SUBCONTRA-CONTINUOUS FUNCTIONS

A weak form of contra-continuity, called subcontra-continuity, is introduced. It is shown that subcontra-continuity is strictly weaker than contra-continuity and stronger than both subweak continuity and sub-LC-continuity. Subcontra-continuity is used to improve several results in the literature concerning compact spaces.


INTRODUCTION
In Dontchev introduced the notion of a contra-continuous function.In this note we develop a weak form of contra-continuity, which we call subcontra-continuity.We show that subcontra- continuity implies both subweak continuity and sub-LC-continuity.We also establish some of the properties of subcontra-continuous functions.In particular it is shown that the graph of a subcontra- continuous function into a Tl-space is closed.Finally, we show that many of the applications of contra-continuous functions to compact spaces established by Dontchev  [1] hold for subcontra- continuous functions.For example, we establish that the subcontra-continuous, nearly continuous image of an almost compact space is compact and that the subcontra-continuous, -continuous image of an S-closed space is compact.

PRELIMINARIES
The symbols X and Y denote topological spaces with no separation axioms assumed unless explicitly stated.The closure and interior of a subset A of a space X are signified by Cl(A) and Int(A), respectively.A set A is regular open (semi-open, nearly open) provided that A Int(Cl(A))(A C_ Cl(Int(A)), A C_ Int(Cl(A))) and A is regular closed (semi-closed) if its complement is regular open (semi-open).A set A is locally closed provided that A U N F, where U is an open set and F is a closed set.DEFINITION 1. Dontchev [1 ].A function f X-, Y is said to contra-continuous provided that for every open set V in Y, f-1(V) is closed in X. DEFINITION 2. Rose [2].A function f X-, Y is said to be subweakly continuous if there isan open base B for the topology on Y such that Cl(f-l(V)) C_ f-(Cl(V)) for every V E B. DEFINITION 3. Ganster and Reilly [3].A function f" X-.Y is said to be sub-LC- continuous provided there is an open base 3 for the topology on Y such that f-(V) is locally closed for every V 6/3.i)EFINITION 4. A function f" X Y is said to be semi-continuous (Levine [4]) (nearly continuous (Ptak [5]), 3-continuous (Abd EI-Monsef et al. [6]) DEFINITION 5. Gentry and Hoyle [7].A function f" X--, Y is said to be c-continuous if, for every :r X and every open set V in Y containing f(z) and with compact complement, there exists an open set U in X containing z such that f(U) C_ V.

