R-Separated Spaces

Liviu Popescu

Abstract In this paper we have generalized the axioms of the separated spaces T_{i}, (i = \overline{0, 4}), by replacing the equality relation on a topological space X, Δ_{X}, by an arbitrary binary relation, R. Many theorems in general topology may be generalized in this way. It will be interesting to study spaces separated by functions, equivalence relations or order relations. In section 1 are presented axioms and caracterizing theorems of R-separation, in section 2 are presented some properties of spaces separeted by equivalence relations and in section 3 we will obtain some results concerning spaces separated by functions.

Mathematics Subject Classification: 54A05 Key words: topological space, normal, regular space, quotient space, equivalence relation.

1 Axioms and theorems of R separation

First let us make the following notations: (X, T ) is a topological space; R ⊂ X × X is a binary relation on X; \overline{R} is the dual of R (i.e.x\overline{R}y ⇔ (x, y) ∉ R); R^{−1} is the inverse of R (i.e. xR^{−1}y ⇔ (y, x) ∈ R); xRA ⇔ xRy, (∀) y ∈ A; x\overline{R}A ⇔ x\overline{R}y (∀) y ∈ A; ARB ⇔ xRy, (∀) x ∈ A, (∀) y ∈ B; A\overline{R}B ⇔ x\overline{R}y, (∀) x ∈ A, (∀) y ∈ B; R (x) = {y| xRy} and R^{−1} (x) = {y| yRx}; R(A) = {y| (∃) x ∈ A so that xRy}; R^{−1}(A) = {y| (∃) x ∈ A so that yRx}; V_{x} is the neighborhood filter of x ∈ X; For any A ⊂ X we note by V_{A} = {B| (∃) D ∈ T so that A ⊂ D ⊂ B} and we note CA = X\A.

We will replace in the classical definitions of the separated spaces x ≠ y by x\overline{R}y; x ∉ A by x\overline{R}A or A\overline{R}x and A ∩ B by A\overline{R}B. Replacing in the following considerations the relation R by Δ_{X} (i.e. xRy ⇔ x = y), we shall find the classical case of T_{i} spaces (i = \overline{0, 4}). Definition 1. X is T_{0}^{R} - space iff (∀)x, y ∈ X with x\overline{R}y, (∃) V_{x} ∈ V_{x} so that V_{x}\overline{R}y or (∃) V_{y} ∈ V_{y} so that x\overline{R}V_{y}. Definition 2. X is T_{1}^{R} - space iff (∀) x, y ∈ X with x\overline{R}y, (∃) V_{x} ∈ V_{x} and (∃) V_{y} ∈ V_{y} so that x\overline{R}V_{y} and V_{x}\overline{R}y. Definition 3. X is T_{2}^{R} - space iff (∀) x, y ∈ X with x\overline{R}y, (∃) V_{x} ∈ V_{x}, (∃) V_{y} ∈ V_{y} so that V_{x}\overline{R}V_{y}.

Balkan Journal of Geometry and Its Applications, Vol.6, No.2, 2001, pp. 81-88 \begin{matrix}c \\ ◯ \end{matrix}Balkan Society of Geometers, Geometry Balkan Press

Remark 1. T_{2}^{R} ⊂ T_{1}^{R} ⊂ T_{0}^{R}. Remark 2. If in the definitions 1,2,3 we take xRy iff x = y, then they become those in the classical case. Theorem 4. X is T_{0}^{R} - space iff for each x, y ∈ X we have: y ∈ \overline{R(x)} and x ∈ \overline{R^{−1}(y)} ⇒ xRy, where by \overline{A} we note the closure of the subset A of X. Proof. V_{x}\overline{R}y or x\overline{R}V_{y} ⇔ V_{x} ∩ R^{−1} (y) = ∅ or R(x) ∩ V_{y} = ∅ ⇔ x ∉ \overline{R^{−1}(y)} or y ∉ \overline{R (x)} for x\overline{R}y ⇔if x ∈ \overline{R^{−1}(y)} and y ∈ \overline{R(x)}, then xRy. □ Theorem 5. X is T_{1}^{R}− space iff R (x) and R^{−1} (x) are closed subsets, for every x ∈ X. Proof. ”⇒”. Suppose that X is T_{1}^{R} - space and x\overline{R}y. Then exists V_{x} ∈ V_{x} an V_{y} ∈ V_{x} so that x\overline{R}V_{y} and V_{x}\overline{R}y. From here: R(x) ∩ V_{y} = ∅ and V_{x} ∩ R^{−1} (y) = ∅ ⇒ y ∉ \overline{R (x)} and x ∈ \overline{R^{−1} (y)}. So:

(a) X is T_{1}^{R}-space ⇒ for each y ∉ \overline{R (x)} we have y ∉ \overline{R (x)} ⇒ \overline{R (x)} ⊂ R(x) ⇒ R(x) = \overline{R (x)};

(b) \overline{R^{−1} (y)} = R^{−1} (y), just like in (a).

