I -Sequential Topological Spaces
Sudip Kumar Pal
yReceived 10 June 2014
Abstract
In this paper a new notion of topological spaces namely, I-sequential topo- logical spaces is introduced and investigated. This new space is a strictly weaker notion than the …rst countable space. Also I-sequential topological space is a quotient of a metric space.
1 Introduction
The idea of convergence of real sequence have been extended to statistical convergence by [2, 14, 15] as follows: If Ndenotes the set of natural numbers andK NthenKn denotes the setfk2K:k ngandjKnjstands for the cardinality of the setKn. The natural density of the subset K is de…ned by
d(K) = lim
n!1
jKnj n ; provided the limit exists.
A sequence fxngn2N of points in a metric space (X; ) is said to be statistically convergent to l if for arbitrary " > 0; the set K(") = fk 2 N : (xk; l) "g has natural density zero. A lot of investigation has been done on this convergence and its topological consequences after initial works by [5, 13].
It is easy to check that the familyId =fA N : d(A) = 0g forms a non-trivial admissible ideal of N (recall that I 2N is called an ideal if (i) A; B 2 I implies A[B 2I and (ii)A2I; B AimpliesB2I. I is called non-trivial ifI6=f gand N2= I: I is admissible if it contains all the singletons, cf. [8]). Thus one may consider an arbitrary idealI of Nand de…neI-convergence of a sequence by replacing a set of density zero in the de…nition of statistical convergence by a member ofI:
In a topological space X; a setA is open if and only if every a2A has a neigh- borhood contained in A: A is sequentially open if and only if no sequence in XnA has a limit inA:In this paper using the idea of ideal convergence in topological spaces (cf. [9]), we de…ne, I-sequentially open set and hence I-sequential topological space.
Though the concept of these two sets, open andI-sequentially open are the same in case of metric spaces. We give an example of a topological space which is notI-sequential.
Mathematics Sub ject Classi…cations: 54A20, 40A35, 54E15..
yDepartment of Mathematics, University of Kalyani, Kalyani-741235, India
236
Next we formulate an equivalent result for a topological space to beI-sequential and show that everyI-sequential topological space is a quotient of a metric space. Finally we give an example of a topological space which isI-sequential but not …rst countable.
Throughout the paper we assumeXto be a topological space andIbe a non-trivial admissible ideal inN:
2 Main Results
We …rst introduce the following de…nitions.
DEFINITION 2.1. A setO X is said to be open inX if and only if everya2O has a neighborhood contained inO:
DEFINITION 2.2. OisI-sequentially open if and only if no sequence inXnOhas anI-limit inO:i.e. sequence can notI-converge out of aI-sequentially closed set.
DEFINITION 2.3. A topological space is I-sequential when any set O is open if and only if it isI-sequentally open.
We …rst show that the concept of these two sets are the same in case of metric spaces.
THEOREM 2.1. IfX is a metric space, then the notion of open andI-sequentially open are equivalent.
PROOF. Let O be open and fxngn2N be a sequence in X nO: lety 2 O: Then there is a neighborhood U ofy which contained in O. HenceU can not contain any term offxngn2N. Soy is not anI-limit of the sequence andO isI-sequentially open.
Conversely, if O is not open then there is an y 2O such that any neighborhood of y intersectsXnO:In particular we can pick an elementxn2(XnO)\B(y;n+11 )for all n2N:Now the sequencefxngn2NinXnO converges and henceI-converges toy2O;
so Ois notI-sequentially open.
The implication from open toI-sequentially open is true in any topological space.
THEOREM 2.2. In any topological spaceX, ifO is open thenO isI-sequentially open.
PROOF. The proof is similar to the …rst part of the Theorem 2.1.
Now we give an example of a topological space which is notI-sequential.
EXAMPLE 2.1. Consider(R; cc), the countable complement topology onR:Thus A R is closed if and only if A = R or A is countable. Suppose that a sequence fxngn2N has anI-limity. Then the neighborhood(Rn fxn :n2Ng)[ fyg ofy must containxn for in…nitely manyn:This is only possible whenxn=yfornlarge enough.
Consequently, a sequence in any setAcan onlyI-converge to an element ofA;so every
subset ofRis I-sequentially open. But as Ris uncountable, not every subset is open.
So(R; cc)is notI-sequential.
The next theorem shows that if the space is …rst countable then it isI-sequential.
THEOREM 2.3. Every …rst countable space isI-sequential.
PROOF. LetA X is not open. Then there exists y 2Asuch that every neigh- borhood of y intersects XnA. Let fUn :n2Ng be a countable basis aty. Now for everyn2Nchoosexn2(XnA)\(Tn
i=1Ui):Then for every neighborhoodV ofythere exists n2Nsuch thatUn V and hence xm2V for every m n:Clearly fxngn2N
isI-convergent toy:ThereforeAis notI-sequentially open.
In the following Lemma we give a necessary and su¢ cient condition for a setA X to beI-sequentially open.
