Almost separable spaces
aa r X i v : . [ m a t h . GN ] F e b Almost separable spaces
Sagarmoy Bag, Ram Chandra Manna, and Sourav Kanti Patra
Abstract
We have defined almost separable space. We show that like separability, almostseparability is c productive and converse also true under some restrictions. Weestablish a Baire Category theorem like result in Hausdorff, Pseudocompacts spaces.We investigate few relationships among separability, almost separability, sequentialseparability, strongly sequential separability.
1. Introduction
Let X be any topological space. C ( X ) be the set of all real valued continuousfunctions on X . A subset A of X is called almost dense in X if for any f ∈ C ( X )with f ( A ) = { } , implies that f ( X ) = { } . And a topological space X is calledalmost separable if it has a countable almost dense subset. A dense subset is alwaysalmost dense. In completely regular space, dense and almost dense sets are identi-cal. But the converse is not true. We give an example of a non completely regularspace in which dense and almost dense sets are same [ Example 2 . . c productive and under some restrictions the con-verse is also true Theorem 3 .
8. In theorem 4.3, we established Baire Category liketheorem. Finally we establish relationships among almost separability, sequentiallyseparability and strongly sequentially separability.
Definition . [ ] A space X is called sequentially separable if there exist acountable set D such that for every x ∈ X there exist a sequence from D convergingto x Definition . [ ] A space X is called strongly sequentially separable if it isseparable and every dense countable subspace is sequentially dense. A subset D of X is called sequentially dense if for every x in X there exists a sequence from D converges to x .
2. Almost dense subsets
Definition . A subset A of a topological space X is called almost dense iffor any f ∈ C ( X ) with f ( A ) = { } , implies f ( X ) = { } . Theorem . Dense subsets are always almost dense.
Proof.
Obvious. (cid:3)
Mathematics Subject Classification.
Primary 54A05; Secondary 54D65.
Key words and phrases.
Almost dense set, Almost separable space, Pseudocompact space,functionally Hausdorff space, sequentially separabile space, Strongly sequentially separabile space.
Theorem . Let X be a completely regular space, then almost dense set aredense in X . Proof.
Let A be almost dense subset of X . If possible let A be not dense in X and x ◦ ∈ X \ A then there exists f ∈ C ( X ) such that f ( x ◦ ) = 0 and f ( A ) = 1.Then f ( A ) = 0 but f ( X ) = { } . This implies that A is not almost dense in X which is a contradiction. Hence A is dense in X . (cid:3) We now show that the converse of the above theorem is not true.
Theorem . Let Y be an open dense subset of a topological space X (mayor may not be a Completely regular space). If Y has a base of cozero subsets of X .Then almost dense set in X is dense in X . A set A of X is called cozero set if A = X \ Z ( f ) , for some f ∈ C ( X ) , where Z ( f ) = { x ∈ X : f ( x ) = 0 } . Proof.
Let A be almost dense subset of X . To show A is dense in X . Suppose A ∩ Y = ∅ . Let y ∈ Y . Then there exist f ∈ C ( X ) such that y ∈ X \ Z ( f ) ⊆ Y . Then f ( A ) = { } but f ( X ) = { } , which is a contradiction as A is almost dense in X . So, A ∩ Y = φ . Claim: A ∩ Y is dense in Y . If not there is a cozero set X \ Z ( g ) ⊂ Y suchthat ( A ∩ Y ) ∩ ( X \ Z ( g )) =, where g ∈ C ( X ). Then Z ( g ) ⊂ ( X \ Y ) ∪ ( A ∩ Y ) ⊃ A .So, that g ( A ) = { } . But g ( X ) = { } . Contradiction shows that A ∩ Y is d4ensein Y . Given that Y is dense in X , A ∩ Y is dense in Y which yields A is dense in X . (cid:3) To find such a spaces we present following examples.
Example . Let K = { n : n ∈ N } and β = { ( a, b ) : a < b, a, b ∈ R } ∪{ ( a, b ) \ K : a < b, a, b ∈ R } be a base for the topology τ K on R which is known as K -topology on R . Then ( R , τ K ) is not regular and hence not completely regular.We would like to show that almost dense set in ( R , τ K ) is dense in ( R , τ K ). Let Y = R \ { } . The set Y is open, dense in ( R , τ K ) and the relative topology of Y is usual topology of Y which is completely regular and , as a result, possess a baseconsisting of cozero sets of ( R , τ K ). We invoke theorem 2 . R , τ K ) are dense in ( R , τ K ). Example . Let X be zero-dimensional, non normal space. Let F, G be twodisjoint closed sets in X which can not be separated by disjoint open sets. Thendefine Y to be the quotient space obtaining by collapsing G to a single point.The following example shows that in a normal space the almost dense set anddense set may not be same. Example . Let X = { a, b } and τ = { φ, X, { a }} be a topology on X . Then X is a normal space. { b } is almost dense set in X but not dense in X . Theorem . Let f : X Y be a continuous onto function and A be almostdense set in X , then f ( A ) is almost dense set in Y . Proof.
