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Home , Induced topology, Product topology, Tychonoff space

THE PRODUCT

GILI GOLAN

Abstract. In this paper we introduce the of an arbitrary number of topological spaces. We define the separation axioms and character- ize the Tychonoff Spaces as those which can be embedded in a cube. We also prove a sufficient condition for a space to be metrizable.

Contents 1. The Product Topology 1 2. Tychonoff’s Theorem 2 3. Separation Properties 3 4. Metrizability 6 References 7

1. The Product Topology

Definition 1.1. Let {(Xi, τi) | i ∈ I} be a family of topological spaces. The Q Q product space, denoted by i∈I (Xi, τi), consists of the product set i∈I Xi (whose elements are the choice functions on {Xi | i ∈ I}) with the topology τ whose basis is the family Y 0 B = { Ui | Ui ∈ τi and Ui = Xi for all but a finite number of i s } i∈I Q Lemma 1.2. Let {(Xi, τi) | i ∈ I} be a family of topological spaces and (Xi, τi) Q i∈I be the corresponding product space. The projections pj : (Xi, τi) → Xj defined Q i∈I by pj( i∈I xi) = xj are continuous open surjective maps.

Proof. Let j ∈ I be. It is clear that pj is onto. We shall prove first that pj is −1 Q continuous. Let U ⊆ Xj be an then pj (U) = i∈I Ui where Uj = U and −1 for all i 6= j Ui = Xi. Therefore, since pj (U) belongs to the basis of the topology Q of (Xi, τi), it is open and pj is continuous. Therefore it remains to show that i∈I Q pj is open. Let V be a set in the basis of (Xi, τi). It suffices to show that Q i∈I pj(V ) is open. But V ∈ B ⇒ V = i∈I Ui where for all i ∈ IUi is open in Xi. Therefore pj(V ) = Uj is open in Xj and pj is an open mapping. 

Proposition 1.3. Let {(Xi, τi) | i ∈ I} be a family of topological spaces and f a Q mapping of a Y into (Xi, τi), then f is continuous if and only i∈I Q if for each j ∈ I, pj ◦ f is continuous, where pj denotes the of i∈I Xi onto Xj.

Date: June 5th, 2011. Key words and phrases. Product Topology, Tychonoff’s Theorem. 1 2

Proof. (⇒) f is continuous, therefore, since by Lemma 1.2 pj is continuous for all j ∈ I, pj ◦ f is continuous for all j ∈ I. Q Q (⇐) Let V = i∈I Ui ∈ B be a set in the basis of i∈I (Xi, τi). It suffices to −1 Q show that f (V ) is open. V = i∈I Ui ∈ B ⇒ Ui ∈ τi for all i ∈ I and there −1 exists a finite J ⊆ I such that Ui = Xi for all i ∈ I \ J. Therefore f (V ) = −1 Q −1 T −1 T −1 −1 T −1 f ( i∈I Uj) = f ( j∈J pj (Uj)) = j∈J f (pj (Uj)) = j∈J (pj ◦f) (Uj). −1 −1 But, for all j ∈ J pj ◦ f is continuous ⇒ (pj ◦ f) (Uj) is open. Therefore f (V ) is open as a finite intersection of open sets. 

Definition 1.4. We say that a family of functions {fi : X → Yi |i ∈ I} from a topological space X to topological spaces Yi separates points if for every x 6= y ∈ X there exists i ∈ I such that fi(x) 6= fi(y).

Definition 1.5. We say that a family of functions {fi : X → Yi | i ∈ I} from a topological space X to topological spaces Yi separates points and closed sets if for every closed subsetset A ⊆ X and any x ∈ Ac, there exists i ∈ I such that fi(x) ∈/ fi(A).

