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.