”⇐”. (a) If R^{−1} (y) is a closed subset of X, then for each x\overline{R}y, we have x ∉ \overline{R^{−1} (y)}. From here: (∃) V_{x} ∈ V_{x} with V_{x} ∩ R^{−1} (y) = ∅ ⇒ V_{x}\overline{R}y. (b) (∃) V_{y} ∈ V_{y} with x\overline{R}V_{y} just like in (a). □ Theorem 6. X is T_{2}^{R} - space iff R is a closed subset of X × X. Proof. X is T_{2}^{R} - space iff (∀) x\overline{R}y, (∃) V_{x} ∈ V_{x} and (∃) V_{y} ∈ V_{y} with V_{x}\overline{R}V_{y} ⇔ (∀) (x, y) ∉ R, (∃) V_{x} ∈ V_{x}, (∃) V_{y} ∈ V_{y}, so that V_{x} × V_{y} ⊂ \overline{R} ⇔ (∀) (x, y) ∈ \overline{R}, then it is interior point of \overline{R} ⇔ \overline{R} is an open subset of X × X. □ Theorem 7. X is T_{2}^{R} - space iff for each x ∈ X we have ⋂_{V ∈V_{x}} \overline{R(V)} = R(x) and for

every y ∈ X we have ⋂_{U∈V_{y}} \overline{R^{−1}(U)} = R^{−1}(y).

Proof. ”⇒”. Suppose X is T_{2}^{R} - space and y ∈ ⋂_{V ∈V_{x}} \overline{R(V)} ⇒ y ∈ \overline{R(V)}, (∀) V ∈ V_{x}.

If y ∉ R(x) ⇒ x\overline{R}y ⇒ (∃) V_{x} ∈ V_{x} and (∃) V_{y} ∈ V_{y}, so that V_{x}\overline{R}V_{y} ⇒ R(V_{x})∩ V_{y} = ∅ ⇒ y ∉ \overline{R(V_{x})}, contradiction, so y ∈ R(x) and from here ⋂_{V ∈V_{x}} \overline{R(V)} ⊂ R(x) ⇒

⋂_{V ∈V_{x}} \overline{R(V)} = R(x).

In the same way we infer ⋂_{U∈V_{y}} \overline{R^{−1}(U)} = R^{−1}(y).

”⇐”. Suppose x\overline{R}y ⇒ y ∉ R(x) ⇒ y ∉ ⋂_{V ∈V_{x}} \overline{R(V)} ⇒ (∃) V_{x} ∈ V_{x} so that

y ∉ \overline{R(V_{x})} ⇒ (∃) V_{y} ∈ V_{y} so that V_{y} ∩ R(V_{x}) = ∅ ⇒ V_{x}\overline{R}V_{y} ⇒ X is T_{2}^{R}. □ Remark 3. It is enough to replace in Theorem 7, V_{x} with a neighborhood basis of x. Definition 8. (a) X is a R_{l} - regular space iff for each F, closed subset so that F \overline{R}y, there exists V_{F} ∈ V_{F} (neinghorhood of F) and there exists V_{y} ∈ V_{y} so that V_{F} \overline{R}V_{y}.

(b) X is a R_{r} - regular space iff for each F, closed subset with x\overline{R}F, there exists V_{F} ∈ V_{F} and there exists V_{x} ∈ V_{x} so that V_{x}\overline{R}V_{F}.

(c) X is a R - regular space iff X is a R_{l} and R_{r} space. Remark 4. If R is a simetric relation (i.e. R ⊂ R^{−1}) then (a) ⇔ (b) ⇔ (c). Remark 5. If in this definition we take xRy iff x = y then they become those in the classical case.

