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Semi-strong continuity.

Abstract.--The class of semi-strongly continuous functions is defined to generalize the notion of strong continuity. A function between topological spaces is semi-strongly continuous provided the preimage of each semi-open set is open. Semistrongly continuous functions are both continuous and irresolute. Equivalences of semi-strong continuity are given for some special cases.

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Levine (1960) introduced strongly continuous functions in the widely read Classroom Notes section of the American Mathematical Monthly. A well-known equivalence of continuity for a function f : X [right arrow] Y is that f([bar.A]) [subset or equal to] [bar.f(A)] for all A [subset or equal to] X. Strong continuity requires the more restrictive condition f([bar.A]) f(A) for all A [subset or equal to] X. It was proven that the following are equivalent: (1) f is strongly continuous, (2) the preimage of every set is closed, and (3) the preimage of every set is open.

Shortly thereafter, Levine (1963) introduced semi-open sets and the notion of semi-continuity. He defined a set A in a topological space to be semi-open if there exists an open set O such that O [subset or equal to] A [subset or equal to] [bar.O]. Semi-continuity is a modification of another established equivalence of continuity. A function f : X [right arrow] Y is continuous if and only if for every open set U [is less than or equal to] Y, [f.sup.-1](U) is an open set in X. Semi-continuity generalizes continuity by requiring that preimages of open sets be semi-open.

Anderson & Jensen (1967) considered functions f such that both f and [f.sup.-1] preserve semi-open sets. Such functions were shown to be both continuous and open. In this Texas Journal of Science, Crossley & Hildebrand (1971) continued the investigation of semiopen sets by defining the semi-closure of a set, and in a subsequent paper (Crossley & Hildebrand, 1972) introduced irresolute functions. An irresolute function is one for which preimages of semi-open sets are semi-open, and a semi homeomorphism is a bijective irresolute function whose inverse is also irresolute. Semi homeomorphisms are the functions studied by Anderson & Jensen (1967).

The notion of semi-continuous generalizes both continuous and irresolute, and each of these generalizes strongly continuous. That is, every strongly continuous function is both continuous and irresolute, every continuous function is semi-continuous, and every irresolute function is semi-continuous. It was shown by example that neither irresolute nor continuous implies the other.

Thus, the following hierarchy was established:

Interest in variations of continuity continued. Piotrowski (1979) further investigated semi homeomorphisms and Noiri & Ahmad (1985) introduced semi-weakly continuous functions and these functions were further characterized by Dorsett (1990). Other weaker forms of continuity and semi-continuity such as faint continuity, subweak continuity, and almost weak continuity were investigated by Noiri (1987).

This article proposes a new property of functions called semistrong continuity. Semi-strongly continuous functions are more restrictive than continuous and irresolute functions by requiring preimages of all semi-open sets to be open. It will be shown in the next section that this class of functions precedes continuous and irresolute and follows strongly continuous in the earlier hierarchy.

IMPLICATIONS AMONG VARIATIONS OF CONTINUITY

All functions in this article are assumed to be between topological spaces.

Definition 1. (Semi-Strongly Continuous) Let f: X [right arrow] Y be a function and p a point of X. Then f is semi-strongly continuous at p provided for each semi-open set B in Y containing f(p), there exists an open set U in X containing p such that f(U) [subset or equal to] B. The function f is semi-strongly continuous provided it is semi-strongly continuous at each point of X.

It follows from this definition that a function is semi-strongly continuous if and only if the preimage of every semi-open set is open.

When the preimage of every set is open, then certainly the preimage of every semi-open set is open. Thus, strong continuity implies semi-strong continuity. Now, if the preimage of every semi-open set is open, then because every open set is semi-open, it follows that the preimage of every open set is open. Thus, semistrong continuity implies continuity. Similarly, if the preimage of every semi-open set is open, it follows that the preimage of every semi-open set is semi-open. Thus, semi-strong continuity implies irresolute. Now, given a constant function, the preimage of any set is either empty or the entire domain space. Therefore, under a constant function, every set has an open preimage. Hence every constant function is strongly continuous.

