Set Known As the Cantor Set. Consider the Clos

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Set Known As the Cantor Set. Consider the Clos COUNTABLE PRODUCTS ELENA GUREVICH Abstract. In this paper, we extend our study to countably infinite products of topological spaces. 1. The Cantor Set Let us constract a very curios (but usefull) set known as the Cantor Set. Consider the 1 2 closed unit interval [0; 1] and delete from it the open interval ( 3 ; 3 ) and denote the remaining closed set by 1 2 G = [0; ] [ [ ; 1] 1 3 3 1 2 7 8 Next, delete from G1 the open intervals ( 9 ; 9 ) and ( 9 ; 9 ), and denote the remaining closed set by 1 2 1 2 7 8 G = [0; ] [ [ ; ] [ [ ; ] [ [ ; 1] 2 9 9 3 3 9 9 If we continue in this way, at each stage deleting the open middle third of each closed interval remaining from the previous stage we obtain a descending sequence of closed sets G1 ⊃ G2 ⊃ G3 ⊃ · · · ⊃ Gn ⊃ ::: T1 The Cantor Set G is defined by G = n=1 Gn, and being the intersection of closed sets, is a closed subset of [0; 1]. As [0; 1] is compact, the Cantor Space (G; τ) ,(that is, G with the subspace topology), is compact. It is useful to represent the Cantor Set in terms of real numbers written to basis 3, that is, 5 ternaries. In the ternary system, 76 81 would be written as 2211:0012, since this represents 2 · 33 + 2 · 32 + 1 · 31 + 1 · 30 + 0 · 3−1 + 0 · 3−2 + 1 · 3−3 + 2 · 3−4 So a number x 2 [0; 1] is represented by the base 3 number a1a2a3 : : : an ::: where 1 X an x = a 2 f0; 1; 2g 8n 2 3n n N n=1 Turning again to the Cantor Set G, it should be clear that an element of [0; 1] is in G if 1 5 and only if it can be written in ternary form with an 6= 1 8n 2 N, so 2 2= G 81 2= G but 1 1 1 3 2 G and 1 2 G. [we denote 2 = 0:1111 ::: , 3 = 0:02222 ::: 1 = 0:2222 ::: ] Thus we have a function f from the Cantor Set into the set of all sequences of the form < a1; a2; a3; : : : ; an; ··· >, where each ai 2 f0; 2g and f is one-to-one and onto. 2. The Product Topology Definition 2.1. Let (X1; τ1); (X2; τ2);:::; (Xn; τn);::: be a countably infinite family of Q1 topological spaces. Then the product i=1 Xi of the sets Xi, i 2 N consists of all the infinite sequences < x1; x2; x3; : : : ; xn; ··· > where xi 2 Xi for all i. The product space Date: May 6, 2011. 1 2 Q1 Q1 i=1(Xi; τi) consists of the product i=1 Xi with the topology τ having as its basis the family 1 Y 0 B = f Oi : Oi 2 τi; and Oi = Xi for all but a finite number of i sg i=1 0 Proposition 2.2. Let (Xi; τi), (Yi; τi ) i 2 N, be countably infinite families of topological Q1 Q1 0 spaces, having product spaces ( i=1 Xi; τ)( i=1 Yi; τ ) respectively. If the mapping hi : 0 Q1 (Xi; τi) ! (Yi; τi ) is continuous for each i 2 N then so is the mapping - h :( i=1 Xi; τ) ! Q1 0 ( i=1 Yi; τ ) given by h(< x1; x2; x3; : : : ; xn; ··· >) =< h1(x1); h2(x2); h3(x3); : : : ; hn(xn); ··· > Q1 −1 Proof. It suffices to show that if O is a basic open set in ( i=1 Xi; τ) then h (O) is open Q1 0 0 in ( i=1 Yi; τ ). Consider the basic open set U1 × U2 × · · · × Un × Yn+1 × ::: where Ui 2 τi for i = 1; 2; : : : ; n. Then −1 −1 −1 −1 h (U1 × U2 × · · · × Un × Yn+1 × ::: ) = h (U1) × h (U2) × · · · × h (Un) × Xn+1 × ::: −1 since the continuity of each hi implies that hi (ui) 2 τi for i = 1; 2; : : : ; n. So h is continu- ous. 3. The Cantor space and the Hilbert cube Proposition 3.1. Let (G; τ) be the Cantor space and for each i 2 N let (Ai; τi) be the Q1 set 0; 2 with the discrete topology, and let ( i=1 Ai; τ) be it's product space. Then the map 0 Q1 P1 an f :(G; τ) ! ( i=1 Ai; τ ) given by f( i=1 3n ) =< a1; a2; a3; : : : ; an; ··· > is a homeomor- phism. Proof. It is easy to see that f is one-to-one and onto. To prove that f is continuous it suffices to show for any basic given set U = U1 × U2 × · · · × Un × An+1 × ::: and any P1 an point a =< a1; a2; a3; : : : ; an; ··· >2 