Uniformity on generalized topological spaces

It was Császár [1] who first initiated the idea of generalized topological space. This opened up a new direction which was pursued by many mathematicians toward generalizations of many topological concepts to this new arena. A generalized topology (GT, for short) μ on a set X is a collection of subsets of X such that $φ$ ∈ μ and arbitrary unions of members of μ belong to μ; and the ordered pair (X, μ) then stands for a generalized topological space (henceforth abbreviated as GTS). The sets in μ are called μ-open sets and their complements μ-closed sets. A GTS (X, μ) is called a strong GTS if X ∈ μ. For any subset A of a GTS (X, μ), the μ-interior i_μ(A) and μ-closure c_μ(A) of A are defined in the usual way as:

$iμ(A)=⋃B⊆X:B⊆A$ and $B∈μ$ and $cμA=⋂B⊆X:A⊆B$ and $X\B∈μ$⁠.

As is expected, μ-interior and μ-closure operators on a GTS (X, μ) obey the following basic properties:

1. i_μ(A) ⊆ A and A ⊆ c_μ(A), for all A ⊆ X.

2. A ⊆ B ⊆ X ⇒ i_μ(A) ⊆ i_μ(B) and c_μ(A) ⊆ c_μ(B).

3. A(⊆ X) is μ-open (μ-closed) if and only if A = i_μ(A) (resp. A = c_μ(A)).

4. i_μ(X \ A) = X \ c_μ(A), for all A ⊆ X.

The notion of uniformity is well-known for a topological space. This article is intended to initiate the study of a uniformity-like structure, termed μ-uniformity, on a generalized topological space.

In what follows in Section 2, we define μ-uniformities on a nonempty set X axiomatically and show that such a μ-uniformity induces a generalized topology on X. Although a μ-uniformity is not necessarily a uniformity. In Section 3, we also prove that a μ-uniform space satisfies a sort of complete regularity condition. Finally in Section 4, we establish that for a μ-uniform space, there exists a μ-proximity relation [2] such that the same generalized topology originates from both the structures.

We now recall the definition of uniformity on a set and some well-known relevant results thereof; related details may be found in [3].

The pair $(X,U)$ is called a uniform space.

Now we state some well-known results for a uniform space $(X,U)$⁠.

Before going into the details we first state two definitions which will be required later on.

In [7] the concept of generalized quasi uniformity was introduced, termed as g-quasi uniformity. In the same manner, we introduce the definition of μ-uniformity as follows.

1. Δ ⊆ U for every $U∈Uμ$⁠,

2. $U∈Uμ$ and $V⊇U∈Uμ⇒V∈Uμ$⁠,

3. $U∈Uμ⇒$ there exists a symmetric $V∈Uμ$ such that V◦V ⊆ U.

The pair $(X,Uμ)$ is called a μ-uniform space.

Proof. Let $U∈Uμ$⁠. Then by axiom (iii), there exists a symmetric $V∈Uμ$ such that V◦V ⊆ U. Again by Result 2.4, V ⊆ V◦V which implies V ⊆ U and so V⁻¹ ⊆ U⁻¹, i.e. V ⊆ U⁻¹ [since V is symmetric]. So by axiom (ii), $U−1∈Uμ$⁠. □

Proof. Axioms (i) and (ii) of Definition 2.3 are obvious from the definition of uniformity given in Definition 1.2. Now for axiom (iii) of Definition 2.3, consider $U∈U$⁠, then by axiom (v) of Definition 1.2 there exists $V∈U$ such that V◦V ⊆ U; we set W = V ∩ V⁻¹. By axioms (ii) and (iv) of Definition 1.2, we see that $W∈U$ , and it is also clear that W is symmetric and W◦W ⊆ U. Hence, $(X,U)$ is a μ-uniform space. □

Proof. Clearly $φ$ ∈ τ_μ. For each p ∈ X, U(p) ⊆ X, for any $U∈Uμ$ so X ∈ τ_μ.

Let G_α ∈ τ_μ, where α ∈ Λ, an index set. Let G = ⋃_α∈ΛG_α and p ∈ G. Then p ∈ G_β for some β ∈ Λ, so there exists $Up∈Uμ$ such that U_p(p) ⊆ G_β ⊆ G. Hence, G ∈ τ_μ.

So, τ_μ is a strong generalized topology on X. □

Proof. Given that the GTS (X, μ) is μ-uniformizable, i.e. there exists a μ-uniformity $Uμ$ on X such that $μ=τ(Uμ)$⁠. Let F be μ-closed and p ∉ F. Thus X \ F = W(say) is μ-open and p ∈ W, so there exists $U∈Uμ$ such that U(p) ⊆ W.

Now we shall show by induction that for every $n∈N∪{0}$⁠, we can construct a symmetric member $Un∈Uμ$ such that U_n ⊆ U and U_n◦U_n ⊆ U_n−1 ⊆ U, when n is positive with U = U₀.