To see that semi-strong continuity implies neither constant nor strong continuity, consider the following example.

Example 1. Let X = {a, b, c} have the topology [tau] = {[empty set], {a}, {a, b}, {a, b, c}} and Y = {0,1,2} have the topology [sigma] = {0, {0}, {0,1,2}}. It is easily shown that the semi-open subsets of Fare the open subsets of Y along with {0,1} and {0,2}. Define the function f: X [right arrow] Y by f = {(a, 0), (b, 0), (c, 2)}. Notice this function is not constant and is not strongly continuous since there are subsets of Ywhose preimages are not open in X. (The preimage of {2} is {c} which is not open.) However, the preimage of each semi-open set in Y is open in X. Thus, semi-strong continuity is not equivalent to strong continuity.

To see that neither continuity nor irresolute implies semi-strong continuity, consider this example.

Example 2. Let f: E [right arrow] E, where both spaces have the usual topology, be defined by f(x) = x. Then f is continuous on E, and since the preimage of any semi-open set is itself, and thus semiopen, f is irresolute. However, f is not semi-strongly continuous at the point p = 1. Observe that (0,1] is a semi-open set in the codomain that contains f(1) = 1. Further, there is no open set in the domain containing p = 1 whose image is a subset of (0,1]. Thus, f is not semi-strongly continuous at p and hence, f is not semi-strongly continuous.

The following is now established:

Constant [??] Strongly Continuous [??] Semi-Strongly Continuous

Furthermore, semi-strongly continuous precedes the double implication to continuous and irresolute as given in the earlier hierarchy diagram.

EQUIVALENCES OF SEMI-STRONG CONTINUITY IN SPECIAL CASES

An investigation of semi-strongly continuous functions into the set of real numbers R reveals a relationship between semi-strong continuity and constant functions as shown in the following example.

Example 3. Let f: R [right arrow] R be a semi-strongly continuous function from a topological space X into the space of real numbers with the usual topology, and let p be any point of the domain X. Construct an open interval of radius 1 about f(p), (f(p), (f(p) - 1 + 1). Note that this interval is the union of the two semiopen sets A = f(p) - 1, f(p)] and B = [f(p), f(p) + 1). Furthermore, A and B intersect in a single point, f(p).

By the semi-strong continuity of f at p, there exist open sets O and U in X, each containing p, such that f(O) [subset or equal to] A and f(U) [subset or equal to] B. Thus, O [intersection] U is an open set containing p such that f(O [intersection] U) [subset or equal to] f(O) [intersection] f(V) = {f(p)}- Therefore, f is constant on O [intersection] V, a neighborhood of p.

Hence, any semi-strongly continuous function into 5f must be constant on some neighborhood of each point of the domain.

It is not the case that all semi-strongly continuous functions behave like the function of Example 3. Example 1 exhibits a semistrongly continuous function which has a point c in its domain such that the function is not constant on any neighborhood of c. The codomain of this function has a topology much different from the real numbers.

Investigating the topological structure of the real numbers to identify the characteristic that was exploited in Example 3 leads to the following definition of a property of spaces that admit semistrongly continuous functions only if they are constant on some neighborhood of each point of the domain. Using this property, some equivalences of semi-strong continuity will be established.

Definition 2. (Pointwise semi-open intersection property) A topological space X is said to have the pointwise semi-open intersection property provided for every x [member of] X, there exist semiopen sets A [subset or equal to] X and B [subset or equal to] X such that {x} = A [intersection] B.

Any set with its discrete topology is a space with the pointwise semi-open intersection property since each singleton set in such a space is open, and hence semi-open. The real line with its usual topology has the pointwise semi-open intersection property, since for any real number x, {x} = (x - 1, x] [intersection] [x, x + 1). It was this structure of the real numbers that was used in Example 3. The following theorem establishes a more general result.

Theorem 1. If f: X [right arrow] Y is a function and Y has the pointwise semi-open intersection property, then f is semi-strongly continuous at a point p [member of] X if and only if there exists a neighborhood of p on which f is constant.