Lemma 1.6 (The Lemma). Let {(Yi, τi) | i ∈ I} be a family of topolog- ical spaces and for each i ∈ I let fi be a continuous mapping of (X, τ) into (Yi, τi). Q Let e :(X, τ) → (Yi, τi) be the evaluation map; that is ∀x ∈ X, e(x) = Q i∈I fi(x). Then if the family {fi | i ∈ I} separates points of X and separates i∈I Q points and closed sets, then e is an embedding of (X, τ) in i∈I (Yi, τi). Proof. It is sufficient to prove that the mapping e :(X, τ) → (e(X), τ 0) is a homeo- morphism where τ 0 is the on e(X). It is clear that e : X → e(X) is onto while the fact that {fi | i ∈ I} separates points of X makes it one-to-one. Since for every i ∈ I, pi◦e = fi is a continuous , Proposition 1.3 implies that e is continuous as well. Therefore, it remains to show that e :(X, τ) → (e(X), τ 0) is open. Let U ∈ τ be, it suffices to prove that e(U) ∈ τ 0. Let y ∈ e(U) be. We need to show that there exists a W ∈ τ 0 such that y ∈ W ⊆ e(U). But y ∈ e(U) ⇒ ∃x ∈ U such that y = e(x). x ∈ U, U is an open set ⇒ U c is closed and x∈ / U c. Therefore, since {fi | i ∈ I} separates points and closed sets, there exists a mapping fj such c c −1 c that fj(x) ∈/ fj(U ). Put W = pj ((fj(U )) ) ∩ e(X). Since pj is continuous and c c (fj(U )) ⊆ Yj is open W is an open subset of e(X). Clearly e(x) ∈ W . It remains to show that W ⊆ e(U). Let y ∈ W . There exists t ∈ X so that y = e(t). Now, c c c c c fj(t) ∈ (fj(U )) ⇒ fj(t) ∈/ fj(U ). In particular, fj(t) ∈/ fj(U ) ⇒ t∈ / U ⇒ t ∈ U. So y = e(t) ∈ e(U) and W ⊆ e(U). 

2. Tychonoff’s Theorem Definition 2.1. Let X be a topological space. A ⊆ X is said to be compact if every open covering of it has a finite subcovering. More explicitly A is compact if S for every family of open sets {Ui | i ∈ I} such that A ⊆ Ui there is a finite S i∈I subset J ⊆ I such that A ⊆ i∈J Ui. If X ⊆ X is compact we say that X is a compact topological space. Lemma 2.2. Let (X, τ) be a and A ⊆ X be a closed subset, then A is compact. 3

c Proof. Let U = {Ui | i ∈ I} be an open covering of A then U1 = {Ui | i ∈ I} ∪ {A } is an open covering of X. Since X is compact there exists a finite subcovering of c X, V1 ⊆ U1. But then V := V1 \{A } ⊆ U is a finite subcovering of A and A is compact. 

Theorem 2.3 (Tychonoff’s Theorem). Let {(Xi, τi) | i ∈ I} be a family of topo- Q logical spaces. The product space i∈I (Xi, τi) is compact if and only if for each i ∈ I (Xi, τi) is compact.

Definition 2.4. Let A be a set and for each a ∈ A let (Xa, τa) be a topological space homeomorphic to [0, 1] with its standard topology, then the product space Q A a∈A(Xa, τa) is denoted I and refered to as a cube. Corollary 2.5. For any set A, The cube IA is compact. Proposition 2.6. Let (X, d) be a . Then (X, d) is homeomorphic to a subspace of the cube IX . Proof. Without loss of generality for all a and b in X d(a, b) ≤ 1. For each a ∈ X let fa : X → [0, 1] be the fa(x) = d(a, x). Let x 6= y ∈ X be. Then fx(y) = d(x, y) 6= 0 while fx(x) = d(x, x) = 0, therefore fx(y) 6= fx(x) and The family {fa | a ∈ X} separates points. Moreover, given a A ⊆ X and an element x ∈ Ac, since Ac ⊆ X is open, there exists an 0 <  < 1 such that c B(x, ) ⊆ A . Therefore for each a ∈ A d(x, a) ≥ . Therefore fx(A) ⊆ [, 1] and fx(x) = 0 ⇒ fx(x) ∈/ fx(A) ⊆ [, 1]. Therefore {fa | a ∈ X} separates points and closed sets as well. By The Embedding Lemma (Lemma 1.6) the evaluation map from X to IX is an embedding which means X is homeomorphic to a subspace of X the cube I . 

3. Separation Properties

Definition 3.1. Let (X, τ) be a topological space. X is said to be a T0 space if for every two distinct elements x, y ∈ X there exists an open set U ∈ τ such that only one of the elements x, y is an element of U.

Definition 3.2. Let (X, τ) be a topological space. X is said to be a T1 space if for each elemnt x ∈ X and any element y 6= x, there exists a set U ∈ τ such that x ∈ U and y∈ / U.