U there exists an open set W 3 i=1 3n such that P1 an 1 P1 an 1 f(W ) ⊆ U. Consider the open interval ( i=1 3n − 3N+2 ; i=1 3n + 3N+2 ) and let W be the P xn intersection of this open intervals with G. Then W is open in (G; τ) and if x = 3n 2 W then xi = ai for i = 1; 2;:::;N. So f(x) 2 U1 × U2 × · · · × Un × An+1 × ::: and thus f(W ) ⊆ U as required. Proposition 3.2. Let (Gi; τi), i 2 N, be a countably infinite family of topological spaces Q1 each of which is homeomorphic to the Cantor space (G; τ) then (G; τ) ' i=1(Gi; τi) ' Qn i=1(Gi; τi) for each n 2 N. Proof. First let us verify that (G; τ) ' (G1; τ1)×(G2; τ2). This is, by proposition 3.1 equiva- Q1 Q1 Q1 lent to showing that i=1(Ai; τi) ' i=1(Ai; τi)× i=1(Ai; τi) where each (Ai; τi) is the set Q1 Q1 0; 2 with the discrete topology. Now we define a function Θ : i=1(Ai; τi) × i=1(Ai; τi) ! Q1 i=1(Ai; τi) by Θ(< a1; a2; a3; ··· >; < b1; b2; b3; ··· >) =< a1; b1; a2; b2; ··· >. It's easy to show that Θ is a homeomorphism, and so (G1; τ1) × (G2; τ2 ' (G; τ). By induction we get Q1 that (G; τ) ' i=1(Ai; τi) for every positive integer n. To show the infinite product case, define the mapping 1 1 1 Y Y Y Φ:[ (Ai; τi) × (Ai; τi) × ::: ] ! (Ai; τi) i=1 i=1 i=1 by Φ(< a1; a2; a3; ··· >; < b1; b2; b3; ··· >; < c1; c2; c3; ··· >; < d1; d2; d3; ··· >; < e1; e2; e3; ··· > ) =< a1; a2; b1; a3; b2; c1; a4; b3; c2; d1; a5; b4; c3; d2; e1; ··· > Again it is easily verified that Φ is homeomorphism and the proof is complete. 3 Proposition 3.3. The topological space [0; 1] is a continuous image of the Cantor space (G; τ). Q1 Proof. It suffices to find a continuous mapping Φ of i=1(Ai; τi) onto [0; 1]. Such a mapping is given by 1 X ai Φ(< a ; a ; : : : ; a ; ··· >) = 1 2 i 2i+1 i=1 It is easy to show that Φ is onto and continuous, and the proof is complete. Definition 3.4. For each positive integer n, let the topological space (In; τn) be homeomor- Q1 phic to [0; 1]. Then the product space n=1(In; τn) is called Hilbert cube and is denoted 1 Qn n by I . The product space i=1(Ii; τi) is called the n-cube, and is denoted by I , for each n 2 N Theorem 3.5. The Hilbert cube is compact. 0 Proof. By proposition 3.3 there is a continuous mapping Φn of (Gn; τn) onto (In; τn where 0 for each n 2 N,(Gn; τn) and (In; τn are homeomorphic to the Cantor space and [0; 1] respec- Q1 Q1 0 1 tively. Therefore, there is a continuous mapping Ψ of n=1(Gn; τn) onto n=1(In; τn = I Q1 1 but i=1(Gn; τn) is homeomorphic to the Cantor space (G; τ). Therefore I is a continuous image of the compact space (G; τ) and hance is compact. Proposition 3.6. Let (Xi; τi) i 2 N be a countably infinite family of matrizable spaces, Q1 then i=1(Xi; τi) is metrizable. Corollary 3.7. The Hilbert cube is metrizable. 4. The Cantor space and the Hilbert cube Definition 4.1. A topological space (X; τ) is said to be separable if it has a countable dense subset. Proposition 4.2. Let (X; τ) be a compact metrizable space. Then (X; τ) is separable. Proof. Let d be a metric on X which induces the topology jtau. For each positive integer n, 1 let Sn be the family of all open balls having centers in X and radius n .Then Sn is an open covering of X and so there is a finite sub covering Un = fUn1 ;Un2 ;:::;Unk g for some k 2 N. 1 Let ynj be the center of Un, j = 1; 2; : : : ; k and YN = fyn1 ; : : : ; ynk g. Put Y = [n=1Yn, then Y is a countable subset of X. Now, if V is any non open set in (X; τ), then for any v 2 V , 1 V contains an open ball B of radius n . As Un is an open cover of X, v 2 Unj for some j. 1 Thus d(v; ynj ) < n , and so ynj 2 B ⊂ V . Hance V \ Y 6= ? and so Y is dense in X. Corollary 4.3. The Hilbert cube is separable space. Definition 4.4. A topological space (X; τ) is said to be a T1space if every singelton set fxg, x 2 X, is a closed set.
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