Theorem 9. (a) X is a R_{l} - regular space iff for each y ∈ X and U ∈ V_{R^{−1}(y)} there exists V ∈ V_{y} so that \overline{R^{−1}(V)} ⊂ U.

(b) X is a R_{r} - regular space iff for each x ∈ X and U ∈ V_{R(x)} there exists V ∈ V_{x} so that \overline{R(V)} ⊂ U.

(c) X is a R - regular space iff (a) and (b) are both trues. Proof. (a) ”⇒”. Suppose X is R_{l} regular space. Let y ∈ X be and U ∈ V_{R^{−1}(y)}. Suppose that U is an open set. Then F = CU is a closed set ⇒ F ∩ R^{−1}(y) = ∅ ⇒ F \overline{R}y ⇒ there exists V_{F} ∈ V_{F} so that V_{F} \overline{R}V_{y} ⇒ V_{F} ∩ R^{−1}(V_{y}) = ∅. Without loss the generality we can suppose V_{F} open set ⇒ CF is a closed set ⇒ \overline{R^{−1}(V_{y})} ⊂ \overline{CV_{F}} = CV_{F} . But F ⊂ V_{F} ⇒ CV_{F} ⊂ CF = U ⇒ \overline{R^{−1}(V_{y})} ⊂ U.

”⇐”. Let F be a closed set and y ∈ X so that F \overline{R}y ⇒ F ∩ R^{−1}(y) = ∅ ⇒ R^{−1}(V_{y}) ⊂ CF; CF = U is an open set ⇒ U ∈ V_{R^{−1}(y)} ⇒ there exists V ∈ V_{y} so that \overline{R^{−1}(V_{y})} ⊂ U ⇒ C\overline{R^{−1}(V_{y})} ⊃ CU = F. But C\overline{R^{−1}(V_{y})} = V_{F} is an open set ⇒ V_{F} ∈ V_{F}. Because R^{−1}(V_{y}) ⊂ \overline{R^{−1}(V_{y})} we have:

CR^{−1}(V_{y}) ⊃ C\overline{R^{−1}(V_{y})} ⇒ CR^{−1}(V_{y}) ⊃ V_{F} ⇒ V_{F} ∩ R^{−1}(y) = ∅ ⇒ V_{F} \overline{R}V_{y}.

(b) In the same way as (a).

(c) Is the consequence of (a) and (b). □ Definition 10. X is a R - normal space iff for each F_{1}, F_{2} closed sets so that F_{1}\overline{R}F_{2}. there exists:

V_{1} ∈ V_{F_{1}}, V_{2} ∈ V_{F_{2}} so that V_{1}\overline{R}V_{2}.

Theorem 11. X is a R - normal space iff for each F closed set and U ∈ V_{R(F)}, there exists V ∈ V_{F} so that (a) \overline{R(V)} ⊂ U and (b) for each U ∈ V_{R^{−1}(y)}, there exists V ∈ V_{F} so that \overline{R^{−1}(V)} ⊂ U. Proof. ”⇒”. Suppose X is a R - normal space. Let F be a closed set and U ∈ V_{R(F)} ⇒ R(F) ⊂ U. Suppose U is a open set ⇒ F_{1} = CU is a closed set. As CR(F) ⊃ F_{1} ⇒ R(F) ∩ F_{1} = ∅ ⇒ F \overline{R}F_{1} ⇒ exists V_{F} ∈ V_{F} and V_{1} ∈ V_{F_{1}} so that V_{F} \overline{R}V_{1}. But V_{1} ⊃ F_{1} ⇒ CV_{1} ⊂ CF_{1}. Suppose V_{1} is an open set ⇒ CV_{1} = \overline{CV_{1}} is a closed set. But V_{F} \overline{R}V_{1} ⇒ R(V_{F}) ⊂ CV_{1} = \overline{CV_{1}} ⇒ \overline{R(V_{F})} ⊂ CV_{1} ⊂ CF_{1} = U. So \overline{R(V_{F})} ⊂ U. In the same way for each U ∈ V_{R^{−1}(y)},there exists V ∈ V_{F} so that \overline{R^{−1}(V_{F})} ⊂ U.

