COUNTABLE PRODUCTS ELENA GUREVICH

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1 COUNTABLE PRODUCTS ELENA GUREVICH Abstract. In this paper, we extend our study to countably infinite products of topological spaces.. The Cantor Set Let us constract a very curios (but usefull) set known as the Cantor Set. Consider the closed unit interval [0, ] and delete from it the open interval (, 2 ) and denote the remaining closed set by G = [0, ] [2, ] Next, delete from G the open intervals ( 9, 2 9 ) and ( 7 9, 8 9 ), and denote the remaining closed set by G 2 = [0, 9 ] [2 9, ] [2, 7 9 ] [8 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 G G 2 G G n... The Cantor Set G is defined by G = n= G n, and being the intersection of closed sets, is a closed subset of [0, ]. As [0, ] 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, that is, ternaries. In the ternary system, would be written as , since this represents So a number x [0, ] is represented by the base number a a 2 a... a n... where x = n= a n n a n {0,, 2} n N Turning again to the Cantor Set G, it should be clear that an element of [0, ] is in G if and only if it can be written in ternary form with a n n N, so 2 / G 5 8 / G but G and G. [we denote 2 = 0...., = = ] Thus we have a function f from the Cantor Set into the set of all sequences of the form < a, a 2, a,..., a n, >, where each a i {0, 2} and f is one-to-one and onto. 2. The Product Topology Definition 2.. Let (X, τ ), (X 2, τ 2 ),..., (X n, τ n ),... be a countably infinite family of topological spaces. Then the product X i of the sets X i, i N consists of all the infinite sequences < x, x 2, x,..., x n, > where x i X i for all i. The product space Date: May 6, 20.

2 2 (X i, τ i ) consists of the product X i with the topology τ having as its basis the family B = { O i : O i τ i, and O i = X i for all but a finite number of i s} Proposition 2.2. Let (X i, τ i ), (Y i, τ i ) i N, be countably infinite families of topological spaces, having product spaces ( X i, τ) ( Y i, τ ) respectively. If the mapping h i : (X i, τ i ) (Y i, τ i ) is continuous for each i N then so is the mapping - h : ( X i, τ) ( Y i, τ ) given by h(< x, x 2, x,..., x n, >) =< h (x ), h 2 (x 2 ), h (x ),..., h n (x n ), > Proof. It suffices to show that if O is a basic open set in ( X i, τ) then h (O) is open in ( Y i, τ ). Consider the basic open set U U 2 U n Y n+... where U i τ i for i =, 2,..., n. Then h (U U 2 U n Y n+... ) = h (U ) h (U 2 ) h (U n ) X n+... since the continuity of each h i implies that h i (u i ) τ i for i =, 2,..., n. So h is continuous.. The Cantor space and the Hilbert cube Proposition.. Let (G, τ) be the Cantor space and for each i N let (A i, τ i ) be the set 0, 2 with the discrete topology, and let ( A i, τ) be it s product space. Then the map f : (G, τ) ( A i, τ ) given by f( an ) =< a n, a 2, a,..., a n, > is a homeomorphism. 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 = U U 2 U n A n+... and any point a =< a, a 2, a,..., a n, > U there exists an open set W an such that n f(w ) U. Consider the open interval ( an, n N+2 an + ) and let W be the n N+2 intersection of this open intervals with G. Then W is open in (G, τ) and if x = x n n W then x i = a i for i =, 2,..., N. So f(x) U U 2 U n A n+... and thus f(w ) U as required. Proposition.2. Let (G i, τ i ), i N, be a countably infinite family of topological spaces each of which is homeomorphic to the Cantor space (G, τ) then (G, τ) (G i, τ i ) n (G i, τ i ) for each n N. Proof. First let us verify that (G, τ) (G, τ ) (G 2, τ 2 ). This is, by proposition. equivalent to showing that (A i, τ i ) (A i, τ i ) (A i, τ i ) where each (A i, τ i ) is the set 0, 2 with the discrete topology. Now we define a function Θ : (A i, τ i ) (A i, τ i ) (A i, τ i ) by Θ(< a, a 2, a, >, < b, b 2, b, >) =< a, b, a 2, b 2, >. It s easy to show that Θ is a homeomorphism, and so (G, τ ) (G 2, τ 2 (G, τ). By induction we get that (G, τ) (A i, τ i ) for every positive integer n. To show the infinite product case, define the mapping Φ : [ (A i, τ i ) (A i, τ i )... ] (A i, τ i ) by Φ(< a, a 2, a, >, < b, b 2, b, >, < c, c 2, c, >, < d, d 2, d, >, < e, e 2, e, > ) =< a, a 2, b, a, b 2, c, a 4, b, c 2, d, a 5, b 4, c, d 2, e, > Again it is easily verified that Φ is homeomorphism and the proof is complete.

