Microsoft Word - 179-185   179 Mathematics | https://doi.org/10.30526/31.2.1956 2018) عام 2العدد( 31المجلد مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 31 (2) 2018 On 𝛉-Totally Disconnected and 𝛉-Light Mappings Haider Jebur Ali Huda Fadel Abass Dept of Mathematics / College of Science / Al-Mustansiriyah University, haiderali89@yahoo.com huda.fadel91@gmail.com Received in:24/January/2018, Accepted:27/March/2018 Abstract In our research, we introduced new concepts, namely 𝜃, 𝜃*and 𝜃**-light mappings, after we knew 𝜃, 𝜃*and 𝜃**-totally disconnected mappings through the use of 𝜃-open sets. Many examples, facts, relationships and results have been given to support our work. Keywords: 𝜃-open set, light mapping, 𝜃-homeomorphism function, 𝜃-totally disconnected set, 𝜃-light mapping.   180 Mathematics | https://doi.org/10.30526/31.2.1956 2018) عام 2العدد( 31لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 31 (2) 2018 Introduction Many researchers studied the light mappings such as the world’s J.J.Charatonic and K.Omiljanowski[2]. In this paper, we provide other types of light mappings namely 𝜃- light open mapping. Other scientists who studied the light mappings are the word M. Wldyslaw [5], M. K. Fort [3] and G. Sh. mohammed [1] and others. In our work, we needed some basic definitions. Let (X, Ʈ) be topological space and A be a subset of X, a point x∈A is said to be 𝜃-interior point to A if x∈ 𝑈 ⊆ 𝐴 for some U∈ 𝜏 containing x. The set of all 𝜃-interior points are called 𝜃-interior set and we denoted by 𝜃 𝑖𝑛𝑡 A , a subset U of topological pace X is 𝜃-open if and only if every point in U is a interior point [7]. Every 𝜃-open set is an open set but the converse may not be true in general. A space X is said to be 𝜃-Hausdorff if for every distinct point x, y∈X there exist 𝜃-open sets Ux, Vy containing x and y respectively such that Ux∩ Vy =∅[4]. A mapping f:X→Y is said to be 𝜃-open(𝜃*-open and 𝜃**-open) if f(V) is 𝜃-open(open and 𝜃-open) in Y, whenever V is open (𝜃-open) in X [6].  Let X and Y be spaces and let f be a mapping from X into Y then f is said to be 𝜃-homeomorphism if f is bijective, continuous and 𝜃-closed (𝜃-open) [6]. A space X is said to be totally disconnected space if for every pair of distinct points, a, b ∈X has a disconnection A∪B to X such that a ∈ A and b ∈ B [8]. A surjective mapping f:X→Y is said to be totally disconnected mapping if and only if for every totally disconnected set U in X, f(U) is totally disconnected set in Y [1]. Definition(1): Let X be topological space, and let A and B are nonempty 𝜃-open sets in X, then A∪B is said to be 𝜃-disconnection in X if and only if A∪B=X and A∩B =∅. Definition(2): Let X be topology space, G⊆X, let A, B are nonempty 𝜃-open sets in X, then A∪B is said to be 𝜃-disconnection in G if and only if satisfy the following: 1- G∩A ∅. 2- G∩B ∅. 3-(G∩A)∩(G∩B)=∅. 