Let g ∈ C ( Y ) such that g ( f ( A )) = { } . Then g ◦ f ∈ C ( X ) such that g ◦ f ( A ) = { } . As A is almost dense in X , then we have g ◦ f ( X ) = { } . Since f is onto we have g ( Y ) = { } . Hence f ( A ) is almost dense in Y . (cid:3) Theorem . Let τ , τ be two topologies on X such that τ is finer than τ .Then if a subset A of X is almost dense in ( X, τ ) , then A is almost dense in ( X, τ ) . Proof.
We denote C ( X, τ i ) for the set of all real valued continuous functionson X with respect to the topology τ i for i = 1 ,
2. Let A be almost dense in ( X, τ ).Let f ∈ C ( X, τ ) with f ( A ) = { } . Since f ∈ C ( X, τ ) as τ is finer than τ . LMOST SEPARABLE SPACES 3
Since A is almost dense in ( X, τ ), then f ( X ) = { } . Hence A is almost dense in( X, τ ). (cid:3) Theorem . If A is almost dense in X and B is almost dense in Y , then A × B is almost dense in X × Y . Proof.
Let f : X × Y R be continuous such that f = 0 on A × B . Fix a ∈ A and look at f a ( y ) = f ( a, y ) , y ∈ Y . f a : Y R is continuous and f a ( y ) = 0for all y ∈ B . By hypothesis f a ( y ) = 0 for all y ∈ Y .Since a is arbitrary we obtain f ( A × Y ) = { } . Take any ( x, y ) ∈ X × Y . then f y : X R defined by f y ( z ) = f ( z, y ) , z ∈ X is a continuous map. Now f y ( a ) = 0for all a ∈ A . Then f y ( x ) = 0 for all x ∈ X i.e. f ( x, y ) = 0. Since y ∈ Y isarbitrary this shows f = 0 on X × Y . Hence A × B is almost dense in X × Y . (cid:3) Corollary . Let A , A , ..., A n be almost dense in X , X , ...., X n respec-tively. Then Q ni =1 A i is almost dense in Q ni =1 X i . Proposition . Let { X α : α ∈ Λ } be a family of topological spaces and A α ⊂ X α , α ∈ Λ be almost dense. Then Q α ∈ Λ A α is almost dense in Q α ∈ Λ X α . Proof.
For a = ( a α ) α ∈ Λ ∈ Q α ∈ Λ A α . Define D = { ( x α ) α ∈ Λ ∈ Q α ∈ Λ X α : { α : x α = a α } is finite } . Let f : Q α ∈ Λ X α R be continuous such that f = 0 in Q α ∈ Λ A α . For any finite subset I ⊂ Λ , Q α ∈ I X α ×{ a α : α ∈ Λ \ I } is homeomorphicto Q α ∈ I X α and Q α ∈ I A α is almost dense in Q α ∈ I X α (by above Corollary). Hence Q α ∈ I A α × { a α : α ∈ Λ \ I } is almost dense in Q α ∈ I X α × { a α : α ∈ Λ \ I } .Consequently f ( Q α ∈ Λ X α × { a α : α ∈ Λ \ I } ) = { } . Now D = ∪ [ Q α ∈ I X α × { a α : α ∈ Λ \ I } : I ⊂ Λ finite ]. This implies that f ( D ) = { } . We now invoke thefollowing important result in product space. (cid:3) We now invoke the following important result on product space.
Theorem . Let { Y α : α ∈ Λ } be a family of topological spaces and and a = ( x α ) α ∈ Λ be a fixed element of Q α ∈ Λ Y α . The set E = { ( y α ) α ∈ Λ ∈ Q α ∈ Λ Y α : { α ∈ Λ : y α = x α } is finite } is dense in Q α ∈ Λ Y α . Because of this theorem D is dense in Q α ∈ Λ X α and f ( D ) = 0 which implies f = 0 on Q α ∈ Λ X α . As a result Q α ∈ Λ A α is almost dense in Q α ∈ Λ X α . Theorem . If a topological space X contains a connected almost densesubset, then X is connected. Proof.