Lemma 3.3. A topological space (X, τ) is a T1 space iff each singleton {x} ⊆ X is a closed subset of X.

Proof. (⇒) Let X be a T1 space and let x be an element of X. It suffices to show that {x}c ∈ τ. let y ∈ {x}c then y 6= x. There exists an open set U ∈ τ such that y ∈ U and x∈ / U which means y ∈ U ⊆ {x}c. Therefore {x}c ∈ τ. c c c (⇐) Let x ∈ X be and let y 6= x be. Then {y} ∈ τ and x ∈ {y} , y∈ / {y} . 

Definition 3.4. A topological space (X, τ) is said to be T2 or a Hausdorff space iff for each x 6= y ∈ X there exist U, V ∈ τ such that x ∈ U, y ∈ V and U ∩ V = ∅.

Remark 3.5. T2 ⇒ T1 ⇒ T0. Lemma 3.6. Let (X, d) be a metric space then (X, τ) is a Hausdorff space, where τ is the . 4

  Proof. Let x 6= y ∈ X be. Denote  = d(x, y) > 0. Then x ∈ B(x, 2 ), y ∈ B(y, 2 ),     B(x, 2 ),B(y, 2 ) ∈ τ and B(x, 2 ) ∩ B(y, 2 ) = ∅ 

Proposition 3.7. Let {(Xi, τi) | i ∈ I} be a family of Hausdorff spaces then the Q product space i∈I (Xi, τi) is a Hausdorff space. Q Q Q Proof. Let x = i∈I xi, y = i∈I yi be two distinct elements in i∈I Xi. There exists a j ∈ I such that xj 6= yj. Since Xj is a Hausdorff space, there exist open −1 U, V ⊆ Xj such that xj ∈ U, yj ∈ V and U ∩V = ∅. Note that x ∈ pj (U), −1 −1 −1 −1 −1 y ∈ pj (V ), pj (U) ∩ pj (V ) = pj (U ∩ V ) = pj (∅) = ∅. Since by Lemma −1 −1 1.2 pj is a continuous function pj (U) and pj (V ) are open sets which separate Q x and y. Therefore i∈I (Xi, τi) is a Hausdorff space.  Definition 3.8. Let (X, τ) be a topological space. X is said to be regular if for each closed subset A ⊂ X and any element x∈ / A there exists a pair of disjoint open sets U, V ∈ τ such that x ∈ U and A ⊆ V . If X is also a T2 space X is said to be a T3 space.

Remark 3.9. T3 ⇒ T2 ⇒ T1 ⇒ T0 Lemma 3.10. Let (X, τ) be a topological space. Then X is regular iff for each U ∈ τ and x ∈ U there exists an open set V such that x ∈ V ⊆ V ⊆ U. Proof. (⇒) Assume X is regular, let U ∈ τ and x ∈ U be. U c is closed and x∈ / U c, therefore there exist disjoint open sets V,W such that x ∈ V and U c ⊆ W . V ∩ W = ∅ therefore V ⊆ W c. U c ⊆ W ⇒ W c ⊆ U. Therefore V ⊆ W c ⊆ U. But W c is closed ⇒ V ⊆ W c therefore x ∈ V ⊆ V ⊆ U. (⇐) Assume that for each U ∈ τ and x ∈ U there exists an open set V such that x ∈ V ⊆ V ⊆ U. Let A ⊂ X be a closed subset of X and x∈ / A be, then x ∈ Ac and Ac ∈ τ. Therefore there exists an open set V such that x ∈ V ⊆ V ⊆ Ac. c Denote U = V then U ∈ τ, A ⊆ U, and U ∩ V = ∅. Therefore U and V separate A and x and X is regular.  Definition 3.11. Let (X, τ) be a topological space. X is said to be completely regular if for each closed subset A ⊂ X and any element x∈ / A there exists a continuous function f : X → [0, 1] such that f(x) = 0 and f(a) = 1 for all a ∈ A. If X is also a T2 space X is said to be a Tychonoff Space or a T 1 space. 3 2

Remark 3.12. T 1 ⇒ T3 3 2 Proposition 3.13. Let (X, d) be a metric space. Then (X, τ) is a Tychonoff space where τ is the induced topology. Proof. Since by Lemma 3.6 X is a Hausdorff space, is suffices to prove that (X, τ) is completely regular. Let A ⊂ X be a closed subset and x ∈ Ac be. It suffices to show that there is a continuous function f : X → [0, 1] such that f(x) = 0 and f(a) = 1 for all a ∈ A. x ∈ Ac and Ac is an open set, therefore there exists an c d(x,y)  > 0 such that B(x, ) ⊆ A Define f : X → [0, 1] by f(y) = min{1,  } for all y ∈ X. Then f is continuous, f(x) = 0 and f(A) ⊆ {1}. 

