## Abstract

In this paper, the α waybelow relation, which is determined by *O*_{2}-convergence, is characterized by the order on a poset, and a sufficient and necessary condition for *O*_{2}-convergence to be topological is obtained.

Show Summary Details# Open Mathematics

### formerly Central European Journal of Mathematics

# A result for *O*_{2}-convergence to be topological in posets

#### Open Access

## Abstract

## 1. Introduction

## 2. Preliminaries

*
**
**x* = sup{*a* ∈ *P* : ∃*a*′ ∈ *P*, *a* ≪_{α*} *a*′ ≪_{α*} *x*} = inf{*b* ∈ *P* : ∃*b′* ∈ *P*, *b* ⊳_{α*} *b′* ⊳_{α*} *x*}.## 3. *O*_{2}-doubly continuous posets

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**It is straightforward to verify that P*_{1} *is α-doubly continuous*, {*x* ∈ *P* : *x* ≪_{α} 1 = {*a*_{i} : *i* ∈ ℕ} *and* {*x* ∈ *P* : *x* ⊳_{α} 1} = {1}. *Notice a*_{1} ≪_{α} 1, *for any finite A* ⊆_{fin} {*x* ∈ *P* : *x* ≪_{α} 1} = {*a*_{i} : *i* ∈ ℕ}, *there exists i*_{0} ∈ ℕ *such that* ${a}_{i0}\in \underset{{a}_{i}\in A}{\cap}[{a}_{i},1]$, *then* ${b}_{i0}\in \underset{{a}_{i}\in A}{\cap}[{a}_{i},1]$, *but a*_{1} ≪_{α} *b*_{i0} *does not hold, so P*_{1} *is not an O*_{2}-*doubly continuous poset*.*Moreover, a*_{1} ≪_{α} *a*_{2} ≤ *b*_{2} *but a*_{1} ≪_{α} *b*_{2} *does not hold, that means P*_{1} *does not satisfy the condition*(*).## 4. A sufficient and necessary condition for *O*_{2}-convergence to be topological

*
*## Acknowledgement

## References

## About the article

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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In this paper, the α waybelow relation, which is determined by *O*_{2}-convergence, is characterized by the order on a poset, and a sufficient and necessary condition for *O*_{2}-convergence to be topological is obtained.

Keywords: O2-convergence; O2-doubly continuous poset; Topological, α-doubly continuous poset; α-doubly continuous poset

The *O*_{2}-convergence on a poset *P*, which is a generalization of *O*-convergence, is defined as follows [1–4]: Let *P* be a poset, a net (*x _{i}*)

- 1.
sup

*M*= inf*N*=*x*; - 2.
For each

*a*∈*M*and*b*∈*N*, there exists*k*∈*I*such that*a*≤*x*≤_{i}*b*holds for all*i*≥*k*.

By saying the *O*_{2}-convergence on a poset *P* is topological, we mean that there exists a topology 𝒯 on *P* such that a net (*x _{i}*)

In [5], Zhao and Li proved that if the *O*_{2}-convergence on a poset *P* is topological, then *P* is an α-doubly continuous poset. Also they give a sufficient condition: If *P* is an *α**-doubly continuous poset, then *O*_{2}-convergence is topological. Moreover, if *P* is a poset satisfying condition (*), *O*_{2}-convergence is topological if and only if *P* is an *α*-doubly continuous poset.

But as pointed in this paper, there exists poset which is *α*-doubly continuous but not *α**-doubly continuous. Then for such a poset, is the *O*_{2}-convergence topological? In other words, can we find a sufficient and necessary condition for *O*_{2}-convergence to be topological for any poset?

In this paper, the *α* waybelow relation is characterized just by the order on a poset. The concept of *O*_{2}-doubly continuous poset is given, which is strictly stronger than *α*-doubly continuous and strictly weaker than *α**-doubly continuous. Most important, we prove that for a poset *P*, the *O*_{2}-convergence on it is topological if and only if *P* is an *O*_{2}-doubly continuous poset.

This section will briefly mention some of the important results that Zhang and Li presented in [5]. For terms which are not defined here, please refer to [9, 10] and related references.

For a poset *P*, *a, b* ∈ *P*, we denote [*a, b*] = {*x* ∈ *P* : *a* ≤ *x* ≤ *b*}. We use *M*_{0} ⊆_{fin}*M* to indicate *M*_{0} is a finite subset of *M*.