3 Proposition.. The topological space [0, ] is a continuous image of the Cantor space (G, τ). Proof. It suffices to find a continuous mapping Φ of (A i, τ i ) onto [0, ]. Such a mapping is given by a i Φ(< a, a 2,..., a i, >) = 2 i+ It is easy to show that Φ is onto and continuous, and the proof is complete. Definition.4. For each positive integer n, let the topological space (I n, τ n ) be homeomorphic to [0, ]. Then the product space n= (I n, τ n ) is called Hilbert cube and is denoted by I. The product space n (I i, τ i ) is called the n-cube, and is denoted by I n, for each n N Theorem.5. The Hilbert cube is compact. Proof. By proposition. there is a continuous mapping Φ n of (G n, τ n ) onto (I n, τ n where for each n N, (G n, τ n ) and (I n, τ n are homeomorphic to the Cantor space and [0, ] respectively. Therefore, there is a continuous mapping Ψ of n= (G n, τ n ) onto n= (I n, τ n = I but (G n, τ n ) 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.6. Let (X i, τ i ) i N be a countably infinite family of matrizable spaces, then (X i, τ i ) is metrizable. Corollary.7. The Hilbert cube is metrizable. 4. The Cantor space and the Hilbert cube Definition 4.. 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 tau. For each positive integer n, let S n be the family of all open balls having centers in X and radius n.then S n is an open covering of X and so there is a finite sub covering U n = {U n, U n2,..., U nk } for some k N. Let y nj be the center of U n, j =, 2,..., k and Y N = {y n,..., y nk }. Put Y = n= Y n, then Y is a countable subset of X. Now, if V is any non open set in (X, τ), then for any v V, V contains an open ball B of radius n. As U n is an open cover of X, v U nj for some j. Thus d(v, y nj ) < n, and so y n j B V. Hance V Y and so Y is dense in X. Corollary 4.. The Hilbert cube is separable space. Definition 4.4. A topological space (X, τ) is said to be a T space if every singelton set {x}, x X, is a closed set. Theorem 4.5. (Urisohn s Theorem) Every separable metric space (X, d) is homeomorphic to a subspace of the Hilbert cube. Proof. We need to find a countably infinite family of mappings f i : (X, d) [0, ] which are a) continuous, and b) separate points and closed sets. Without loss of generality we can assume that d(a, b) for all a and b in X, since every metric is equivalent to such a metric. As (X, d) is separable, there exists a countable dense subset Y = {y i, i N}. For

4 4 each i N, define f i : X [0, ] by f i (x) = d(x, y i ). It is clear that each f i continuous. To see that the mappings {f i : i N} separate points and closed sets, let x X and A be any closed set not containing x. Now X \ A is an open set, and so contains an open ball B of radius ɛ and center x, for some ɛ > 0. As Y is dense in X, there exists a y n, such that d(x, y n ) < ɛ 2, thus d(y n, a) ɛ 2 for all a A. So [0, ɛ 2 ) is an open set in [0, ] which contains f n (x), but f n (a) / [0, ɛ 2 ), for all a A. This implies f n(a) [ ɛ 2, ]. As the set [ ɛ 2, ] is closed, this implies f n (A) [ ɛ 2, ]. Hance f n(x) / f n (A) and thus the family {f i : i N} separates points and closed spaces. Corollary 4.6. Every compact metrizable space is homeomorphic to a closed subspace of the Hilbert cube. Corollary 4.7. If for each i N, (X i, τ i ) is compact metrizable space, then X i, τ i is compact and metrizable. Definition 4.8. A topological space (X, τ) is said to satisfy a the second axiom of countability (or to be second countable) if there exists a basis B for τ such that B consists of only a countable number of sets. Proposition 4.9. Let (X, d), be a matric space and τ the induced topology. Then (X, τ) is a separable space if and only if it satisfies the second axiom of countability. Proof. Let (X, τ) be separable. Then it has a countable dense subset Y = {y i : i N}. Let B consists of all the balls (in the metric d) with center y i, and radius n, for some positive integer n. Clearly B is countable, and we will show that it is a basis for τ. Let V τ. Then for any v V, V contains an open ball, B, of radius n about r. As Y is dense in X, there exists a y m Y such that d(y m, v) < 2n. Let B be the open ball with center y m and radius 2n. Then the triangle inequality implies B B V Also B B. Hance B is a basis for τ. Conversely, let (X, τ) second countable, having a countable basis B = {B i : i N}. For each B i let b i be any element of B i, and put Z equal to the set of all such b i. Then Z is countable set. Further, if V τ, then B i V for some i, and so b i V. Thus V Z Hance Z is dense in X. Consequently (X, τ) is separable. Theorem 4.0. (Urisohn s theorem and its converse) Let (X, τ) be a topological space. Then (X, τ) is separable and metrizable if and only if it is homeomorphic to a subspace of the Hilbert cube. Proof. If (X, τ) is separable and metrizable then Urisohn s theorem says that it is homeomorphic to a subspace of the Hilbert cube. Conversely, let (X, τ) be homeomorphic to the subspace (Y, τ ) of the Hilbert cube I, I is separable and metrizable, hance it is second countable and metrizable (since any subspace of a second countable and metrizable space is second countable and metrizable) (Y, τ ) is second countable and metrizable. Therefor it is separable. Hance (X, τ) is also separable and metrizable. 5. Peano s Theorem Proposition 5.. Every separable metrizable space (X, τ ) is a continuous image of a subspace of the cantor space (G, τ). Further if (X, τ ) is compact, then the subspace can be chosen to be closed in (G, τ). Proposition 5.2. Let (Y, τ ) be a (non empty) closed subspace of the Cantor space (G, τ). Then there exists a continuous mapping of (G, τ) onto (Y, τ ). Theorem 5.. Every compact metrizable space is a continuous image of the Cantor space.

5 Proposition 5.4. let f be a continuous mapping of a compact metric space (X, d) onto a Hausdorff space (Y, τ ). Then (Y, τ ) is compact and metrizable. Theorem 5.5. (Peano) For each positive integer n, there exists a continuous mapping φ n of [0, ] onto the n-cube I n. Proof. There exists a continuous mapping Φ n of the Cantor space (G, τ) onto the n-cube I n. As (G, τ) is obtained from [0, ] by successively dropping out middle thirds, we extend Φ n to a continuous mapping Ψ n : [0, ] I n by defining Ψ n to be linear on each omitted interval, that is, if (a.b) is one of the open intervals comprising [0, ] G, then Ψ n is defined on (a.b) by Ψ n (αa + ( α)b) = αφ n (a) + ( αφ n (b)) 0 α It is easily verified that Ψ n is continuous. 5. S. Morris, Topology without tears, References