SUBCONTRA-CONTINUOUS FUNCTIONS
We define a function f X-Y to be subcontra-ontinuous provided there exists an open base B for the topology on Y such that f-(V) is closed in X for every V /3.Obviously contra- continuity implies subontra-ominuity.The following example shows that the reverse implication does not hold.EXAMPLE 1.Let X be a nondiscrete Tx-space and let Y be the set X with the discrete topology.Finally let f X-Y be the identity mapping.If/3 is the ollection of all singleton subsets of Y, then/3 is an open base for the topology on Y. Since X is T, f is subcontra-ontinuous with respect to/3.Obviously f is not contra-continuous.Subcontra-ontinuity is independent of continuity.The function in Example is subcontra- continuous but not continuous.The next example shows that ontinuity does not imply subcontra- continuity.
EXAMPLE 2 Let X {a, b} be the Sierpinski space with the topology T {X, , {a} } and let f X--, X be the identity mapping.Obviously f is continuous.However, any open base for the topology on X must contain {a} and f-({a}) is not closed.It follows that f is not subcontra- continuous.
Since closed sets are locally closed, subcontra-continuity implies sub-LC-continuity.We see from the following theorem that subcontra-cominuity also implies subweak continuity.THEOREM 1.Every subcontra-continuous function is subweakly continuous.
PROOF.Assume f X--, Y is subcontra-continuous. Let/ be an open base for the topology on Y for which f-'(V) is closed in X for every V /.Then for V /, Cl(f-X(V)) f-(V) C_ f-(l(v)) and hence f is subweakly continuous.I-! Since a subweakly continuous function into a Hausdorff space has a closed graph (Baker [8]), a subcontra-continuous function into a Hausdorff space has a closed graph.However, the following stronger result holds for subcontra-continuous functions.THEOREM 2. If f X--, Y is a subcontra-continuous function and Y is T, then the graph of f, G(f), is closed.PROOF.Let (x,y) 6 X x Y G(f).Theny # f(x).LetB be an open base for the topology on Y for which f-x (V) is closed in X for every V 6 B. Since Y is T, there exists V 6 B such that y e v and f(x) : V. Then we see that (x, y) 6 (X -/-a(v)) x V c_ x x Y G(I).t follows that G(/') is closed, l'-! COROLLARY 1.If f" X Y is contra-continuous and Y is T,, then the graph of f is closed.
Long and Hendrix [9] proved that the closed graph property implies c-continuity.Therefore we have the following corollary.COROLLARY 2. If f X-Y is subcontra-continuous and Y is T, then f is c-continuous.
The next two results are also implied by the closed graph property (Fuller 10]).COROLLARY 3. If f X-Y is subcontra-continuous and Y is T1, then for every compact subset C of Y, f-l(u) is closed in X. COROLLARY 4. If f X Y is subcontra-eontinuous and I/is T1, then for every compact subset C of X, f(C') is closed.
For a function f" X--, I/, the graph function of f is the function g" X-,X x I/given by g(x) (z,f(x)).We shall see in the following example that the graph function of a subeontra- continuous function is not necessarily subcontra-continuous.
EXAMPLE 3. Let X {a,b} be the Sierpinski space with the topology T {X,0, and let f" X--, X be given by f(a) b and f(b) a. Obviously f is subeontra-continuous, in fact eontra-continuous.Let/3 be any open base for the product topology on X x I,'.Then there exists V E/3 for which (a,b) V C_ {(a,a),(a,b)}.We see that V= {(a,a), (a,b)) and that, ifg" X-, X x X is the graph function for f, then g-l(V) {a} which is not closed.Thus the graph function of f is not subeontra-continuous.
However, the following result does hold for the graph function.
THEOREM 3. The graph function of a subcontra-continuous function is sub-LC-continuous.
PROOF.Assume f" X-Y is subcontra-continuous and let g" X -X x I/be the graph function of f.Let/3 be an open base for the topology on Y for which f-(V) is closed in X for every V /3.Then {U x V U is open in X, V /3} is an open base for the product topology on X Y.
Since g-(U x V) U n f-l(V), we see that g is sub-LC-continuous.El The graph function of a subweakly continuous function is subweakly continuous (Baker [8]) and the graph function of a sub-LC-continuous function is sub-LC-continuous (Ganster and Reilly [3]).
It follows that the graph function in Example 3 is subweakly continuous and sub-LC-continuous but not subcontra-continuous.Therefore subcontra-continuity is strictly stronger than sub-LC-continuity and subweak continuity.THEOREM 4. If Y is a Tl-space and f X-, Y is a subcontra-continuous injection, then X is T.
PROOF.Let x and x2 be distinct points in X. Let/3 be an open base for the topology on for which f-(V) is closed in X for every V /3.Since I/is T1 and f(:rl) 4 f(:r2), there exists V /3 such thatf(x) V and f(x) E V. Then aq .X-f-(V) which is open and 2 X-/-I(v).D THEOREM 5. Let A C_ X and f-X-X be a subcontra-continuous function such that f(X) A and flA is the identity on A. Then, if X is TI, A is closed in X.
PROOF.Suppose A is not closed.Let z Ul(A) A. Let/3 be an open base for the topology on Y for which f-I(V) is closed for every V /3.Since :r A, we have that a: f(:r).
Since X is T 1, there exists V /3 such that x E V and f(x) V. Let U" be an open set containing z.
Then :r U f V which is open.Since z UI(A), (U f V) q A .Let y (U f V) q A. Since y A, f(v) V V.So V f-l(V) Thus V U f f-I(V) and hence U f-l(V) :/: .We see The next result follows easily for the definition.THEOREM is a union of closed sets in X.
Obviously every function with a Tl-domain satisfies the above condition.However, as we see in the following example, a function with a Tl-domain can fail to be subcontra-continuous.It follows that the converse of Theorem 6 does not hold.. C.W. BAKER EXAMPLE 4. Let X R with the usual topology and let f X X be the identity mapping.Since X isconnected, f is not subcontra-continuous.However, since X is Tx, f has the property that the inverse image of every (open) set is a union of closed sets.