**Definition 2.1:** *([5]). Let P be a poset. For x, y, z ∈ P, define x ≪_{α} y if for every net (x_{i})_{i ∈ I} in P which O_{2}-converges to y, x ≤ x_{i} holds eventually; z ⊳_{α} y if for every net (x_{i})_{i ∈ I} in P which O_{2}-converges to y, x_{i} ≤ z holds eventually.It follows from Definition 2.1 that x ≪_{α} y ⟹ x ≤ y and z ⊳_{α} y ⟹ z ≥ y. If P has a bottom element ⊥ (top element ⊤), then ⊥ ≪_{α} x (⊤ ⊳_{α} x) for each x ∈ P*

**Definition 2.2:** *([5]). A poset P is called an α-doubly continuous poset if a = sup {x ∈ P : x ≪_{α} a} = inf {y ∈ P: z ⊳_{α} a} for every a ∈ P.*

**Remark 2.3:** *
*

*(1)**Every finite lattice is α-doubly continuous, every chain or antichain is α-doubly continuous*.*(2)**If P is α-doubly continuous, then for every a*∈*P, the set*{*x*∈*P*:*x*≪}_{α}a*and*{*y*∈*P*:*z*⊳}_{α}a*are both nonempty*.

**Definition 2.4:** *([5]). Let P be a poset. For x, y, z ∈ P, define x ≪_{α*} y if for every net (x_{i})_{i ∈ I} in P which O_{2}-converges to some ω ∈ P with y ≤ ω, x ≤ x_{i} holds eventually. We define the order dual relation z ⊳_{α*}y if z ≪_{α*}y in P^{op}, where P^{op} denotes P endowed with the reverse order.*

**Definition 2.5:** *([5]). A poset P is called an α ^{*}-doubly continuous poset if a= sup {x ∈ P : x ≪_{α*}a} = inf {y ∈ P: z ⊳_{α*}a} for every a ∈ P.*

**Remark 2.6:** *([5]).
*

*(1)**Obviously, x*≪_{α*}*y*⟹*x*≪⊳_{α}y and z_{α*}*y*⟹*z*⊳._{α}y, then an α*-doubly continuous poset is α-doubly continuous*(2)**If P is an α*-doubly continuous poset, then for each x*∈*P*,

**Theorem 2.7:** *([5]). For any poset P, if O_{2}-convergence is topological, then P is α-doubly continuous.*

**Theorem 2.8:** *([5]). If P is an α*-doubly continuous poset, then O_{2}-convergence is topological.That is to say, α*-doubly continuous ⟹ O_{2}-convergence is topological⟹ α-doubly continuous.Condition (*). Let P be a poset and x, y, z ∈ P with x ≪_{α} y ≤ z, then x ≪_{α} z. Let w, s, t ∈ P with s ⊳_{α} t ≥ w, then s ⊳_{α} w.*

**Corollary 2.9:** *([5]). For any poset P which satisfies condition (*), O_{2}-convergence is topological if and only if P is α-doubly continuous.*

In this section, we give a characterization of the relations ≪_{α} and ⊳_{α}, then introduce the concept of *O*_{2}-doubly continuous poset which is strictly stronger than *α*-doubly continuous and strictly weaker than *α**-doubly continuous.