Proof. Let f: X [right arrow] Y be a function, let Y have the pointwise semi-open intersection property and let p [member of] X.

First, suppose that there exists a neighborhood of p on which f is constant. Then there exists an open set O [subset or equal to] X such that p [member of] O and f(O) = {f(p)}. For any semi-open set B [subset or equal to] Y, we have the open set O containing p such that f(O) [subset or equal to] B. Therefore f is semistrongly continuous at p.

Now let f be semi-strongly continuous at p. Since Y has the pointwise semi-open intersection property, there exist semi-open sets A, B [subset or equal to] Y such that {f(p)} = A [intersection] B. Thus f(p) [member of] A and f(p) [member of] B. Since f is semi-strongly continuous at p, there exist neighborhoods O and U of p such that f(O) [subset or equal to] A and f(U) [subset or equal to] B. Therefore, O [intersection] U is an open set in X containing p such that f(O [intersection] U) [intersection] A and f(O [intersection]) [subset or equal to] B. Thus f(O [intersection] U) [subset or equal to] A [intersection] B.

Since {f(p)} = A [intersection] B, f(O [intersection] U) [subset or equal to] {f(p)}. Therefore, there exists a neighborhood of p on which f is constant.

Certainly a function can be constant on some neighborhood of each point of the domain and not be a constant function. The example below gives a semi-strongly continuous function into the real numbers which is not a constant function.

Example 4. Let X be any disconnected space. Then X is the union of two disjoint nonempty open sets, U and V. Define f : X [right arrow] R by f(x) = 0 for each x [member of] U and f(x) = 1 for each x [member of] V. Thus, f is not a constant function, but since f is constant on a neighborhood of each point of X, then by Theorem 1, f is semistrongly continuous on X.

A comparison of the semi-strongly continuous functions in Examples land 4 shows the domain in Example 4 is not connected while the domain in Example 1 is connected. However, the codomain of the function in Example 1 does not have the pointwise semi-open intersection property as is a required in order to apply Theorem 1. Thus, Example 4 leads to the next result by suggesting that if Theorem 1 is altered by including the additional assumption that the domain is connected, then semi-strong continuity is equivalent to being a constant function on the entire domain.

Theorem 2. Let X and Y be topological spaces such that X is connected and Y has the pointwise semi-open intersection property. Then a function f: X [right arrow] Y is semi-strongly continuous on X if and only if f is a constant function on X.

Proof. Suppose f: X [right arrow] Y is a function from a connected space X to a space Y where Y has the pointwise semi-open intersection property. Assume first that f is semi-strongly continuous on X. By way of contradiction, assume that f is not a constant function. It will be shown that X is not connected by showing X is the union of two disjoint, nonempty, open sets. To this end, let a be any point of X and let S = {x [member of] X | f(x) = f(a)}. Since f is not a constant function, then S [not equal to] X. Let [x.sub.0] be an arbitrary point of S. By Theorem 1, there exists an open set U containing [x.sub.0] on which f is constant. Thus, each point of It has the same image as [x.sub.0] and hence the same image as a. Therefore, [x.sub.0] [member of] [subset or equal to] S. This proves that S is an open set.

It will now be shown that X - S, the complement of S, is also an open set. Recall that X [not equal to] S and thus X - S is nonempty. Let [y.sub.0] be an arbitrary point of X - S. By Theorem 1, there exists an open set V containing [y.sub.0] on which f is constant. Since [y.sub.0] [not member of] S then f([y.sub.0]) [not equal to] f(a), and thus no point of V has an image equal to f(a). Therefore, no point of V belongs to S. Hence, [y.sub.0] [member of] V [subset or equal to] (X - S). This proves that X - S is an open set.

Since X is the union of the two disjoint, nonempty, open sets S and X - S, then X is a disconnected space. This is a contradiction, thus proving that if f is semi-strongly continuous, then f is a constant function.

Now, since every constant function is strongly continuous, and strong continuity implies semi-strong continuity, then the converse follows.