APPLICATIONS TO COMPACT SPACES
In [1] Dontchev establishes that the image of an almost compact space under a contra- continuous, nearly continuous mapping is compact and that the contra-continuous image of a strongly S-closed space is compact.In this section, we strengthen both of these results by replacing contra- continuity with subcontra-continuity.The proofs mostly follow Dontchev's.DEFINITION 6. Dontchev ].A space X is almost compact provided that every open cover of X has a finite subfamily the closures of whose members cover X. THEOREM 7. The image of an almost compact space under a subcontra-continuous, nearly continuous mapping is compact.
PROOF.Let f X Y be subcontra-continuous and nearly continuous and assume that X is almost compact.Let B be an open base for the topology on Y for which f-l(v) closed in X for every V 6 B. Let C be an open cover of f(X).For each z 6 X, let C C such that f(z) C. Then let Vz B for which f(z) Vx c_ C'x.Now f-X(Vz) is closed and nearly open.It follows that f-l(V) is clopen and hence that {f-X(Vx) :c 6 X} is a clopen cover of X.Since X is almost compact, there is a finite subfamily {f-X(Vx,): i= 1,...,n} forwhich X U Cl(f-l(vz,)) U f-x(Vx,) c_ i=I i=I [3 f-1(Cz,).Thus we have tha f(X) c_ [3 C, and therefore tha f(X) is compact [3 i=1 i=1 DEFINITION 7. Dontchev [1].A space X is strongly S-closed provided that every closed cover of X has a finite subcover.THEOREM $.The subeontra-ontinuous image of a strongly S-closed space is compact.
PROOF.Let f X -Y be subcontra-continuous and assume that X is strongly S-closed.Let   B be an open base for the topology on Y for which f-(V) is closed in X for every V B. Let C be an open cover of f(X).For each z X, let 6' C with f(z) 6 C. Then let V 6 B for which f(z) V C_ C. Since {f-X(Vx) z X} is a closed cover of X and X is strongly S-closed, there is a finite subcover {f-X(V,) l n} of X.Then we see that f(X) f ( i=1 I,J f (f-1(Vz,)) c_ U Vx, C_ U Cx, and hence that f (X) is compact.E! i=1 i=1 i=1 In [1] Dontchev also shows that the contra-continuous,/-continuous image of an S-closed space is compact.We extend this result by replacing contra-continuity with subcontra-continuity.The proof parallels that of Dontchev's.DEFINITION 8. Mukherjee and Basu [11 ].A space X is S-closed provided that every semi- open cover of X has a finite subfamily the closures of whose members covers X.
From Herrmann [12], a space X is S-closed if and only if every regular closed cover of X has a finite subcover.
THEOREM 9.The subcontra-continuous, /-continuous image of an S-closed space is compact.
PROOF.Assume that f X--, Y is subcontra-continuous and/-continuous and that X is S- closed.Let B be an open base for the topology on Y for which f-X(V) is closed in X for every V B. Let C be an open cover of f(X).Then for each z X there exists C 6 for which f(z) E Cz.For each z e X, let Vx B such that f(z) 6 V C_ Cx.Since f is subcontra-continuous, {f-(Vx):xX} is a closed cover of X.The ]-continuity of f implies that f-X(V) C_ Cl(Int(Ol(f-X(V,:)))) and therefore we see that f-l(vz)= Cl(Int(f-(V))) or that f-X(Vx) is regular closed.Since X is S-closed, the regular closed cover {f-X(Vx) :z E X} has a finite subcover {f-l(V,) 1 n}.Then we have I(X) I U J'-l(Vx,) -U V:, _ i=1 i=1 U C, and therefore J'(X) is compact.KI i=1 In the above proof we showed that, if J' X Y is subcontra-continuous and -continuous, then there exists an open base B for the topology on Y such that for every V E B, f-l(v) is regular closed and hence semi-open.Since unions of semi-open sets are semi-open (Arya and Bhamini [13]), it follows that inverse images of open sets are semi-open.Therefore we have the following theorem which strengthens the corresponding result for contra-continuous functions established by Dontchev [11.TI-IEOREM 10.Every subcontra-continuous, /-continuous function is semi-continuous.