**Proposition 3.1:** *Let P be a poset. y* ≪* _{α} x if and only if for every subset M and N such that* sup

**Proof:** *Sufficiency, if there exists a net ( x_{i})_{i ∈ I} in P which O_{2}-converges to x, then there exist subsets M and N of P such that sup M = inf N = x, and for each pair a ∈ M and b ∈ N, x_{i} ∈ [a, b] holds eventually. By the condition there exist M_{0} ⊆_{fin} M and N_{0} ⊆_{fin} N such that $\underset{a\in {M}_{0}}{\cap}\underset{b\in {N}_{0}}{\cap}[a,b]\subseteq \uparrow y$. Since M_{0} and N_{0} are finite, ${x}_{i}\in \underset{a\in {M}_{0}}{\cap}\underset{b\in {N}_{0}}{\cap}[a,b]\subseteq \uparrow y$ holds eventually, thus y ≤ x_{i} holds eventually, we conclude that y ≪_{α} x.Necessary, suppose y ≪_{α} x and sup M = inf N = x, denote ${\mathcal{B}}_{x}=\{\underset{a\in A}{\cap}\underset{b\in B}{\cap}\phantom{\rule{thinmathspace}{0ex}}[a,b]:A{\subseteq}_{f\phantom{\rule{thinmathspace}{0ex}}in}\phantom{\rule{thinmathspace}{0ex}}M$ and 𝒟 = {(r, U) ∈ (⋃𝓑_{x}) × 𝓑_{x}: r ∈ U}, we define a preorder " ≤ " on 𝒟 by: (r_{1}, U_{1}) ≤(r_{2}, U_{2}) ⇔ U_{2} ⊆ U_{1}. Arbitrarily take (r_{1}, U_{1}) and(r_{2}, U_{2}) ∈ 𝒟, suppose ${U}_{1}=\underset{a\in {A}_{1}}{\cap}\underset{b\in {B}_{1}}{\cap}\text{\hspace{0.17em}}[a,b]$ and ${U}_{2}=\underset{a\in {A}_{2}}{{\displaystyle \cap}}\underset{b\in \text{\hspace{0.17em}\hspace{0.17em}}{B}_{\text{2}}}{{\displaystyle \cap}}\text{[}a,b\text{]}$ where A_{1}, A_{2} ⊆_{fin} M and B_{1}, B_{2} ⊆_{fin} N. Let A_{3} = A_{1} ⋃ A_{2}, B_{3} = B_{1} ⋃ B_{2}, ${U}_{3}=\underset{a\in {A}_{3}}{\cap}\underset{b\in {B}_{3}}{\cap}\text{\hspace{0.17em}}[a,b]$, then (x, U_{3}) ∈ 𝒟 and (r_{1}, U_{1}) (r_{2}, U_{2}) ≤ (x, U_{3}), hence (𝒟, ≤) is directed. We construct a net (x_{d})_{d∈𝒟} = x_{(r, U)} = r. We know sup M = inf N = x, and for each m ∈ M and n ∈ N, [m, n] ∈ 𝓑_{x}, when (r, U) ≥ (x, [m; n], we have U ⊆ [m, n] and x_{(r, U)} = r ∈ U ⊆[m, n], so m ≤ (x_{d})_{d∈𝒟} ≤ n holds eventually. This implies that (x_{d})_{d ∈ 𝒟} O_{2}-converges to x. Then y ≪ _{α} x implies that y ≤ x_{(r, U)} holds eventually, i.e., there exists (r_{o}, U_{0}) ∈ 𝒟 such that y ≤ x_{(r, U)} = r for every (r, U) ≥ (r_{o}, U_{0}). Since (r_{o}, U_{0}) ∈ 𝒟, there exist M_{0} ⊆_{fin} M and N_{0} ⊆_{fin} N such that ${U}_{0}=\underset{a\in {M}_{0}}{\cap}\underset{b\in \text{}\text{\hspace{0.17em}\hspace{0.17em}}{N}_{\text{0}}}{\cap}\text{[a,b]}$, for each r ∈ U_{0}, (r, U_{0}) ≥ (r_{o}, U_{0}), so y ≤ x_{(r, U0)} = r, we conclude that ${U}_{0}=\underset{a\in {M}_{0}}{\cap}\underset{b\in {N}_{0}}{\cap}[a,b]\subseteq \uparrow y$.*

**Definition 3.2:** *A poset P is called an O*_{2}-*doubly continuous poset if it satisfies the following condition*:

*(1)**P is α-doubly continuous, i.e. x*= sup {*y*∈*P*:*y*≪_{α*}*x*} = inf {*z*∈*P*:*z*⊳_{α*}*x*}*for each x*∈*P*.*(2)**If y*≪⊳_{α}x and z⊆_{α}x, then there exist finite A_{fin}{*a*∈*P*:*a*≪}_{α}x*and B*⊆_{fin}{*b*∈*P*:*b*⊳}_{α}x*such that y*≪⊳_{α}c and z._{α}$c\in \underset{m\in A}{\cap}\underset{m\in B}{\cap}[m,n]$

