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§20, 21 Metric Topology

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§20, 21 Metric Topology Recall Product topology Xl topological spaces Hl Î LL ÛlÎL Xl = 8HxlLlÎL : xl Î Xl< pm : ÛlÎL Xl ® Xm xm : HxlLlÎL ® xm Product topology on ÛXl is the coarsest topology such that all the pm ' s are continuous. we saw the basis ÛlÎL Ul where Ul Ì Xl open " l and Xl = Ul for l outside for some finite subset of L Y ÛlÎL Xl continuous fl is continuous for all l y Ì H flHyLL. Box topology : Basis ÛlÎL Ul, Ul Ì Xl open " l * finer than product topology §20, 21 Metric Topology Recall Metric space: HX, dL, X set, d : X X ® R³0 (i) d Hx, yL = 0 x = y (ii) d Hx, yL = dHy, xL (iii) d Hx, zL £ d Hx, yL + dHy, zL – triangular inequality Open ball (or e–ball) Be, dHxL = 8y Î X : dHx, yL < e< or just BeHxL Definition The metric topology is the topology on X generated by the basis 8Be, dHxL : e > 0, x Î X< Check basis: (i) union in X : clear Î H L Ý H L (ii) x Be1 x1 Be2 x2 H L Ì H L Ì H L £ He - H L e - H LL Br x Be1 x1 Be2 x2 , provided r min 1 d x1, x2 , 2 d x1, x2 . This follows by triangular inequality. Theorem 23 Let HX, dL be metric space U Ì X open " x Î U $ e > 0 such that BeHxL Ì U Proof: “Ü” trivial. “Þ” By definition, $ e > 0, y Î X such that x Î BeHyL Ì U. But BrHxL Ì BeHyL provided r £ e - dHx, yL. Examples n n 2 (1) X = R dHx, yL = Úi=1 Hxi - yiL (Euclidean metric) metric topology = standard topology (2) X arbitrary set Printed by Mathematica for Students 1 if x ¹ y dHx, yL = : 0 if x = y metric topology = discrete topology If X¤ > 1, ± d : metric s.t. metric topology of HX, dL is the trivial topology. Why ? Lemma 24 Any metric topology is T2. 1 Proof: If x ¹ y, then BeHxL, BeHyL disjoint nbds provided e £ dHx, yL. 2 Definition A topological space HX, tL is metrizable if $ metric d on X s.t. metric topology of HX, dL equals t. Other basic properties of the metric topology. (1) X, Y metric spaces. f : X ® Y in continuous for metric topology continuous in e–d sense. Has in lecture 1L (2) If Y Ì X subset of a metric space HX, dL, then the two natural topologies on Y coincide. - subspace topology in metric topology on X. - metric topology of HY, d¤YY L This justifies why S2 \ 8N< ® R2 continuous where S2 \ 8N< has subspace topology in R3. Ha, b, cL I a , b M Ì 1-c 1-c (3) If HXi, d iL for 1 £ i £ n metric spaces, then the product topology on ÛXi is the metric topology of HÛXi, dL, where n n dHHxiLi=1, HxiLi=1L = maxHdiHxi, yiLL. n \ the product topology on R is the metric topology of the metric dHx, yL = max1£i£n 8 xi - yi¤< Now we can see in a nicer way that product topology = standard topology on Rn. Thereom 25 Suppose d, d¢ are metrics on a set X inducing metric topology t, t¢. Then t Ì t¢ " x Î X, " e > 0 $ d > 0 such that Bd, d¢ HxL Ì Be, dHxL. Proof: Be, dHxL x U ¢ Let U Ì X open in t. We need U open in t . U open in t Þ $ e > 0 such that Be, dHxL Ì U. By assumption, $ d > 0, such that Bd, d¢ HxL Ì Be, dHxL. Examples 2 1) X = R , d = Euclidean metric, dHHx1, x2L, Hy1, y2LL = maxH x1 - y1¤, x2 - y2¤L n 2 2) X = R , dHx, yL = ÚHxi - yiL (standard topology), dHx, yL = max 8 xi - yi¤< (product topology) We want to apply theorem 25 to see these are the same topology dHx, yL £ dHx, yL £ n dHx, yL Þ B Ì Be, d Ì Be, d e n , d Theorem 26 Let HX, dL metric space then there exist a metric d¢ on X that induces the same topology as d. s.t. d¢Hx, yL £ 1. " x, y. Examples d¢Hx, yL = minH1, dHx, yLL = dHx, yL dHx, yL d¢Hx, yL = 1+dHx, yL Proof: We need to show that d and d induce the same topology. d is a metric: clearly 0 £ d £ 1. (i) dHx, yL = 0 Þ dHx, yL = 0 Þ x = y (ii) dHx, yL = dHx, yL (iii) triangle inequailty, (we want minH1, dHx, zLL £ minH1, dHx, yLL + minH1, dHy, zLL) This is true since if dHx, yL ³ 1, then LHS £ 1 £ dHx, yL £ RHS, for dHy, zL ³ 1 