Because strongly continuous functions lie between constant and semi-strongly continuous functions in the now-established hierarchy, Theorem 2 establishes the equivalence of semi-strongly continuous, strongly continuous, and constant for functions from connected spaces into spaces with the pointwise semi-open intersection property.

The following theorem relates the pointwise semi-open intersection property to a class of familiar topological spaces. This will lead to a characterization of semi-strongly continuous functions f: X [right arrow] Y where Y is a first countable Hausdorff space.

Theorem 3. If X is a first countable Hausdorff space, then X has the pointwise semi-open intersection property.

Proof. Let X be a first countable Hausdorff space, and let x be a point of X. In case {x} is an open set, then it is also semi-open and is thus the intersection of the semi-open sets {x} and {x}.

Assume then, that {x} is not an open set. It must be the case that every open set containing x contains some point of the space distinct from x. Consider now an arbitrary open set G containing x. Then there exists a point p of G distinct from x. Since X is Hausdorff, there exist disjoint open sets V and W containing p and x, respectively. Now, since W [intersection] G is an open set containing x, there exists a point q [member of] W [intersection] G distinct from x. Using the

Hausdorff properly again, q and x can be separated by disjoint open sets H and K containing q and x, respectively.

The sets V [intersection] G, H [intersection] W [intersection] G and K [intersection] W [intersection] G are mutually disjoint open subsets of G containing p, q and x, respectively. Therefore, for any open set G containing x, there exist two points p and q and three mutually disjoint open subsets of G containing p, q and x, respectively.

Since X is first countable, then x has a countable neighborhood base {[G.sub.n]}.sub.n[member of]N] of open sets. By the preceding argument, there exist points [p.sub.1] and [q.sub.1] and mutually disjoint open sets [U.sub.1], [V.sub.1], and [W.sub.1] such that [U.sub.1] [subset] [V.sub.1] [subset] [W.sub.1] [subset or equal to] [G.sub.1], [p.sub.1] [member of] [U.sub.1], [q.sub.1] [member of] [V.sub.1], and x [member of] [W.sub.1].

Focusing now on the open set [W.sub.1] [intersection] [G.sub.2] containing x, the procedure can be repeated to produce points [p.sub.2] and [q.sub.2] and mutually disjoint open sets [U.sub.2], [V.sub.2], and [W.sub.2] such that [U.sub.2] [subset] [V.sub.2] [subset] [W.sub.2] [subset or equal to] ([W.sub.1] [intersection] [G.sub.2]), [p.sub.2] [member of] [U.sub.2], q [member of] [V.sub.2], and x [member of] [W.sub.2]. Sequences of points {[p.sub.n]} and {[q.sub.n]} and collections of open sets {[U.sub.m]}, {[V.sub.n]} and {[W.sub.n]} can be defined inductively as follows.

Let n be a positive integer and assume {[p.sub.1], ..., [p.sub.n]}, {[q.sub.1], ..., [q,sub.n]}, {[U.sub.1], ..., [U.sub.n]}, and {[W.sub.1] ..., [W.sub.n]} have been defined such that for 2 [less than or equal to] i [less than or equal to] n, [U.sub.i] [V.sub.1], and [W.sub.i] are mutually disjoint open sets such that [U.sub.i] [subset] [V.sub.i] [subset] [W.sub.i] [subset or equal to] ([W.sub.i-1] [intersection] [G.sub.i]), [p.sub.i] [member of] [U.sub.i], [q.sub.i] [member of] [V.sub.i], and x [member of] [W.sub.n+1]. The points [p.sub.n+1], [q.sub.n+1], and the open sets [U.sub.n+1], [V.sub.n+1] and [W.sub.n+1] can be defined using the already established procedure as follows. Since [G.sub.n+1] [intersection] [W.sub.n] is an open set containing x, there exist points [p.sub.n+1] and [q.sub.n+1] as well as mutually disjoint open sets [U.sub.n+1], [V.sub.n+1], and [W.sub.n+1], subsets of [G.sub.n+1] [intersection] [W.sub.n], such that [p.sub.n+1] [member of] [U.sub.n+1], [q.sub.n+1] [member of] [V.sub.n+1] and x [member of] [W.sub.n+1].