**Remark 3.3:** *Every finite lattice is O _{2}-doubly continuous, every chain or antichain is O_{2}-doubly continuous*.

**Proposition 3.4:** *If P is an α**-*doubly continuous poset, then P is an O*_{2}-*doubly continuous poset*.

**Proof:** *By Remark 2.6 we know an α*-doubly continuous poset is α-doubly continuous. In an α*-doubly continuous poset P, we denote M_{x} = {a ∈ P : ∃a′ ∈ P; a ≪_{α*} a′ ≪_{α*} x} and N_{y} = {b ∈ P : ∃ b′ ∈ P : ⊳_{α*} b′ ⊳_{α*} x} for each x ∈ P, then sup M_{x} = inf N_{x} = x. If y ≪_{α} x and z ⊳_{α} x, from Proposition 3.1 there exist finite M_{0} ⊆_{fin} M_{x} and N_{0} ⊆_{fin} N_{x} such that $\underset{a\in {M}_{0}}{\cap}\underset{b\in {N}_{0}}{\cap}[a,b]\subseteq [y,z]$. Denote A_{0} = {a′ ∈ P : a ∈ M_{0}} and B_{0} = {b′ ∈ P : b ∈ N_{0}}, we have $\underset{{a}^{\prime}\in {A}_{0}}{\cap}\underset{{a}^{\prime}\in {B}_{0}}{\cap}[{a}^{\prime},{b}^{\prime}]\subseteq \underset{a\in {M}_{0}}{\cap}\underset{b\in {M}_{0}}{\cap}[a,b]\subseteq [y,z]$. Obviously A_{0} ⊆_{fin} {a ∈ P : a ≪_{α} x} and B_{0} ⊆_{fin} {b ∈ P : b ⊳_{α} x}, so we only need to prove y ≪_{α} c and z ⊳_{α} c hold for each $c\in \underset{{a}^{\prime}\in {A}_{0}}{\cap}\underset{{b}^{\prime}\in {B}_{0}}{\cap}[{a}^{\prime},{b}^{\prime}]$.If there exists a net (x_{i})_{i ∈ I} O_{2}-convergence to c, for each a ∈ M_{0} and b ∈ N_{0}, a ≪_{α} c and b ⊳_{α} c, then x_{i} ∈ [a, b] holds eventually. Because M_{0} and N_{0} are both finite, thus ${x}_{i}\in \underset{a\in {M}_{0}}{\cap}\underset{b\in {N}_{0}}{\cap}[a,b]\subseteq [y,z]$ eventually, we conclude that y ≪_{α} c and z ⊳_{α} c.An α*-doubly continuous poset implies it is O_{2}-doubly continuous. An O_{2}-doubly continuous poset implies it is α-doubly continuous. But conversely both are untrue – see the following examples:*

**Example 3.5:** *Let P_{1} = {1} ⋃ {a_{j} i ∈ ℕ} ⋃ {b_{i} : i ∈ ℕ} ⋃ {c_{i}: i ∈ ℕ}, ⋃ {b_{ij}: i, j ∈ ℕ} ⋃ {c_{ij}: i, j ∈ ℕ} where ℕ denotes the set of all positive integers. The order ≤ on P_{1} is constructed as follows (see Fig. 1 for the Hasse diagram of P_{1}:
*

*(1)**a*≤_{i}*a*_{i+1};*b*≤_{i}*b*_{i+1};*c*≤_{i}*c*_{i+1};*for all i*∈ ℕ.*(2)**a*≤_{i}*b*;_{i}*a*≤_{i}*c*∈ ℕ_{i}for all i*(3)**b*≤_{ij}*b*_{i,j+1}≤*b*≤_{i}and c_{ij}*c*_{i,j+1}, ≤*c*∈ ℕ._{i}for all i, j*(4)**1**is the greatest element in P*.

**Example 3.6:** *Let P*_{2} = {1}⋃{*a*}⋃{*b*_{1}, *b*_{2}, *b*_{3}, ...}. *The order* “≤” *on P*_{2} *is defined as follows: (see Fig. 1 for the Hasse diagram): a* ≤ 1; *b*_{i} ≤ 1 *for all i* ∈ ℕ; *b*_{i} ≤ *b*_{j} , *whenever i* ≤ *j, for all i, j* ∈ ℕ.*It is obvious that P is an α-doubly continuous poset and satisfies the condition (2) in Definition 3.2, then P is an O _{2}-doubly continuous poset. But it is not α*

In this section, we obtain the main result that *O*_{2}-convergence on a poset *P* is topological if and only if *P* is an *O*_{2}-doubly continuous poset.