Proposition 3.14. Let {(Xi, τi) | i ∈ I} be a family of completely regular spaces Q then the product space X = i∈I (Xi, τi) is completely regular. 5

Q Q Proof. Let A ⊂ X be a closed subset and x = i∈I xi ∈ i∈I Xi be an element of Ac. x ∈ Ac and Ac is open, therefore by Definition 1.1 of the topology of Q Q c i∈I Xi, there exists a finite subset J of I such that x ∈ i∈I Ui ⊆ A , Ui ∈ τi for all i ∈ I and Ui = Xi for all i ∈ I \ J. As (Xj, τj) is completely regular for all j ∈ J there exist continuous mappings fj : Xj → [0, 1] such that fj(xj) = 0 c and fj(Uj ) ⊆ {1}. Since the projection pj is continuous for all j, the mappings Q fj ◦ pj : Xi → [0, 1] are continuous. Put f = max{fj ◦ pj(x) | j ∈ J} then f i∈I Q Q is continuous. for x = i∈I xi, f(x) = 0. For every a ∈ A, a∈ / i∈I Ui ⇒ there exists j1 ∈ J such that a∈ / Uj1 . Therefore fj1 (a) = 1 and f(a) = 1 as well. 

Corollary 3.15. Let {(Xi, τi) | i ∈ I} be a family of Tychonoff spaces then the Q product space i∈I (Xi, τi) is a Tychonoff space. Corollary 3.16. For every set X the cube IX is a Tychonoff space. Proposition 3.17. Let (X, τ) be a Tychonoff space, Then X can be embedded in a cube. Proof. Let F be the family of all continuous finctions f : X → [0, 1]. Since (X, τ) is a Tychonoff Space the family F separates points and closed sets. Since X is Hausdorff and therefore T1 every singleton {x} is a closed subset. Therefore the family F separates points as well. According to the Embedding Lemma (Lemma F 1.6), (X, τ) can be embedded in the cube [0, 1] .  Lemma 3.18. Every subspace of a Tychonoff space is also a Tychonoff space. Proof. Let (X, τ) be a Tychonoff space and (Y, τ 0) be a subspace of it. Let A ⊂ Y be a closed subset of Y and x ∈ Y \ A be. A is closed in Y ⇒ there exists a closed set P ⊆ X such that A = P ∩ Y . x ∈ Y and x∈ / A ⇒ x∈ / P . Since (X, τ) is a Tychonoff space there exists a continuous function f : X → [0, 1] such that f(x) = 0 and f(a) = 1 for all a ∈ P . Therefore f|Y : Y → [0, 1] is also a continuous function and f|Y (x) = 0, f|Y (a) = 1 for all a ∈ A.  Corollary 3.19. (X, τ) is a Tychonoff space iff it is homeomorphic to a subspace of a cube. Definition 3.20. Let (X, τ) be a topological space. X is said to be normal if for each two disjoint closed sets P,S ⊆ X there exist two disjoint open sets U, V ∈ τ such that P ⊆ U and S ⊆ V . If X is also a T2 space X is said to be a T4 space.

Remark 3.21. T4 ⇒ T3 ⇒ T2 ⇒ T1 ⇒ T0 Lemma 3.22 (Urysohn’s Lemma). Let (X, τ) be a topological space. Then X is normal iff for each pair of disjoint closed sets P,S ⊆ X, there exists a continuous function f : X → [0, 1] such that f(p) = 0 for each p ∈ P and f(s) = 1 for each s ∈ S.