In fact, let U = U₀; then there exists a symmetric $U1∈Uμ$ such that U₁◦U₁ ⊆ U₀, where U₁ = U₁◦Δ ⊆ U₁◦U₁ ⊆ U₀. Let U_n−1 have been constructed in this way, then there exists a symmetric $Un∈Uμ$ such that U_n◦U_n ⊆ U_n−1 and similarly U_n = U_n◦Δ ⊆ U_n◦U_n ⊆ U_n−1 ⊆ U. So, we get a decreasing sequence {U_n : n ≥ 0} with each member being a subset of U.

Next for every diadic rational [A diadic rational number r is of the form $r=12n1+12n2+⋯+12nm=p2nm$⁠, where p is some positive integer] r ∈ (0, 1], we define $Vr=Un1◦Un2◦…◦Unm$⁠, where $r=Σi=1m2−ni$ with 0 ≤ n₁ < n₂ < … < n_m; since every diadic rational number has unique expression, V_r is well-defined. We define V₀ = Δ, though it may not be in $Uμ$ and also note that V₁ = U₀. Then it can be shown that (Lemma 3.3 below)

$Vk2−n⊆Vk2−n◦Un⊆V(k+1)2−n$ …(⋆)

which holds for every non-negative n and all k = 0, 1, …, 2ⁿ − 1. Also for two diadic rational numbers r, s with 0 ≤ r ≤ s ≤ 1, there exists positive integer n such that r = i ⋅ 2⁻ⁿ and s = j ⋅ 2⁻ⁿ, where i, j are positive integers satisfying 0 ≤ i ≤ j ≤ 2ⁿ.

Hence, we have $Vr=Vi⋅2−n⊆V(i+1)2−n⊆⋯⊆Vj⋅2−n=Vs$⁠. Thus if 0 ≤ r ≤ s ≤ 1 and r, s are diadic rationals then V_r ⊆ V_s.

Next, we define a function g: X → [0, 1] by taking

Since V₀ = Δ, V₀(p) = {p}. For each x( ≠ p) ∈ X, x∉V₀(p) ⇒ 0 ∈{r : x∉V_r(p)}⇒{r : x∉V_r(p)} ≠ $φ$⁠. Also, r ≤ 1 ⇒{r : x∉V_r(p)} is bounded above and so its supremum exists.

Now for any point q ∈ F, i.e. q ∈ X \ W, we have q∉V₁(p), as U(p) ⊆ W and V₁ = U₀ ⊆ U. Again, q∉V₁(p) ⇒ 1 ∈{r : q∉V_r(p), r ≤ 1}⇒ g(q) = 1.

Finally, we shall show that g is μ-continuous in (X, μ). For this it is enough to show that g⁻¹([0, t)) and g⁻¹((t, 1]) are μ-open [since [0, t), (t, 1] are the basic μ-open sets of [0, 1] where t ∈ (0, 1), when it is considered as a subspace of the GTS $(R,ν)$ defined previously]. Let x ∈ g⁻¹([0, t)), then g(x) ∈ [0, t); let us take g(x) = s then s < t ≤ 1. We set r = t − s > 0, now there exists $n∈N$ such that $2n>2r$⁠. We show that U_n(x) ⊆ g⁻¹([0, t)), consequently $g−1([0,t))∈τ(Uμ)=μ$⁠.

Now let k be the uniquely determined positive integer satisfying k − 1 ≤ s ⋅ 2ⁿ < k i.e. (k − 1)2⁻ⁿ ≤ s < k ⋅ 2⁻ⁿ, then g(x) = s < k ⋅ 2⁻ⁿ. Now, $x∉Vk2−n(p)⇒k⋅2−n∈{r:x∉Vr(p)}⇒s=sup{r:x∉Vr(p)}≥k⋅2−n$⁠, which is a contradiction. So $x∈Vk2−n(p)⇒(p,x)∈Vk2−n$⁠. Also for y ∈ U_n(x) we get (x, y) ∈ U_n. Hence, $(p,y)∈Vk2−n◦Un⊆V(k+1)2−n$⁠, by (a), and so $y∈V(k+1)2−n(p)$⁠, and hence g(y) ≤ (k + 1)2⁻ⁿ. Therefore, $g(y)−s≤(k+1)2−n−(k−1)2−n=22n<r=t−s$ i.e. g(y) < t ⇒ y ∈ g⁻¹([0, t)). Hence, U_n(x) ⊆ g⁻¹([0, t)), so $g−1([0,t))∈τ(Uμ)=μ$⁠.