Let U = U{[U.sub.n] | n [member of] N} and V = U{[V.sub.n] |n [member of] N}. Then U and V are disjoint open sets neither containing x. Furthermore, x is a limit point of each of U and V. To see this, consider an arbitrary open set O containing x. Since [([G.sub.n]}.sub.n[member of]N] is a neighborhood base at x, there exists some positive integer m such that x [member of] [G.sub.m] [subset or equal to] O. According to the preceding construction, [p.sub.m] and [q.sub.m] are points of O distinct from x such that [p.sub.m] [member of] [U.sub.m] [subset or equal to] [G.sub.m] [subset or equal to] O and [q.sub.m] [member of] [V.sub.m] [G.sub.m] [subset or equal to] O. Since [p.sub.m] [member of] U and [q.sub.m] [member of] V, then O contains a point of U distinct from x, and O contains a point of V distinct from x. Hence x is a limit point of both U and V. This guarantees that the sets U [subset] {x} and V [subset] {x} are semi-open since U [subset or equal to] U [subset] {x} [subset or equal to] [bar.U] and V [subset or equal to] V [subset] {x} [subset or equal to] [bar.V]. Since U and V are disjoint, then (U [subset] {x}) [intersection] (V [subset] {x}) = (x). This establishes that {x} is the intersection of two semi-open sets. Therefore X has the pointwise semi-open intersection property.

The preceding theorem now yields the following corollaries to Theorems 1 and 2, respectively.

Corollary 1. If f: X [right arrow] Y is a function and Y a first countable Hausdorff space, then f is semi-strongly continuous at a point x [member of] X if and only if f is constant on some open set containing x.

Corollary 2. If f is a function from a connected space X to a first countable Hausdorff space Y, then the following are equivalent:

(1) f is semi-strongly continuous, (2) f is strongly continuous, and (3) f is a constant function.

CONCLUSIONS

Semi-strong continuity has been established to be more general than strong continuity, yet more restrictive than continuity, semi-continuity and irresoluteness. None of these implications is reversible, but it may be possible to determine additional reasonable conditions on the functions under which these various types of continuity are equivalent.

Imposing conditions on the domain and codomain of the function allowed the establishment of some equivalences of semi-strong continuity. To this end, the pointwise semi-open intersection property was defined. This new property should be investigated further to more precisely determine how it relates to other well-known topological properties. This could lead to other conditions on the domain or codomain of a function, which would imply semi-strong continuity.

LITERATURE CITED

Anderson, D. R. & J. A. Jensen. 1967. Semi-continuity on topological spaces. Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur. 8:782-783.

Crossley, S. G. & S. K. Hildebrand. 1971. Semi-closure. Texas J. Sci., 22:99-112.

Crossley, S. G. & S. K. Hildebrand. 1972. Semi-topological properties. Fund. Math., 74(3):233-254.

Dorsett, C. 1990. Semi-continuous and semi-weakly continuous functions. Kyungpook Math. J., 30(1):95-100.

Levine, N. 1960. Strong continuity. Amer. Math. Monthly, 67:269.

Levine, N. 1963. Semi-open sets and semi-continuity in topological spaces. Amer. Math. Monthly, 70:36-41.

Noiri, T. & B. Ahmad. 1985. On semiweakly continuous mappings. Kyungpook Math. J., 25(2): 123-126.

Noiri, T. 1987. Properties of some weak forms of continuity. Internat. J. Math. Math. Sci., 10(1):97-111.

Piotrowski, Z. 1979. On semi-homeomorphisms. Boll. Un. Mat. Ital. B, 5( 16-A):501 509.

PDR at: roberson@sfasu.edu

Pamela D. Roberson and Jennifer M. Scheers

Department of Mathematics & Statistics, Stephen F. Austin State University, Nacogdoches, Texas 75962
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Author:Roberson, Pamela D.; Scheers, Jennifer M.
Publication:The Texas Journal of Science
Date:Feb 1, 2012
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