Corollary 3.23. If (X, τ) is a T4 space then it is a T 1 space as well. 3 2

Remark 3.24. T4 ⇒ T 1 ⇒ T3 ⇒ T2 ⇒ T1 ⇒ T0 3 2 Proposition 3.25. Let (X, τ) be a compact Hausdorff space, then (X, τ) is normal. 6

Proof. We shall show first that X is regular. Let A ⊂ X be a closed subset and x∈ / A be. For each a ∈ A x 6= a ⇒ there exists a pair of disjoint open sets S Ua,Va such that x ∈ Ua and a ∈ Va. A ⊆ a∈A Va therefore {Va | a ∈ A} is an open covering of A. Since A is a closed subset of a compact space X, by

Lemma 2.2 A is compact. Therefore there exists a finite subcovering {Vai | 1 ≤ Tn Sn Tn Sn i ≤ n} of A. i=1 Uai , i=1 Vai are open sets, x ∈ i=1 Uai ,A ⊆ i=1 Vai Tn Sn Sn Sn Tn and i=1 Uai ∩ i=1 Vai ⊆ i=1 Uai ∩ Vai = i=1 ∅ = ∅. Therefore i=1 Uai and Sn i=1 Vai separate x and A and X is regular. Now, we can prove that X is normal. Let A, B be disjoint closed sets. For each a ∈ A there exist a pair of disjoint open sets Ua,Va such that a ∈ Ua and B ⊆ Va. S A ⊆ a∈A Ua therefore {Ua | a ∈ A} is an open covering of A. Since A is a closed subset of a compact space X, by Lemma 2.2 A is compact. Therefore there exists Sn Tn a finite subcovering {Uai | 1 ≤ i ≤ n} of A. i=1 Uai , i=1 Vai are open sets, A ⊆ Sn Tn Sn Tn Sn Sn i=1 Uai ,B ⊆ i=1 Vai and i=1 Uai ∩ i=1 Vai ⊆ i=1 Uai ∩ Vai = i=1 ∅ = ∅. Sn Tn Therefore i=1 Uai and i=1 Vai separate A and B and X is a . 

Proposition 3.26. Let (X, d) be a metric space, then (X, τ) is a T4 space, where τ is the induced topology. Proof. By Lemma 3.6 X is a Hausdorff space. Therefore it remains to prove that X is normal. Let A, B be disjoint closed subsets of X. A ⊆ Bc and Bc is open c c ⇒ for each a ∈ A there exists an a > 0 such that B(a, a) ⊆ B . B ⊆ A and c c A is open ⇒ for each b ∈ B there exists an b > 0 such that B(b, b) ⊆ A . Put S a S b U := a∈A B(a, 2 ),V := b∈B B(b, 2 ). Then U, V ∈ τ, A ⊆ U and B ⊆ V . It remains to show that U ∩ V = ∅. Assume by contradiction that U ∩ V 6= ∅ a and let x ∈ U ∩ V be. There exist a ∈ A and b ∈ B such that x ∈ B(a, 2 ) and b a b a x ∈ B(b, 2 ). WLOG a ≥ b, then d(a, b) ≤ d(a, x) + d(x, b) < 2 + 2 ≤ 2 2 = a c ⇒ b ∈ B(a, a) by contradiction to B(a, a) being a subset of B .  4. Metrizability Definition 4.1. A topological space (X, τ) is said to be metrizable if there is a metric d on X such that the topology induced by d is equal to τ. Definition 4.2. A topological space (X, τ) is said to be second countable if τ has a countable basis.