Let A be connected almost dense subset of X . If possible let X bedisconnected. Then there exists a onto continuous function f : X
7→ { , } . Let Y = { x ∈ X : f ( x ) = 0 } . Then X \ Y = f − { } . Since A is connected, then either A ⊆ Y or A ⊆ X \ Y .Case 1: If A ⊆ Y . Then f ( A ) = { } but f ( X ) = { } , which contradicts thefact that A is almost dense set in X .Case 2: If A ⊆ X \ Y . Consider g = 1 − f . Then g ( A ) = { } but g ( X ) = { } ,which is a contradiction as A is almost dense set in X .Hence X is connected. (cid:3) Of course the converse of 2 .
14 is not true i.e., X is a connected space butan almost dense subset need not be connected. As an illustration consider thefollowing: X = R and the topology consist of all subsets of R containing 0. X is obviouslyconnected. The subspace A = { , } is almost dense in X but not connected. S. BAG, R.C.MANNA, AND S. K. PATRA
Theorem . A subset A of X is almost dense in X if and only if every nonempty cozero set intersects A . Proof.
Let A be an almost dense subset of X and X − Z ( f ) be a non-emptycozero set in X . Let us assume that A ∩ ( X − Z ( f )) = ∅ . Then f ( A ) = { } . Since A is almost dense in X , then f ( X ) = { } ,Which contradicts the fact that X − Z ( f )is non-empty. So A ∩ ( X − Z ( f )) = ∅ .Conversely, let A intersect every non-empty cozero sets of X . Suppose A is notalmost dense subset of X . Then there exists a continuous function f : X R suchthat f ( A ) = { } but f ( X ) = { } . Then X − Z ( f ) is a non empty cozero set and A does not intersect the non-empty cozero set X − Z ( f ). (cid:3) As is well known that in a topological space a non-empty subset is dense if andonly if it intersect every non-empty open set. Theorem 2 .
15 presents the analogousresult in the case of almost denseness property. We have seen that in a topologicalspace X if A ⊂ X is dense and U is non-empty open set, then not only A ∩ U = ∅ , A ∩ U is dense in U also. Now the question arises as follows:Let A be almost dense in a topological space X and ∅ 6 = U ⊂ X is a cozero set.Then A ∩ U = ∅ is no doubt, but A ∩ U is almost dense in U ? The answer does notseem to be clear!
3. Almost separable spaces
Definition . A space X is called almost separable if it contains a countablealmost dense subset. Theorem . Each separable space is almost separable.
The converse of the above theorem is not true.
Example . Let X = R and τ c be cocountable topology on X . For anycountable set A in X , X \ A is open. Therefore A is not dense in X . Therefore X is not separable.We want to show that Q is almost dense in X . Let f ∈ C ( X ) and f ( Q ) = { } .Since each real valued continuous function on X is constant, then f ( X ) = { } .Therefore Q is almost dense in X . Therefore X contains a countable almost densesubset. Hence X is an almost separable space. In fact, any countable subset of X is almost dense in X . Theorem . Finite product of almost separable spaces is almost separable.
Proof.
It follows from the Corollary 2 . (cid:3) In the case of infinite products, like the case of separable spaces, the followingresult is true.
Theorem . Let { X α : α ∈ Λ } be a family of almost separable topologi-cal spaces, Λ , with card( Λ) = c , Then Q α ∈ Λ X α with product topology is almostseparable space. Proof.