4-(G∩ A)∪(G∩B)=G. Example (3): Let X={a, b, c} and let ƮD is discrete topology define to X. Then {a}, {b, c} are 𝜃-disconnection to X and {a}, {b, c} are 𝜃-disconnection to subset {a, b} to X. *Its known that every 𝜃 open set s is open but the converse may be not true. Example (4): (R, Ʈcof ) the open subsets of R is open set but not 𝜃-open. Definition(5): A topology space X is said to be 𝜃-totally disconnected if for every two distinct point p & q there exist 𝜃-disconnection G∪H to X such that P∈G & q∈H.   181 Mathematics | https://doi.org/10.30526/31.2.1956 2018) عام 2العدد( 31لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 31 (2) 2018 Example (6): The rational numbers with relative usual topology is a 𝜃-totally disconnected. Since if we take q1&q2 ∈Q where q1 q2 there exist r∈Qc such that q1 r q2 G={x∈Q:x 𝑟} and H={ x∈Q:x r} Then G∪H is 𝜃-disconnection to Q such that q1∈ 𝐺 & q2∈H 𝐺inG=G & 𝐻inH=H So Q is a 𝜃-totally disconnected. Proposition (7): Every 𝜃-totally disconnected set is totally disconnected. Proof: Let X be 𝜃-totally disconnected space to prove X is totally disconnected space. Let x,y∈X with x y. So there exist a 𝜃-totally disconnection to X (I mean there exist G and H which are 𝜃-open sets and G, H ∅ and G∪H=X , G∩H=∅ with x∈G, y∈H). But every 𝜃-open set is open set soX is totally disconnected space. Remark (8): The converse of above proposition is not true in general but in discrete space it is availed. Definition (9): A surjective mapping f:X→Y is said to be 𝜃-light mapping if for every y∈Y, f-1(y) is 𝜃-totally disconnected set. Example(10): Let(Q, ƮD) to topological space such that ƮD is the discrete topology define to the rational number Q and let (Q, Ʈind) is the indiscrete topology such that k∈R.Let f:(Q, ƮD)→(Q, Ʈind) is a mapping define the following: f(x)=0.5 for each x∈Q note that f-1(x)=Q if x =0.5 and f-1(x)=∅ when x 0.5 where ∅ and Q are 𝜃-totally disconnected. Then f is 𝜃-light mapping. Remark (11): Every 𝜃-totally disconnected is 𝜃-hausdorff but the converse may be not true in general for example: Example (12): (R, Ʈu) is 𝜃-hausdorff but not 𝜃-totally disconnected, where R is the set of real number .To show that (R, Ʈu) is not 𝜃-totally disconnected. Let x &y ∈Q⊆R such that x y, x y. Then ∃ p∈Qc such that x p y, (p, ∞)&(-∞, p) are 𝜃-open sets in R since P-1∈(-∞, p) there exist (-∞, p-1], p-1 ∈ (-∞, p-1] ⊆ (-∞, p) where ∞, 𝑝 =(-∞, p] the set (p, ∞) is similar. (p, ∞)∩(-∞, p)=∅, but (p,∞)∪(-∞, p) R (R has no 𝜃-disconnection)   182 Mathematics | https://doi.org/10.30526/31.2.1956 2018) عام 2العدد( 31لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 31 (2) 2018 So (R, ƮU) is not 𝜃-totally disconnected. Definition (13): A surjective mapping f:X→Y is said to be 𝜃-totally disconnected if and only if for every totally disconnected set U⊆X then f(U) is 𝜃-totally disconnected in Y. Definition (14): A surjective mapping f:X→Y is said to be 𝜃*-totally disconnected mapping if and only if for every 𝜃- totally disconnected set U⊆X then f(U) is totally disconnected Examples (15):1-Let f: (R, Ʈu)→(R, ƮD) such that f(x)=x for each x∈R . Since (Q, Ʈu) is totally disconnected set in (R, Ʈu) and f(Q)=Q⊆(R,ƮD) For each x, y∈Q there exist p∈Qc such that xp} are two open sets in (Q, Ʈu) such that G∪H=Q, G∩H=∅ Now to prove (Q, ƮD) is 𝜃-totally disconnected in (R, ƮD) where f(Q)=Q. G={x∈Q:x≤0} is 𝜃-open set in (Q, ƮD) H={x∈Q:x>0} is 𝜃-open set in (Q, ƮD) H∪G=Q, H∩G=∅ So (Q, ƮD) is 𝜃-totally disconnected in (R, ƮD) . 2- If we replace Q by (a, b] then the sets G={x ∈(a, b]:x p} and H={x∈(a, b]:x>p} where p∈Qc such that a