Next, for g⁻¹((t, 1]), let x ∈ g⁻¹((t, 1]), then g(x) = s > t ≥ 0. Let r = s − t > 0 and $n∈N$ so that $2n>2r$⁠. We shall show that U_n(x) ⊆ g⁻¹((t, 1]). Let k be the uniquely determined positive integer satisfying (k − 1)2⁻ⁿ ≤ t < k ⋅ 2⁻ⁿ. If possible, let y ∈ U_n(x) and y∉g⁻¹((t, 1]). Then g(y) ≤ t < k ⋅ 2⁻ⁿ and so $y∈Vk2−n(p)$ (in fact otherwise, $y∉Vk2−n(p)⇒g(y)≥k⋅2−n$⁠). Therefore $(p,y)∈Vk2−n$ and since y ∈ U_n(x), (x, y) ∈ U_n and hence, as U_n is symmetric, (y, x) ∈ U_n. Thus $(p,x)∈Vk2−n◦Un⊆V(k+1)2−n$ [by (⋆)]. So, $x∈V(k+1)2−n(p)$⁠. Consequently, g(x) ≤ (k + 1)2⁻ⁿ. Now $g(x)−t≤(k+1)2−n−(k−1)2−n=22n<r⇒s−t<r$⁠, a contradiction to the equality.

Hence U_n(x) ⊆ g⁻¹((t, 1]), so $g−1((t,1])∈τ(Uμ)=μ$⁠. Hence, g is μ-continuous and so (X, μ) is μ-completely regular. □

Proof. This relation holds for n = 0, since for n = 0, k = 0 and V₀ = Δ so that V₀◦U₀ = U₀ = V₁. Let n > 0 and we assume that the inclusions hold for n − 1. We shall prove the inclusions for n. Since $Vk⋅2−n=Vk⋅2−n◦Δ⊆Vk⋅2−n◦Un$ is always true, it remains only to prove $Vk⋅2−n◦Un⊆V(k+1)2−n$⁠, for k = 0, 1, 2, …, 2ⁿ − 1.

If k is an even integer, say k = 2m, we have k ⋅ 2⁻ⁿ = (2m) ⋅ 2⁻ⁿ = m ⋅ 2⁻⁽ⁿ⁻¹⁾, i.e. (k + 1) ⋅ 2⁻ⁿ = m ⋅ 2⁻⁽ⁿ⁻¹⁾ + 2⁻ⁿ = (2m + 1) ⋅ 2⁻ⁿ.

It then follows from the definition of the sets V_r, given in Theorem 3.2, that $V(k+1)⋅2−n=Vm⋅2−(n−1)◦Un=Vk⋅2−n◦Un$⁠, thus the inclusion is proved in this case.

If k is an odd integer, say k = 2m + 1, then k ⋅ 2⁻ⁿ = (2m + 1) ⋅ 2⁻ⁿ = m ⋅ 2⁻⁽ⁿ⁻¹⁾ + 2⁻ⁿ and (k + 1) ⋅ 2⁻ⁿ = (2m + 2) ⋅ 2⁻ⁿ = (m + 1) ⋅ 2⁻⁽ⁿ⁻¹⁾. By our induction hypothesis, we get $Vm⋅2−(n−1)◦Un−1⊆V(m+1)⋅2−(n−1)$⁠. …(*)

Since U_n◦U_n ⊆ U_n−1, it implies that $Vk⋅2−n◦Un=Vm⋅2−(n−1)+2−n◦Un=Vm⋅2−(n−1)◦Un◦Un⊆Vm⋅2−(n−1)◦Un−1$ and by using (*) we get $Vk⋅2−n◦Un⊆V(m+1)⋅2−(n−1)=V(k+1)⋅2−n$⁠. Thus, the inclusion also holds for odd integers. □

In a uniform space $(X,U)$⁠, there is a result that a uniformity always induces a proximity on X which generates the same topology as is induced by $U$ on X. In the following theorem, we also have a similar result for a GTS. First we state the definition of μ-proximity.

Now δ_μ generates a generalized topology on X which is given below:

Proof. Let U(A) ∩ B ≠ $φ$⁠. Since B ⊆ U(B) (as Δ ⊆ U), we get U(A) ∩ U(B) ≠ $φ$ for all $U∈Uμ$⁠.Conversely, let U(A) ∩ U(B) ≠ $φ$ for all $U∈Uμ$ and if possible let there exist $V∈Uμ$ such that V(A) ∩ B = $φ$⁠. Now there exists a symmetric $W∈Uμ$ such that W◦W ⊆ V. By the given condition, W(A) ∩ W(B) ≠ $φ$ and let p ∈ W(A) ∩ W(B), i.e. (a, p) ∈ W and (b, p) ∈ W for some a ∈ A, b ∈ B. Since W is symmetric, we get (a, b) ∈ W◦W ⊆ V which implies b ∈ V(a) ⊆ V(A). Thus V(A) ∩ B ≠ $φ$⁠, a contradiction. □

Aδ_μB if and only if for every $U∈Uμ,U(A)∩U(B)≠φ$

is a μ-proximity structure on X such that $τ(Uμ)=τ(δμ)$⁠.

The authors are thankful to the referee for certain comments towards the improvement of the paper.