LEMMA 2.1. Let X be a topological space. Then A X isI-sequentially open if and only if every sequence with I-limit in A has all but …nitely many terms in A:
Where the index set of the part inAof the sequence does not belong toI:
PROOF. IfAis notI-sequentially open, then by de…nition there is a sequence with terms in X nA but I-limit in A. Conversely, suppose fxngn2N is a sequence with in…nitely many terms inXnAsuch thatI-converges toy2Aand the index set of the part in A of the sequence does not belong toI. Thenfxngn2N has a subsequence in XnAthat must still converges toy2A;so Ais not sequentially open.
THEOREM 2.4. The following are equivalent for any topological spaceX:
(i) X isI-sequential.
(ii) For any topological spaceY and functionf :X !Y; f is continuous if and only if it preservesI-convergence.
PROOF. SupposeXisI-sequential. Any continuous function preservesI-convergence of sequences [1], so we only need to prove that if f :X !Y preservesI-convergence, thenf is continuous. Suppose to the contrary that f is not continuous. Then there is an open setU Y such that f 1(U)is not open inX:AsX is I-sequential, f 1(U) is also not I-sequentially open, so there is a sequence fxngn2N in X nf 1(U) that I-converges to an y 2f 1(U). Howeverff(xn)gn2N is then a sequence in the closed setYnU, so it can not havef(y)as anI-limit. Sof does not preservesI-convergence, as required. Thus assertions (ii) holds.
Suppose that the topological space(X; )is notI-sequential. Let(X; Iseq)be the topological space whereA Xis open if and only ifAisI-sequentially open in(X; ).
SinceX is notI-sequential, the topology Iseq is strictly …ner than . Hence the identity map from to Iseq is not continuous. Suppose fxngn2N is I-convergent to y in (X; ). Then every open neighborhood A of y in (X; Iseq) is I-sequentially open in (X; ), soA contains all but …nitely many terms offxngn2N. ThereforeX is I-sequential.
We now give some result to prove the fact that all I-sequential topological spaces are the quotients of some metric spaces [3, 4].
First we recall the de…nition of a quotient space. LetX be a topological space and let be an equivalence relation onX. Consider the set of equivalence classesX=
and the projection mapping :X !X= : Now we considerX= as a topological space by de…ningA X= to be open if and only if 1(A)is open inX [10].
PROPOSITION 2.1. Any quotient spaceX= of anI-sequential topological space X isI-sequential.
PROOF. Suppose that A X= is not open. By de…nition of quotient space
1(A) is not open in X. As X is I-sequential, there is a sequence fxngn2N in X n 1(A) that I-converges to some y 2 1(A): As is continuous it preserve convergence. Hence f (xn)gn2N is a sequence in (X= )nA with I-limit (y)2 A:
ThusAis notI-sequentially open. HenceX= isI-sequential.
PROPOSITION 2.2. EveryI-sequential spaceXis a quotient of some metric space.
PROOF. LetM be the set of all sequencesfxngn2NinX thatI-converges to their
…rst term, i.e. xn !I x0:Consider the subspaceY =f0g [ fn+11 ; n2NgofRwith the standard metric. ThusA Y is open if and only if02=AorAcontains all but …nitely many elements ofY. Now consider the disjoint sum
S= M
fxngn2N2M
fxngn2N Y:
A S is open if and only if for everyfxngn2N2M the setfy2Y : (fxngn2N; y)2Ag is open inY. Consider the mapf :S!Xby(fxngn2N;0)!x0and(fxngn2N;i+11 )! xi for all i 2 N: Here f is clearly surjective as for all x 2 X the constant sequence I-converges tox, sox=f(fxg;0):
Suppose thatA X is open. AsX is I-sequential, every sequence fxngn2N inX thatI-converging to somea2Amust have all but …nitely many terms inAwhere, the index set of the part inAof the sequence does not belong toIby De…nition 2.2. Hence if (fxngn2N;0) 2 f 1(A) we have f 1(A) contains all but …nitely many elements of fxngn2N Y. So for each fxngn2N 2 M, the set fy 2 Y : (fxngn2N; y) 2 f 1(A)g is open in Y. Hence f 1(A) is open in S. Conversely, if A is not open in X then there is a sequence fxngn2N in X n A that I-converges to some a 2 A: But then fy2Y : (fxngn2N; y)2f 1(A)g=f0g is not open inY, sof 1(A)is not open inS:
We can now easily prove that I-sequential topological space is a strictly weaker notion than …rst countable topological space: There exists an I-sequential space X which is not …rst countable.
EXAMPLE 2.2. Consider Rwith standard topology and the quotient relation on R, the equivalence classes areNand fxg for every x2RnN: The quotient space R= is I-sequential as a quotient of a metric space. Suppose that fUn : n 2 Ng
is any countable collection of neighborhood of N: Then for all n 2 N; 1(Un) is a neighborhood of n in R with the standard topology, so there is a "n > 0 such that B(n; "n) 1(Un): Now consider (S
n2NB(n;"2n)), this is a neighborhood of N in R= ; but it does not containUn for anyn 2N: SofUn : n2Ng is not a countable basis atN:
Acknowledgement:This work is funded by University Grants Commission, Govt.
of India through the D. S. Kothari Post Doctoral Fellowship. The author is grateful to the referee for his valuable suggestions which considerably improved the presentation of the paper. The author also acknowledge the kind advice of Prof. Indrajit Lahiri for the preparation of this paper.
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