Let A α ⊂ X α be a countable almost dense subset of X α . To avoidtriviality we assume A α to be countably infinite. Let f α : N A α be a bijection.Define Q α ∈ Λ f α : N Λ Q α ∈ Λ A α as follows: Q α ∈ Λ f α ( n α , α ∈ Λ) = ( f α ( n α ) , α ∈ Λ) where ( n α , α ∈ Λ) ∈ N Λ . We write f Λ for Q α ∈ Λ f α . We know N Λ = { g : Λ N } and f Λ ( g ) = ( f α ( g ( α )) : α ∈ Λ) ∈ Q α ∈ Λ A α . It is easy to see that f Λ is onto. Let p α : Q α ∈ Λ A α A α be the projection to α -th coordinate. Then p α ◦ f Λ ( g ) = f α ◦ g ( α ) , α ∈ Λ. This shows that f Λ : N Λ Q α ∈ Λ A α is continuous and onto. It iswell known that N Λ is separable and, hence almost separable. Therefore Q α ∈ Λ A α LMOST SEPARABLE SPACES 5 is almost separable. Since Q α ∈ Λ A α is almost dense in Q α ∈ Λ X α , Q α ∈ Λ X α isalmost separable. (cid:3) Definition . A topological space X is called functionally Hausdorff if forany two distinct points a, b ∈ X there exists f ∈ C ( X ) such that f ( a ) = 0 and f ( b ) = 1. Lemma . If X a functionally Hausdorff space, then for any two distinctpoints a, b ∈ X there exist two distinct cozero sets C, D in X such that a ∈ C and b ∈ D . Proof. X is functionally Hausdorff. Then for a, b ∈ X with a = b there exists f ∈ C ( X ) such that f ( a ) = 0 and f ( b ) = 1. Let C = { x ∈ X : f ( x ) < } =(( f − ) ∧ − ( R \ { } ) and D = { x ∈ X : f ( x ) > } = (( f − ) ∨ − ( R \ { } ),where ( g ∨ h )( x ) = max { g ( x ) , h ( x ) } and ( g ∧ h )( x ) = min { g ( x ) , h ( x ) } for all x ∈ X .Thus C, D are disjoint cozero sets in X and a ∈ C, b ∈ D . This completes theproof. (cid:3) Theorem . Let X = Q α ∈ Λ X α be almost separable space, where each X α is functionally Hausdorff and contains at least two points. Then each X α is almostseparable and card (Λ) ≤ c . Proof.
Let D be a countable ,almost dense subset of X . Consider the projec-tion function p α : X X α to the α -th coordinate. Then the function p α : X X α is continuous. Since continuous image of an almost separable space is almost sepa-rable, therefore each X α is almost separable.For every α ∈ Λ, let a α , b α ∈ X α with a α = b α . Since each X α is functionallyHausdorff, therefore there exist disjoint cozero sets C α , D α in X α such that a α ∈ C α and b α ∈ D α . p − α ( C α ) is a non-empty cozero set in X . By Theorem 2 . K α = D ∩ p − α ( C α ) = ∅ .Define a function φ : Λ
7→ P ( D ) by φ ( α ) = K α . Now p − α ( C α ) ∩ p − α ( D β ) isa non-empty cozero set in X . Then there exist x ∈ D ∩ p − α ( C α ) ∩ p − α ( D β ) byTheorem 2 .
15. So, x ∈ K α and x / ∈ K β . Therefore K α = K β . Thus φ is injective.Hence card (Λ) ≤ card ( P ( D )) = c . This completes the proof. (cid:3) Theorem . For an almost separable space X , the cardinality of C ( X ) isless than or equal to c . Proof.
Let A be countable almost dense subset of X . Define a map φ : C ( X ) C ( A ) by φ ( f ) = f | A . We sow that φ is injective mapping. Let φ ( f ) = φ ( g ), where f, g ∈ C ( X ). Then f | A = g | A . Let h = f − g . Then h ∈ C ( X ) and h ( A ) = { } . Since A is almost dense in X , then h ( X ) = { } . Therefore φ isinjective. Since the cardinality of C ( A ) is less that or equal to c , therefore thecardinality of C ( X ) is less than or equal to c . (cid:3) Corollary . If an almost separable space has a uncountable closed dis-crete subspace, then it is not normal.
Theorem . Let X be functionally Hausdorff, almost separable space. Thenthe cardinality of X is atmost c . Proof.
Consider the function ψ : X
7→ P ( C ( X )) by ψ ( x ) = { f ∈ C ( X ) : f ( x ) = 0 } . Use the functionally Hausdorff property of X to conclude that ψ isinjective. By the Theorem 3 .
9, cardinality of C ( X ) is c . Thus cardinality of X isatmost 2 c .Alternative proof of the above theorem: Let X be a functionally Hausdorff,almost separable space. Let τ w be the weak topology on X induced by C ( X ). S. BAG, R.C.MANNA, AND S. K. PATRA
Then (
X, τ w ) is completely regular Hausdorff space show that almost dense setbecomes dense. Thus ( X, τ w ) is a separable Hausdorff space. It is known that thecardinality of a separable Hausdorff space is atmost 2 c . (cid:3) Theorem . Let Y be almost dense in X and Y be almost separable as asubspace. Then X is almost separable. Proof.
