On θ-Totally Disconnected and θ-Light Mappings

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.


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 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 -interior point to A if x∈  ⊆  for some U∈  containing x.The set of all -interior points are called -interior set and we denoted by   A , a subset U of topological pace X is -open if and only if every point in U is a interior point [7].Every -open set is an open set but the converse may not be true in general.A space X is said to be -Hausdorff if for every distinct point x, y∈X there exist -open sets Ux, Vy containing x and y respectively such that Ux∩ Vy =∅ [4] [6].Let X and Y be spaces and let f be a mapping from X into Y then f is said to be -homeomorphism if f is bijective, continuous and -closed (-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 -open sets in X, then A∪B is said to be -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 -open sets in X, then A∪B is said to be -disconnection in G if and only if satisfy the following: 1-G∩A ∅.

Definition(5):
A topology space X is said to be -totally disconnected if for every two distinct point p & q there exist -disconnection G∪H to X such that P∈G & q∈H.

Example (6):
The rational numbers with relative usual topology is a -totally disconnected.Since if we take q1&q2 ∈Q where q1 q2 there exist r∈Q c such that q1 r q2 G={x∈Q:x } and H={ x∈Q:x r}

Proof:
Let X be -totally disconnected space to prove X is totally disconnected space.Let x,y∈X with x y.So there exist a -totally disconnection to X (I mean there exist G and H which are -open sets and G, H ∅ and G∪H=X , G∩H=∅ with x∈G, y∈H).

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 -light mapping if for every y∈Y, f -1 (y) is -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 -totally disconnected.Then f is -light mapping.

Remark (11):
Every -totally disconnected is -hausdorff but the converse may be not true in general for example: where R is the set of real number .To show that (R, Ʈu) is not -totally disconnected.
Let x &y ∈Q⊆R such that x y, x y.So (R, ƮU) is not -totally disconnected.

Definition (13):
A surjective mapping f:X→Y is said to be -totally disconnected if and only if for every totally disconnected set U⊆X then f(U) is -totally disconnected in Y.

Definition (16):
A surjective mapping f:X→Y is said to be  ** -totally disconnected mapping if and only if for every totally disconnected set U⊆X then f(U) is -totally disconnected Proposition (17): 1-Every -totally disconnected mapping is totally disconnected mapping.
3-Every  ** -totally disconnected mapping is  * -totally disconnected mapping. Proof: 1-Let U be totally disconnected set in X, but f is -totally disconnected mapping then f(U) is -totally disconnection set in Y, but every -totally disconnected set is totally disconnected so f (U) is totally disconnected in Y, then f is totally disconnected mapping.The proof of 2 and 3 are similar.https://doi.org/10.30526/31.2.1956

Corollary (19):
A property of space being -totally disconnected a topological property.

Proposition (20):
Let X and Y be topological space, let f:X→Y be homeomorphism.So if X is totally disconnected then Y is totally disconnected set.

Corollary (22):
Let X and Y be topological space, let f:X→Y be -homeomorphism.So if X is totally disconnected then Y is -totally disconnected set again.3-Every -Inversely totally disconnected mapping is  * -Inversely totally disconnected mapping. Proof: Then U is totally disconnected set in Y(proposition 7).Since f is  * -Inversely totally disconnected mapping.But f -1 (U) is totally disconnected in X so f -1 (U) is totally disconnected in X.Then f is  * -Inversely totally disconnected mapping Theorem (25): If f:X→Y is -Inversely totally disconnected mapping then f is -light mapping .

Proof:
Since f is -Inversely totally disconnected mapping to prove f is -light mapping.Let y∈Y to prove f -1 (y) is -totally disconnected set.Since f is -Inversely totally disconnected mapping, and {y} is totally disconnected in Y, then f -1 ({y}) is -totally disconnected set in X so f is -light mapping.Proposition (26): let f:X→Z and g:Z→Y be surjective mapping if f is **-inversely totally disconnected and g is -light mappings, then h:X→Y is -light mapping Proof: Let c∈Y so h -1 (c)=(g∘f) -1 (c)=(f -1 ∘g -1 )(c)= f -1 (g -1 (c)).As g is -light mapping so g -1 (c) is -totally disconnected.Also As f is  ** -Inversely totally disconnected mapping so f -1 (g - 1 (c)) is -totally disconnected.h -1 (c) is -totally disconnected then h is -light mapping.
. A mapping f:X→Y is said to be -open( * -open and **-open) if f(V) is -open(open and -open) in Y, whenever V is open (-open) in X

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 -disconnection to X and {a}, {b, c} are -disconnection to subset {a, b} to X. *Its known that every  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 -open.
A surjective mapping f:X→Y is said to be  * -totally disconnected mapping if and only if for every 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∈Q c such that x<p<y G={x ∈Q:x<p} and H={x∈Q:x>p} are two open sets in (Q, Ʈu) such that G∪H=Q, G∩H=∅ Now to prove (Q, ƮD) is -totally disconnected in (R, ƮD) where f(Q)=Q.G={x∈Q:x≤0} is -open set in (Q, ƮD) H={x∈Q:x>0} is -open set in (Q, ƮD) H∪G=Q, H∩G=∅ So (Q, ƮD) is -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∈Q c such that a<p b then ((a, b], Ʈu) is totally disconnected set in (R, Ʈu) Let f:X→Y be bijective -open mapping.Then Y is -totally disconnected set whenever X is totally disconnected Proof: Let y1, y2 ∈Y with y1 y2 since f is bijective, then there exist two distinct points x1, x2 ∈ X such that f(x1)=y1, f(x2)=y2.But X is totally disconnected space, then there exist disconnection G∪H to X such that x1∈G & x2∈H.also f is open mapping and G, H are open sets in X.So f(G) and f(H) are open sets in Y.But f(G)∪f(H)=f(G∪H)=f(X)∩f(H)=f(G∩H)=f(∅)=∅ Such that y1∈f(G), y2∈f(H) So f(G)∪f(H) is disconnection to Y. therefor Y is -totally disconnected set.
G and H are -open sets in X.So f(G) and f(H) are open sets in Y.But f(G)∪f(H)=f(G∪H)=f(X)=Y. Since f is bijective mapping.So f(G)∩f(H)=f(G∩H)=f(∅)=∅ Such that y1∈f(G), y2∈f(H) which implies f(G)∪f(H) is disconnection to Y. Therefor Y is totally disconnected set.Proposition (21): Let f:X→Y be bijective  ** -open mapping.Then Y is -totally disconnected set whenever X is -totally disconnected Proof: