ON ULTRACONNECTED SPACES

In this paper, we study some properties of ultraconnected spaces and show that ultraconnected T spaces are maximal ultraconnected and minimal T We also introduce the notion 2 of F-connected spaces, topological spaces wich are both Nyperconnected and ultraconnected and characterize compact maximal F-connected topologies on a set.


I. INTRODUCTION.
A topological space is ultraconnected if the intersection of any two nonempty closed sets is nonempty (Steen and Seebach [I]).Each topology on a set X may be associated with a pre-order relation p() on X, defined by (a,b) E ) if every open set containing b contains a.
In 1978 Andlma and Thron [2] defined a topological space (X,) to be upward directed if any two elements in (X,p()) have an upper bound, and it can easily be seen that the notion of upward directed and that of ultraconnected are equivalent.Let (X,R) be a pre-ordered set.Define {} E X x R y} and {x} X y R x}, for each x E X.
(R), the point closure topology of R, is the smallest topology in which all sets {x}, x E X, are closed and V(R), the kernel topology of R, is the topology with basis {{} x E X}.A topology on X induces a pre-order R as described above iff (R)CCV(R) [2].
In [2], it is proved that a topological space (X,z) is maximal upward directed iff (X,p(z)) is a partially ordered set of length I, with a greatest element and P.M. MATHEW T V(p(T)).
If (X,R) is a partially ordered set of length I, with a greatest element, say a, then V(R) P(X{a}) {X}.Thus the maximal ultraconnected topologies on a set X are precisely P(X{a }) {X }, where a c X. DEFINITION 2.1.A topological space is T if each singleton subset is either open or closed (Levine [3]). 2 REMARK 2. I. Any T space is T and Dunham [4] characterized the minimal O T topologies on a set2X as those of the form {OX OA or AGO and O'finite }, for 2 some proper subset A of X. (When X is finite with more than one element, A must also be nonempty.)Obviously, any maximal ultraconnected space is minimal T 2 THEOREM 2. I.
Any ultraconnected T space is maximal ultraconnected and minimal T 2 PROOF. 2 Let (X, ) be an ultraconnected T space.Since (X, ) is T the induced order p() is a partial order.Suppose there e2xist x,y,z c X such that x2o()y and y D(T)z.If {y} is open, then x D()y---> x }; i.e., x y.On the other hand, if } is closed, then y D()z =--> z E {y} {y}; i.e., z y.Since the singletons are either open or closed, it is evident that the length of (X, p()) is at most I.
If {x} is open and y D()x, then y x and hence x is minimal in (X, (T)).
Similarly if {x} is closed, then x is maximal in (X, 0()).Since (X, ) is ultraconnected any two minimal elements have an upper bound and there exists only one maximal element which will be the greatest element in (X, O(T)).Moreover, if x is minimal in (X, p(T)), then {x} is open and not closed.Hence T V(p(T)).Thus (X, T) is maximal ultraconnected, and by the above remark it is minimal T too.
NOTE 2.1.Though every maximal ultraconnected space is minimal T here are mlnmal T spaces which are not even ultraconnected.However, every mnlmal T space 2 2 is connected [4].
Let X be a set with 3 or more elements and AX such that IX%AI )2.
Then T {0XI0A or A0 and 0' finite is a minimal T topology, which is not ultraconnected.
For if x,y X%A, then {x} and } are c21osed subsets of (X,T)with empty intersection.
DEFINITION 2.2.A subset of a topological space is called ultraconnected if it is ultraconnected as a subspace.
REMARK 2.2.We will call two subsets A and B of a topological space (X, T)   equivalent (A --B) if every open set containing A contains B and conversely.
A --{0 E I0A} is the largest subset of X equivalent to A. Note that, if ABC and A C, then A B and B C.
THEOREM 2.2.Let A and B be subsets of a topological space (X, ) and A --B. Then A is ultraconnected iff B is ultraconnected. PROOF.
Suppose A is ultraconnected, but B is not.Then there exist two nonempty Since A E B, ACD I' D 2 and hence DI D2A @.But DI A , for otherwise A=D; ==> BD ==> C DlB .Similarly D2A .Since DIO A, D2OA are nonempty disjoining closed sets in A, we get a contradiction.Hence the result.DEFINITION

