα}. Notice that the product top. on Aα makes each projection function πα cont. The product topology is the coarsest top. on Aα where each πα is cont. Basis for prod. top. is B = {π 1 (Vα1) π 1 (Vαn)} αn α1
5 Proposition 134 (Basis for the product topology) Let {Aα} be a family of topological spaces indexed by the set J and let πα : Aα be the αth projection function. A basis for the product topology on Aα is { } B = Vα open in Aα and Vα = Aα Vα. for all but fin. many α Proof. Let S be as in definition. Finite intersections of elts. of S are exactly elts. of B. Example 135 Let A be {0, 1} with the discrete topology. Consider A ω with the product topology. Does have the A ω the discrete topology? NO! Only open sets are of the form Bn where all but fin. many n Z+ Bn = {0, 1}.
6 Proposition 136 Let {Aα} be a family of topological spaces indexed by the set J and let X be a space. Suppose we have a functions f : X Aα and functions fα : X Aα such that for all x X we have f (x) = (fα(x)). Then f is cont. iff each fα is cont. Proof. If f is cont. then fa = πα f is cont. Suppose each fα is cont. Then f 1 (π 1 α (Vα)) = fα 1 (Vα) is open. Much less important is: Definition 137 (The box topology) Let {Aα} be a family of topological spaces indexed by the set J and let πα : Aα Aα be the αth projection function. The box topology on Aα has basis { } B box = Vα Vα open in Aα Notice that the box topology is finer than the product topology. For finite products of top. spaces they agree.
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