”⇐”. Let F_{1,} F_{2} be closed sets so that F_{1}\overline{R}F_{2} ⇒ R(F_{1}) ∩ F_{2} = ∅ ⇒ R(F_{1}) ⊂ CF_{2}; CF_{2} = U is a open set ⇒ U ∈ V_{R(F_{1})} ⇒ there exists V_{1} ∈ V_{F_{1}} so that \overline{R(V_{1})} ⊂ U ⇒ C\overline{R(V_{1})} = V_{2}, is an open set and V_{2} ⊃ F_{2} ⇒ V_{2} ∈ V_{F_{2}}. Observe that CV_{2} = \overline{R(V_{1})} ⊃ R(V_{1}) ⇒ R(V_{1}) ∩ V_{2} = ∅ ⇒ V_{1}\overline{R}V_{2}. □ Remark 6. Definition 5 ⇔ condition (a) ⇔ condition (b), as we can see from the proof. The R - separated spaces can be characterized by using sequences. First we define the T_{3}^{R} and T_{4}^{R} spaces. Dfinition 12. (a) X is T_{3}^{R} space iff X is a T_{1} space and R - regular space.

(b) X is T_{4}^{R} space iff X is a T_{1} space and an R - normal space. Theorem 13. T_{4}^{R} ⊂ T_{3}^{R} ⊂ T_{2}^{R} ⊂ T_{1}^{R} ⊂ T_{0}^{R}. Proof. Observe that if X is T_{1} space then \{x\} = \overline{\{x\}} for each x ∈ X ; using this condition results the first and the second inclusion of theorem 7. □ Remark 7. A naturally condition for T_{3}^{R} and T_{4}^{R} spaces would to be T_{1}^{R} space, but it is not good enough to Theorem 7.

Theorem 14. X is T_{0}^{R} space iff for the generalized sequences (x_{α})_{α∈I} and (y_{β})_{β∈J} we have

\begin{matrix}x_{α} → x \\ y_{β} → y \\ x_{α}Ry \\ xRy_{β} \end{matrix} \} ⇒ xRy.

Proof. We will use Theorem 1.

” ⇒ ”. Suppose y ∈ \overline{R(x)} and x ∈ \overline{R^{−1}(y)} ⇒ xRy. See that:

\begin{matrix}x_{α} → x \\ x_{α}Ry ⇒ x ∈ R^{−1}(y) \end{matrix} \} ⇒ x ∈ \overline{R^{−1}(y)}.

Also:

\begin{matrix}y_{β} → y \\ xRy_{β} ⇒ y ∈ R(x) \end{matrix} \} ⇒ y ∈ \overline{R(x)}.

By using Theorem 4 we have xRy.

”⇐”. y ∈ \overline{R(x)} ⇒ there exists (y_{β})_{β∈J} so that y_{β} → y and y_{β} ∈ R(x) ⇒ xRy_{β}, x ∈ \overline{R^{−1}(y)} ⇒ there exists (x_{α})_{α∈I} so that x_{α} → x and x ∈ R^{−1}(y) ⇒ x_{α}Ry. From here it follows xRy. □ Theorem 15. X is T_{1}^{R} space iff for the generalized sequences (x_{α})_{α∈I} and (y_{β})_{β∈J} we have

\begin{matrix}x_{α} → x \\ x_{α}Ry \end{matrix} \} ⇒ xRy and \begin{matrix}y_{β} → y \\ xRy_{β} \end{matrix} \} ⇒ xRy.

Proof. We use Theorem 5.

”⇒”. x_{α} → x; x_{α}Ry ⇒ x_{α} ∈ R^{−1}(y) = \overline{R^{−1}(y)} ⇒ x ∈ R^{−1}(y) ⇒ xRy, y_{β} → y; xRy_{β} ⇒ y_{β} ∈ R(x) = \overline{R(x)} ⇒ y ∈ \overline{R(x)} ⇒ xRy

”⇐”. y ∈ \overline{R(x)} ⇒ there exists (y_{β})_{β∈J} a generalized sequence so that y_{β} ∈ R(x) and y_{β} → y. Observe that xRy_{β} ⇒ xRy ⇒ y ∈ R(x) ⇒ R(x) = \overline{R(x)}. x ∈ \overline{R^{−1}(y)} ⇒ there exists (x_{α})_{α∈I} a generalized sequence so that x_{α} ∈ R^{−1}(y) and x_{α} → x. Observe that x_{α}Ry ⇒ xRy ⇒ x ∈ R^{−1}(y) ⇒ xRy. □ Theorem 16. X is T_{2}^{R} space iff for each generalized sequences (x_{α})_{α∈I} and (y_{α})_{α∈I} so that x_{α}Ry_{α}, x_{α} → x and y_{α} → y, we have xRy. Proof. We use Theorem 6.