2.3.
A subset A of a topological space is generalized closed if AO and 0 whenever A0 and 0 is closed [3].
If A is a generalized closed subset of (X, T), then A is ultraconnected iff A is ultraconnected. PROOF.
In view of Theorem 2.2, it is sufficient to show that A m A. Since A is generalized closed, if A0 x, then AO.The other implication is trivial.COROLLARY 2.2.If A and B are subsets of a space (X, ) such that ABA then A is ultraconnected iff B is ultraconnected.

PROOF.
Since A BA and A A it follows that A E B (see the previous remark).Thus the conclusion is an immediate consequence of Theorem 2.2.DEFINITION

A subset A of a space X is called seml-open if there exists an
open set 0 such that OAO (Levlne [5]).A semi-homeomorphlsm is a blJection under which both images and inverse images of sem-open sets are seml-open.
A topological property invariant under semi-homeomorphisms is called a seml-topologlcal property by Crossley and Hildebrahd [6].

F-CONNECTED SPACES.
A topological space in which the intersection of any two nonempty open sets is nonempty is called hyperconnected [I].
We define a topological space to be F- connected if it is both hyperconnected and ultraconnected.
In the above remark (X, i is F-connected while (X, 2 is not.
Hence F-connectedness is not a seml-topologlcal property.Neither the Join nor the product of two F-connected topologies on a set are F-connected.Let T1 {' A,X} and 2 {,B,X} where AB .Then IV 2 and I x T2 are not F-connected but I and 2 are F-connected.
THEOREM 3.1.Every subspace of a topological space (X,) is F-connected iff is nested.
PROOF.Necessity: Assume T is not nested.Then there exist A,BX such that AB and BA.Choose x AB and y BA.Then the subspace {x,y} has the discrete topology which is obviously not F-connected.
P.M. MATHEW Sufficiency: Let be nested and A.X. Let 01, 02 be nonempty open sets in A. Then there exist BI, B 2 E such that 01 AB and 02 AB 2. Since is nested, BIC B 2 or B2 B I. Assume Blab2.Then OICO2 and hence 0102 .Similarly, the intersection of any two nonempty closed sets in A is also nonempty.
Thus A is F- connected.THEOREM 3.2.
If U is an ultrafilter on X {a}, for some a X, then z {,X}L) U is a maximal F-connected topology on X.
PROOF.Obviously, (X,z) is F-connected.Suppose (X,z I) is F-connected and I > " Let A e lZ.Since A e i, a A. For if a e A, then {a}(XA) , a contradiction since (X, zl) is ultraconnected.Now a A and A II implies (X\ {a}) A Thus A and(X{a})A are two nonempty disjoint open sets in (X, zl), a contradiction.Hence the result.THEOREM 3.3.Any compact, maximal F-connected topology on a set X is of the form {,X} II where II is an ultrafilter on X{a}, for some a e X.
PROOF.Let (X, z) be compact and maximal F-connected.Since the family of all the nonempty closed sets has finite intersection property and (x,z) is compact, it has nonempty intersection.Choose a {CX[C is closed and nonempty }.Thus the proper open sets are subsets of X\{a} and they form a filter base F.  Let be an a ultrafilter on X{a} containing F.
Then z{,X} U z Since (X, z) is maximal a a F-connected in view of Thoerem 3.2, z z a ACKNOWLEDGMENT.The author wishes to thank Professor T. Thrlvlkraman for his guidance during the preparation of this paper.He also wishes to thank the referee for the valuable comments which improved the presentation considerably.