Let A be countable almost dense subset of Y . Let f ∈ C ( X ) such that f ( A ) = { } . Then f | Y ∈ C ( Y ) and f | Y ( A ) = { } . This implies that f | Y ( Y ) = { } as A is almost dense in Y . Since Y is almost dense in X , then f ( X ) = { } .Therefore A is countable almost dense subset of X . Hence X is almost separablespace. (cid:3)
4. Baire Category Like Theorem
Theorem . For a topological space X the following are equivalent: ( i ) X is pseudocompact ( ii ) If { F n : n ∈ N } is a sequence of zero sets of X with finite intersection property, Then T ∞ n =1 F n = ∅ . ( iii ) If { U n : n ∈ N } is a countable cover of X consisting of cozero sets , thereexists a finite subcover. Theorem . Let X be a Hausdorff space. Given a non-empty cozero-set U and x ∈ U there exists a cozero-set V and a zero-set F such that x ∈ V ⊂ F ⊂ U . Proof.
Let f : X −→ [0 ,
1] such that U = f − (0 ,
1] = X − f − ( { } ) . Since x ∈ U, f ( x ) >
0. Choose δ > < f ( x ) − δ < f ( x ). Let V = f − ( f ( x ) − δ, , F = f − [ f ( x ) − δ, x ∈ V ⊂⊂ F ⊂ U . Now ( f ( x ) − δ, ⊂ [0 ,
1] isan open set and hence a cozero-set and [ f ( x ) − δ,
1] is a closed subset of [0 ,
1] andhence a zero-set. Then V = f − ( f ( x ) − δ,
1] is a cozero-set and F = f − [ f ( x ) − δ, X . (cid:3) Theorem . (Baire Category Like Theorem) Let X be a Hausdorff, pseudo-compact space. If { U n : n ∈ N } is a sequence of almost dense cozero-sets of X ,then T ∞ n =1 U n is a non-empty almost dense subset of X . Proof.
Write D = T ∞ n =1 U n . To show D = ∅ and D intersects every non-empty cozero-set. Let V be a non-empty cozero-set and let x ∈ V . Since U isalmost dense , V ∩ U = ∅ and is a cozero-set . Let x ∈ V ∩ U . By theorem (4 . ∃ cozero-set V and a zero-set F such that x ∈ V ⊂ F ⊂ V ∩ U . Now V = ∅ cozero-set = ⇒ V ∩ U = ∅ and is a zero-set. Let x ∈ V ∩ U . ∃ V a non emptyco-zero set and F , a zero-set , such that x ∈ V ⊂ F ⊂ V ∩ U ⊂ V ∩ U ∩ U .Now V = ∅ cozero-set, V ∩ U = ∅ and is a cozero -set. Let x ∈ V ∩ U . ∃ V , acozero-set, and F , a zero set , such that x ∈ V ⊂ F ⊂ V ∩ U ⊂ V ∩ U ∩ U ∩ U .Proceeding in this way we obtain non-empty cozero-set V n +1 , zero-set F n +1 suchthat x n +1 ∈ V n +1 ⊂ F n +1 ⊂ V n ∩ U n +1 ⊂ V ∩ U ∩ U ∩ U ∩ ...... ∩ U n +1 forall n ≥
0. Note that F n +1 ⊂ F n and F ′ n s are non-empty zero-sets. Since X ispseudocompact , in virtue of the theorem (4 . T ∞ n =1 F n = ∅ . Hence T ∞ n =1 F n ⊂ T ∞ n =1 ( V ∩ U ∩ U ∩ U ∩ ....... ∩ U n ) = V T ∞ n =1 U n so that V ∩ ( T ∞ n =1 U n ) = ∅ . i.e, V ∩ D = ∅ . Since V is an arbitary non-empty cozero-set , D is almost dense. (cid:3) We know that separability is not a hereditary property. Niemytzky’s plane is awell known example. Same is true about almost separability as well. Niemytzky’splane provides an example in this case also.We now finish our paper giving relations among the different types of separa-bility notions which are defined earlier.
LMOST SEPARABLE SPACES 7
Strong sequentially separable ⇒ Sequentially separable ⇒ Separable ⇒ Almostseparable.
Acknowledgments:
The authors would like to thank to Professor Alan Dowfor Theorem 2.4. Also the authors wish to thank Professor Asit Baran Raha fortheir valuable suggestions that improved the article.
References [1] A. Bella, M. Bonanzinga, M. Matveev, Sequential+ sparable vs. sequntially separable andanother variation on sective separability, Cent. Eur. J. Math., Department of Pure Mathematics, University of Calcutta, 35, Ballygunge Circu-lar Road, Kolkata 700019, West Bengal, India
E-mail address : [email protected] Ramakrishna Mission Vidyamandir, Belur Math, Howrah-711202, West Bengal, In-dia
E-mail address : [email protected] Ramakrishna Mission Vidyamandir, Belur Math, Howrah-711202, West Bengal, In-dia
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