X is T_{2}^{R} space iff R ⊂ X × X is a closed set ⇔ for each (x_{α}, y_{α}) ∈ R so that (x_{α}, y_{α}) → (x, y) we have (x, y) ∈ R ⇔ for each x_{α} → x; y_{α} → y, x_{α}Ry_{α} it follows xRy. □ Example 17. If ” ≤ ” is an order relation on X and if X is a T_{1}^{R} space then x_{α} ≤ y; x_{α} → x ⇒ x ≤ y; x ≤ y_{β}; y_{β} → y ⇒ x ≤ y.

If X is a T_{2}^{R} space, then from x_{α} ≤ y_{α}; x_{α} → x; y_{α} → y results x ≤ y. Example 18. Let ” ≺ ” be an order relation on R : x ≺ y iff y − x ∈ N. R is T_{i}^{≺} space for i ∈ {0, 1, 2} but it is not T_{i}^{≺} space for i ∈ {3, 4}. Proof. (a) It is obious that ” ≺ ” is an order relation on R and more: ≺ (x) = x + N, ≺^{−1} (y) = y − N, for each x, y ∈ R. x_{n} ≺ y_{n} ⇒ y_{n} − x_{n} ∈ N, where (x_{n})_{n∈N} and (y_{n})_{n∈N} are real sequences. x_{n} → x, y_{n} → y ⇒ y_{n} − x_{n} → y − x. From here exists n_{0} ∈ N so that y_{n} − x_{n} ∈ (y − x − 1/2, y − x + 1/2) for each n ≥ n_{0}. Because y_{n} − x_{n} ∈ N it follows y_{n} − x_{n} = m ∈ N , so y_{n} − x_{n} is constant for n ≥ n_{0} ⇒ y_{n} − x_{n} = y − x ∈ N ⇒ x ≺ y. Using Theorem 16 results R is T_{2}^{≺} space, so T_{i}^{≺} space for every i ∈ {0, 1, 2}.

(b) R is not ≺_{r} regular space. From Theorem 9 we have that for x = 0, if V =

⋃_{n=0}^{∞} (n − \frac{1}{2^{n}}; n + \frac{1}{2^{n}} ) ∈ V_{≺(0)}, because ≺ (0) = N. R is ≺_{r} regular space ⇒ exists

U ∈ V_{0} so that \overline{≺ (U)} ⊂ V. Let ε > 0 be so that U_{0} = (−ε, ε) ⊂ U. We have

≺ (U_{0}) ⊂ \overline{≺ (U_{0})} ⊂ \overline{≺ (U)} ⊂ V and then U_{0} + N ⊂ V. So ⋃_{n=0}^{∞}(n − ε; n + ε) ⊂

⋃_{n=0}^{∞} (n − \frac{1}{2^{n}}; n + \frac{1}{2^{n}} ). It follows 0 < ε ≤ \frac{1}{2^{n}} for each n ∈ bn ⇒ ε = 0, contradiction.

Therefore R is not T_{i}^{≺} space for i ∈ {3, 4}. □

2 Spaces separated by equivalence relations

In this section we will consider the case of an equivalence relation R = ρ. First we define a notion of continuity of a binary relation. Definition 19. (a) A binary relation R on a topological space (X, T ) is continuous iff for every D ∈ T an open set, R^{−1}(D) ∈ T is an open set.

(b) R is an open relation iff for each D ∈ T , R(D) ∈ T is an open set. Remark 8. If R = ρ is an equivalence relation, then ρ = ρ^{−1} so (a)⇔(b). Theorem 20. Let ρ be an equivalence relation on X and (X̂, T̂) be the quotient space. If X̂ is T_{i} space then X is T_{i}^{R} space for each i = \overline{0, 4}. Proof. Let x̂ = ρ(x) be the equivalence class of each x ∈ X, and p : X → X̂, p(x) = x̂ be the canonical projection. Suppose X̂ is T_{2} space and let x, y ∈ X be so that x\overline{ρ}y ⇔ x̂ ≠ ŷ, more, ρ (x)∩ρ(y) = ∅. There exists V_{x} ∈ V_{x} and V_{y} ∈ V_{y} neighborhoods of x̂, ŷ so that V_{x} ∩V_{y} = ∅.