is similarly done. If dHx, yL < 1 and dHy, zL < 1. We need min H1, d Hx, zLL £ d Hx, yL + dHy, zL. True b/c LHS £ d Hx, zL £ RHS. Hence d is metric. Now we will show that d, d induce the same topology. This is true b/c H L = H L " e £ Be, d x Be, d x , 1 and by Thm 25. Infinite Products HXl, dlL metric spaces Hl Î LL. In general, ÛlÎL Xl not metrizable Hcounterexample in §21L. But can at least define a natrual metric on it. rHHxlLlÎL, HylLlÎLL = sup 9dlHxl, ylL¥ l Î L= : A real number in @0, 1D b/c 0 £ dl £ 1, " l. “uniform metric” L Just consider the case Xl = R, " l Þ ÛXl = R , dlHx, yL = minH1, x - y¤L, " l. Check this is a metric (exercise). Theorem 27 On RL, this metric topology of r Huniform topologyL is coarser than the box topology and finer than the product topology. Proof: Compare with box topology: Fix U open in the uniform topology, pick x Î U. U is open Þ $ e > 0 s.t. Be, rHxL Ì U. L Be, rHxL = 9y Î R : sup dHxl, ylL < e=. x Be, rHxL U e e If dHxl, ylL < " l Þ sup dHxl, ylL £ < e y Î ÛB HxlL (open set in box topology.) 2 2 e2, d proof continued next lecture. Recall Product topology Xl topological spaces Hl Î LL ÛlÎL Xl = 8HxlLlÎL : xl Î Xl< pm : ÛlÎL Xl ® Xm xm : HxlLlÎL ® xm Product topology on ÛXl is the coarsest topology such that all the pm ' s are continuous. we saw the basis ÛlÎL Ul where Ul Ì Xl open " l and Xl = Ul for l outside for some finite subset of L Y ÛlÎL Xl continuous fl is continuous for all l y Ì H flHyLL. Box topology : Basis ÛlÎL Ul, Ul Ì Xl open " l * finer than product topology §20, 21 Metric Topology Recall Metric space: HX, dL, X set, d : X X ® R³0 (i) d Hx, yL = 0 x = y (ii) d Hx, yL = dHy, xL (iii) d Hx, zL £ d Hx, yL + dHy, zL – triangular inequality Open ball (or e–ball) Be, dHxL = 8y Î X : dHx, yL < e< or just BeHxL Definition The metric topology is the topology on X generated by the basis 8Be, dHxL : e > 0, x Î X< Check basis: (i) union in X : clear Î H L Ý H L (ii) x Be1 x1 Be2 x2 H L Ì H L Ì H L £ He - H L e - H LL Br x Be1 x1 Be2 x2 , provided r min 1 d x1, x2 , 2 d x1, x2 . This follows by triangular inequality. Theorem 23 Let HX, dL be metric space U Ì X open " x Î U $ e > 0 such that BeHxL Ì U Proof: “Ü” trivial. “Þ” By definition, $ e > 0, y Î X such that x Î BeHyL Ì U. But BrHxL Ì BeHyL provided r £ e - dHx, yL. 2 MAT327-lecturenotes09.nb Examples n n 2 (1) X = R dHx, yL = Úi=1 Hxi - yiL (Euclidean metric) metric topology = standard topology (2) X arbitrary set 1 if x ¹ y dHx, yL = : 0 if x = y metric topology = discrete topology If X¤ > 1, ± d : metric s.t. metric topology of HX, dL is the trivial topology. Why ? Lemma 24 Any metric topology is T2. 1 Proof: If x ¹ y, then BeHxL, BeHyL disjoint nbds provided e £ dHx, yL. 2 Definition A topological space HX, tL is metrizable if $ metric d on X s.t. metric topology of HX, dL equals t. Other basic properties of the metric topology. (1) X, Y metric spaces. f : X ® Y in continuous for metric topology continuous in e–d sense. Has in lecture 1L (2) If Y Ì X subset of a metric space HX, dL, then the two natural topologies on Y coincide. - subspace topology in metric topology on X. - metric topology of HY, d¤YY L This justifies why S2 \ 8N< ® R2 continuous where S2 \ 8N< has subspace topology in R3. Ha, b, cL I a , b M Ì 1-c 1-c (3) If HXi, d iL for 1 £ i £ n metric spaces, then the product topology on ÛXi is the metric topology of HÛXi, dL, where n n dHHxiLi=1, HxiLi=1L = maxHdiHxi, yiLL. n \ the product topology on R is the metric topology of the metric dHx, yL = max1£i£n 8 xi - yi¤< Now we can see in a nicer way that product topology = standard topology on Rn. Thereom 25 Suppose d, d¢ are metrics on a set X inducing metric topology t, t¢. Then t Ì t¢ " x Î X, " e > 0 $ d > 0 such that Bd, d¢ HxL Ì Be, dHxL. Proof: Be, dHxL x U ¢ Let U Ì X open in t.
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