Proposition 4.3. Let (X, τ) be a second countable T4 space, then X metrizable. Proof. Since Hilbert’s cube I∞ is metrizable it suffices to show that X can be embedded in it. By the Embedding Lemma (Lemma 1.6) it is enough to find a countable family F of continuous functions f : X → [0, 1] such that F separates points and closed sets. (Note that since X is Hausdorff and therefore T1, if F separates points and closed sets it separates points as well). Let B be a countable basis of τ and let S be the set of all the pairs (V,U) such that U, V ∈ B and V ⊆ U. Since B is countable, S ⊆ B2 is countable as well. Now, for each pair (V,U) ∈ S, by Urysohn’s Lemma (Lemma 3.22) there is a continuous function fVU : X → [0, 1] such that f(x) = 0 for each x ∈ V and f(y) = 1 for each y ∈ U c. Let F be the family of functions thus obtained, then F is countable. It suffices to show that F separates points and closed sets. Therefore, let x ∈ X and A be a closed set such that x∈ / A. x ∈ Ac ⇒ ∃U ∈ B such that x ∈ U ⊆ Ac. Since X is regular, by Lemma 3.10 there exists an open set P such that x ∈ P ⊆ P ⊆ U. P ∈ τ ⇒ 7 there exists a set V ∈ B such that x ∈ V ⊆ P . But V ⊆ P ⇒ V ⊆ P . Therefore c x ∈ V ⊆ V ⊆ U and (V,U) ∈ S. Now, fVU (A) ⊆ fVU (U ) ⊆ {1} while fVU (x) = 0. Therefore, F separates x and A.  Lemma 4.4. Let (X, τ) be a second countable . Then X is normal. Proof. Let A, B be two disjoint closed subsets of X and B be a countable basis of c τ. Since B is open and X is regular, for each a ∈ A there exists a set Ua ∈ B such c that a ∈ Ua ⊆ Ua ⊆ B . B is countable, therfore we can list the elements of the set S∞ {Ua | a ∈ A} thus obtained by Ui, i ∈ N. That is, A ⊆ i=1 Ui and Ui ∩ B = ∅ for all i ∈ N. WLOG we can assume that the {Ui}i∈N is an ascending sequence, since otherwise we can switch our attention to the ascending sequence U1 ⊆ U1 ∪ U2 ⊆ U1 ∪ U2 ∪ U3... which has the same properties. Similarly, there is S∞ an ascending sequence of open sets {Vi}i∈N such that B ⊆ i=1 Vi and Vi ∩ A = ∅. Now, define two other of open sets: Wn := Un \ Vn and Sn := Vn \ Un S∞ S∞ and put U = n=1 Wn and V = n=1 Sn. It suffices to show that the open sets U and V separate A and B. For each a ∈ A, there exists i ∈ I such that a ∈ Ui and Vi ∩ B = ∅. Therefore a ∈ Wi ⊆ U and A ⊆ U. Similarly B ⊆ V . Therefore, it remains to show that U ∩ V = ∅. Assume by contradiction that there exists an element x ∈ U ∩ V . x ∈ U ⇒ x ∈ Wn for some n ∈ N. Therefore x ∈ Un and x∈ / Vn. In particular x∈ / Vn. Since {Vi}i∈N is an ascending sequence x∈ / Vi for all i ≤ n. x ∈ V ⇒ x ∈ Sm for some m ∈ N. Therefore x ∈ Vm and x∈ / Um. Since for all i ≤ n x∈ / Vi, m > n. But then since x ∈ Un, x∈ / Um is a contradiction to {Ui} being an ascending sequence. 

Corollary 4.5 (Urysohn’s Metrization Theorem). Every second countable T3 space is metrizable. Proposition 4.6. Let (X, τ) be a compact space. X is metrizable iff X is a second countable Hausdorff space.

Proof. (⇐) X is compact Hausdorff and therefore by Lemma 3.25 X is a T4 space. Therefore since X is also second countable, by Lemma 4.3 X is metrizable. (⇒) Since X is metrizable, by Proposition 3.6 X is a T2 space. Therefore it suffices to prove that X is second countable. For this we will construct a countable basis of τ. Led d be a metric which induces τ, then for each n ∈ N and a ∈ X the open 1 1 ball B(a, n ) ∈ τ. Note that for each n ∈ N, Cn = {B(a, n ) | a ∈ X} is an open covering of X. Since X is compact, for each n ∈ N Cn has a finite subcovering 1 S∞ FCn = {B(ani , n ) | 1 ≤ i ≤ kn}. Let B = n=1 FCn, then B is countable as a countable of finite sets. It suffices to prove that B is a basis of τ. Note that each set in B is open. Let U ∈ τ be and let x be an element of U. It suffices to show there exists a set W ∈ B such that x ∈ W ⊆ U. But x ∈ U, U ∈ τ ⇒ 1 ∃n ∈ N such that B(x, n ) ⊆ U. Since FC2n covers X there exists a2ni ∈ X such 1 1 1 that x ∈ B(a2ni , 2n ) ∈ FC2n. But x ∈ B(a2ni , 2n ) ⇒ d(a2ni , x) < 2n ⇒ for 1 1 1 1 all z ∈ B(a2ni , 2n ) d(x, z) ≤ d(x, a2ni ) + d(a2ni , z) < 2n + 2n = n . Therefore 1 1 1 B(a2ni , 2n ) ⊆ B(x, n ) and for W = B(a2ni , 2n ), W ∈ B and x ∈ W ⊆ U.  References [1] Sydney A. Morris, Topology Without Tears, 2011. [2] D. Leibovich, Set Topology. The Open University 2007.