If V_{x} = p^{−1}(V_{x}) and V_{y} = p^{−1}(V_{y}), then, as p is continuous we have V_{x} ∈ V_{x} and V_{y} ∈ V_{y} and more because V_{x} ∩V_{y} = ∅ ⇒ V_{x}\overline{ρ}V_{y} ⇒ X is T_{2}^{ρ} space. The cases of i ∈ {0, 1, 2, 3, 4} can be proved analogously. □

This theorem has a converse given by: Theorem 21. If ρ is an equivalence relation on X and if ρ is continous then X is T_{i}^{ρ} space ⇒ X̂ is T_{i} space, for each i = \overline{0, 4}. Proof. We will prove this result only in the case of i = 2. Suppose X is T_{2}^{ρ} space. Fisrt, observe that p is an open map ⇔ ρ is an open relation. Let be x̂, ŷ ∈ X̂ so that x̂ ≠ ŷ ⇒ x\overline{ρ}y ⇒there exists V_{x} ∈ V_{x} and V_{y} ∈ V_{y} so that V_{x}\overline{ρ} V_{y}. Because p is open map we have that V_{x} = p(V_{x}) and V_{y} = p(V_{y}) are neighborhoods of x̂ and ŷ and more

V_{x}\overline{ρ} V_{y} ⇒ V_{x} ∩V_{y} = ∅ so X̂ is T_{2} space. □

Example 22. 1) Let X = R be with usualy topology and ρ the equivalence on R defined by: xρy ⇔ x − y ∈ Q. Observe that for each D open set, ρ(D) = D + Q = R, so ρ is a continuous relation. From here we can see that R is not T_{0}^{ρ}space because for each x\overline{ρ}y, V_{x} ∈ V_{x}, V_{y} ∈ V_{y} results ρ(V_{x}) = ρ(V_{y}) = R so y ∈ ρ(V_{x}) and x ∈ ρ(V_{y}).

If R̂ = R/_{ρ} would be T_{0} space, then R would be T_{0}^{ρ} space, so R/_{ρ} is not T_{i} space, i = \overline{0, 4}.

2) Some surfaces can be obtained as quotient spaces by identifying points of the border of a plane quadrate P ⊂ R^{2}. For example the 2-sphere, S^{2}. Construction of S^{2}: We note by Int P the interior of P and by bP the border of P. We consider P as a topological subspace of R^{2}. See that if D is an open set of P, then D ∪ bP is also an open set of P. We define on P the relation ρ : xρy ⇔ x = y or x, y ∈ bP. We see that ρ is a continuous equivalence relation, since if D is an open subset of P :

a) If D ∩ bP = ∅ then ρ(D) = D is an open subset of P.

b) If D ∩ bP ≠ ∅ then ρ(D) = D ∪ bP is also an open subset of P.

We define S^{2} = P/_{ρ} the quotient space of P. It is not difficult to see that S^{2} is homeomorphe with any sphere of R^{3}. P is a T_{2}^{ρ} space, using Theorem 16 of Section 1.

3 Function relations

Let f : X → X be a function and (X, T ) be a topological space. We will establish some properties of T_{i}^{f} spaces, i = \overline{0, 4}. Theorem 23. If X is T_{0} space and f is continuous, then X is T_{0}^{f} space. Proof. Observe that iff is continuous then f(\overline{A}) ⊂ \overline{f(A)} and \overline{f^{−1}(B)} ⊂ f^{−1}(\overline{B}) for each A, B subsets of X. X is T_{0} space iff:

x ∈ \overline{\{y\}} and y ∈ \overline{\{x\}} ⇒ x = y. (1)

We shall prove:

y ∈ \{\overline{f(x)}\} and x ∈ \{\overline{f^{−1}(y)}\} ⇒ f(x) = y. (2)

x ∈ \{\overline{f^{−1}(y)}\} and \{\overline{f^{−1}(y)}\} ⊂ f^{−1}(\overline{\{y\}}) ⇒ x ∈ f^{−1}(\overline{\{y\}}), because f is continuous.

From here f(x) ∈ \overline{\{y\}} and \begin{matrix}f(x) ∈ \overline{\{y\}} \\ y ∈ \overline{\{f(x)\}} \end{matrix} \} ⇒^{(1)} y = f(x), so X is T_{0}^{f} space (2). □

Theorem 24. If X is T_{0}^{f} space and if f is bijection, having the inverse f^{−1} continuous, then X is T_{0} space. Proof. Let x, y ∈ X be. We shall prove that: x ∈ \overline{\{y\}} and y ∈ \overline{\{x\}} ⇒ x = y, i.e. X is T_{0}^{f} space. Since f is bijection ⇒exists z ∈ X so that y = f(z). Suppose that x ∈ \overline{\{f(z)\}} and f(z) ∈ \overline{\{x\}} ⇔ x ∈ \overline{\{y\}} and y ∈ \overline{\{x\}}. Because f^{−1}is continuous we have: z ∈ f^{−1}(\overline{\{x\}}) ⊂ \overline{\{f^{−1}(x)\}} and so z ∈ \overline{\{f^{−1}(x)\}}. From here x ∈ \overline{\{f(z)\}} and z ∈ \overline{\{f^{−1}(x)\}} ⇒ x = f(z), because X is T_{0}^{f} space. So we have x = f(z) = y, and then X is T_{0} space. □ Theorem 25. If X is T_{1} space and f is continuous, then X is T_{1}^{f} space. Proof. Observe that X is T_{1}^{f} space iff {f(x)} and f^{−1}(y) are closed sets, for each x, y ∈ X. X is T_{1} space ⇒ {f(x)} is closed set (1). Because {y} is a closed set and f is continuous, we have that f^{−1}(y) is a closed set (2). From (1) and (2) we have that X is T_{1}^{f} space. □ Theorem 26. If X is T_{1}^{f} space and f is onto, then X is T_{1} space. Proof. X is T_{1}^{f} space ⇒ {f(x)} is a closed set for each x ∈ X. Because for each y ∈ X,exists x ∈ X so that y = f(x), we have that {y} = {f(x)} is a closed set so X is T_{1} space. □ Theorem 27. If X is T_{2} space and f is continuous, then X is T_{2}^{f} space. Proof. Note G_{f} the graph of f. We shall prove that CG_{f} is an open subset of X ×X. Let be (x, y) ∈ CG_{f}. Then f(x) ≠ y. Because X is T_{2} space, there exists V ∈ V_{y} and V_{1} ∈ V_{f(x)} so that V ∩ V_{1} = ∅. But f is continuous and from here1 there exists U ∈ V_{x} so that f(U) ⊂ V_{1} ⇒ f(U) ∩ V = ∅ ⇒ U × V ⊂ CG_{f} ⇒ CG_{f} is a open set ⇒ G_{f} is a closed set ⇒ X is T_{2}^{f} (see Theorem 4 Section 1). □ Remark 9. Theorem 27 states that each continuous function on a T_{2} space has a closed graphical. Theorem 28. If X is T_{2}^{f} space and f is bijection having the inverse f^{−1} continuous, then X is T_{2} space. Proof. Let y ∈ X be ⇒ there exists x ∈ X so that y = f(x) ⇒ x = f^{−1}(y). Because f^{−1} is continuous we have: For each U ∈ V_{x} there exists V ∈ V_{y} so that f^{−1}(V) ⊂ U ⇒ V ⊂ f(U) ⇒ \overline{V} ⊂ \overline{f(U)}. From here we infer

y ∈ ⋂_{V ∈V_{y}} \overline{V} = ⋂_{V ∈V_{f(x)}} \overline{V} ⊂ ⋂_{V ∈V_{y}} \overline{f(U)} = \{f(x)\} = \{y\},

because X is T_{2}^{f} (Theorem 4 Section 1). So we have: ⋂_{V ∈V_{y}} ⋂\overline{V} = \{y\} for each y ∈ X,

and then X is T_{2}^{f} space. □ Theorem 29. If X is T_{2}^{f} space and compact, then f is continuous.

Proof. X is T_{2}^{f} space ⇒ ⋂_{U∈V_{x}} \overline{f(U)} = \{f(x)\} for each x ∈ X. Hence ⋃_{U∈V_{x}} C\overline{f(U)} =

X \\{(f(x)\}, for an arbitrary point x of X. Let V ∈ V_{f(x)}; then X = ⋃_{U∈V_{x}} C\overline{f(U)} ⋃ V.

Because X is a compact space there exist U_{1}, U_{2}, ..., U_{n} ∈ V_{x} such that

X = C\overline{f(U_{1})} ∪ C\overline{f(U_{2})}∪, ..., ∪C\overline{f(U_{n})} ∪ V = C [\overline{f(U_{1})} ∪ \overline{f(U_{2})}∪, ..., ∪\overline{f(U_{n})}] ∪ V.

Let U ∈ V_{x} be so that U ⊂ ⋂_{i=1}^{n} U_{i}. We have \overline{f(U)} ⊂ ⋂_{i=1}^{n} \overline{f(U_{i})}. Hence

C\overline{f(u)} ⊃ C [⋂_{i=1}^{n} \overline{f(U_{i})}] ⇒ C\overline{f(U)} ∪ V = X ⇒ C\overline{f(U)} ∩ CV = ∅ ⇒ \overline{f(U)} ⊂ V.

We have proved that f is continuous in x. So f is continuous on X. □ Remark 10. This theorem is in fact an alternative of the principle of the closed graphical. Theorem 30. If X is f_{r}−regular space, then f is continuous. Proof. It is a consequence of Theorem 9 Section 1. □ Theorem 31. If X is f−regular space and f is bijection, then f is homeomorphism. Proof. It is a consequence of Theorem 9 Section 1. □ Theorem 32. If X is a regular space and f is continuous, then X is f_{r}−regular space. Proof. Let x ∈ X be and V ∈ V_{f(x)} be ⇒ there exists V_{1} = \overline{V_{1}} ∈ V_{f(x)} so that V_{1} ⊂ V, because X is regular space. As f is continuous there exists U ∈ V_{x} so that

f(U) ⊂ V_{1} ⇒ \overline{f(U)} ⊂ V_{1} = \overline{V_{1}} ⇒ \overline{f(U)} ⊂ V.

Using now Theorem 9 Section 1, it follows that X is f_{r}−regular space. □

Theorem 33. If X is regular space and f is homeomorphism, then X is f regular space. Proof. It is a consequence of Theorem 9 Section 1. □ Theorem 34. If X is T_{4}^{f} space(i.e. f −normal and T_{1}) then f is continuous. If f is bijection, then it is a homeomorphism. Proof. It is a consequence of Theorem 11 Section 1, if we remark that X is T_{1} space ⇒ {x} is a closed set for each x ∈ X. □ Theorem 35. (a) If f is continuous and X is regular space, then X is f_{r}−regular.

(b) If f is homeomorphism and X is regular space, then X is f − regular. Proof. (a) Let x ∈ X be; for each V ∈ V_{f(x)}, there exists U ∈ V_{x} so that f(U) ⊂ V. Suppose V = \overline{V}, because X is regular space. From here \overline{f(U)} ⊂ \overline{V}; because V = \overline{V} ⇒ X is f_{r}−regular space.

(b) It is a consequence of (a), observing that f^{−1} is continuous. □ Theorem 36. If f is homeomorfism and X is normal space, then X is f-normal. Proof. Let F be a closed subset of X ⇒ f(F) is a closed set, because f is homeomorfism. If V ∈ V_{f(F)} then V ∈ V_{f(x)} for each x ∈ F. But f is continuous ⇒ there exists U_{x} ∈ V_{x} so that f(U_{x}) ⊂ V. Let U = ⋃_{x∈F} U_{x} be; notice U is neighbor-

hood of F it follows f(U) = ⋃_{x∈F} f(U_{x}) ⊂ V. Because X is normal space, suppose

V = \overline{V} ⇒ \overline{f(U)} ⊂ V (1).

In the same way replacing f with f^{−1} that is continuous too, we will have : for each F a closed subset of X, and for each U ∈ V_{f^{−1}(x)} there exists V ∈ V_{F} so that f^{−1}(V) ⊂ U, supposing U = \overline{U} (2) From (1), and (2), using Theorem 11 Section 1 we have that X is f-normal space. □

References

[1] J. Kelley, General Topology, Van Nostrand, Princeton, 1950.

[2] R. Engelking, General Topology, Monografie Matematyczne, vol. 60, Warszawa, 1977.

[3] Ch. Christenson and W. Voxman, Aspects of Topology, Marcel Dekker Inc. New York and Basel, 1977.

[4] M. Megan, Topologie generală, Universitatea din Timişoara, 1987.

[5] I. Pop, Elemente de topologie algebrică, Universitatea din Iaşi, 1984.

University of Oradea

Department of Mathematics

Oradea, Str. Armatei Romane, nr. 5

Jud. Bihor, 3700, ROMANIA