We recall that saying the *O*_{2}-convergence on a poset *P* is topological means that there exists a topology 𝒯 on *P* such that a net (*x _{i}*)

**Theorem 4.1:** *For any poset P, if the O*_{2}-*convergence is topological, then P is an O*_{2}-*doubly continuous poset*.

**Proof:** *For a poset P, if the O_{2}-convergence is topological, then there exists a topology 𝒯 on P such that a net (x_{i})_{i∈I} O_{2}-converges to x ∈ P if and only if it converges to x with respect to the topology 𝒯. For every x ∈ P, let I_{x} = {(W,k,r) ∈ 𝒩(x)× ℕ× P : r ∈ W}, where 𝒩(x) consists of all open sets containing x in the topology 𝒯, i.e., 𝒩(x) = {W ∈ 𝒯: x ∈ W}. Now we define the lexicographic order on the first two coordinates on I_{x}, this means for (W_{1}, k_{1}, r_{1}), (W_{2}, k_{2}, r_{2}) ∈ I_{x}, (W_{1}, k_{1}, r_{1}) ≤. (W_{2}, k_{2}, r_{2}) if and only if W_{2} is a proper subset of W_{1} or W_{1} = W_{2} and k_{1} ≤ k_{2}. This is a preorder. Arbitrarily take (W_{1}, k_{1}, r_{1}), (W_{2}, k_{2}, r_{2}) ∈ I_{x}, then (W_{1} ⋂ W_{2}, max{k_{1}, k_{2}}, x) ≤ (W_{1}, k_{1}, r_{1}) and (W_{1} ⋂ W_{2}, max{k_{1}, k_{2}},x) ≤ (W_{2}, k_{2}, r_{2}). Hence I_{x} is directed. Let x_{i} = r for every i = (W,k,r) ∈ I_{x}. Then it is straightforward to verify that the net (x_{i})_{i∈Ix} converges to x with respect to the topology 𝒯. Thus (x_{i})_{i∈Ix} O_{2}-converges to x. By the definition of O_{2}-convergence, there exist subsets M_{0} and N_{0} such that sup M_{0} = inf N_{0} = x, and for each m ∈ M_{0} and n ∈ N_{0} there exists i_{0} = (W_{0}, k_{0}, r_{0}) ∈ I_{x} such that x_{i} = r ∈ [m, n] holds for every i = (W,k,r) ≥ i_{0}. In particular, we have (W_{0}, k_{0} + 1; w) ≥ (W_{0}, k_{0} + 1, w) ≥ (w_{0}, k_{0}, r_{0}) for all w ∈ W_{0}, hence x(W_{0}, k_{0}+1, w) = w ∈ [m, n] for all w ∈ W_{0}, this means W_{0} ⊆ [m, n]. Since W_{0} is related to m and n, we denoted it by W_{mn}.For arbitrary m ∈ M_{0} and n ∈ N_{0}, if there exists a net (x_{j})_{j∈J} O_{2}-convergence to x, then (x_{j})_{j∈J} converges to x with respect to the topology 𝒯, this means x_{j} is eventually in W_{mn} ⊆ [m, n] (W_{mn} is an open neighborhood of x in the topology 𝒯), thus m ≪_{α} x and n ⊳_{α} x. We conclude that M_{0} ⊆ {y ∈ P : y ≪_{α} x} and N_{0} ⊆ {z ∈ P : z ⊳_{α} x}, since sup M_{0} = inf N_{0} = x, we have sup {y ∈ P : y ≪_{α} x} = inf{z ∈ P : z ⊳_{α} x} = x, i.e., P is an α-doubly continuous poset.Suppose y ≪_{α} x and z ⊳_{α} x, by Proposition 3.1 we know there exist M_{1} ⊆_{fin} M_{0} and N_{1} ⊆_{fin} N_{0} such that $\underset{m\in {M}_{1}}{\cap}\underset{n\in {N}_{1}}{\cap}[m,n]\subseteq [y,z]$. For arbitrary m ∈ M_{1} and n ∈ N_{1}, there exists W_{mn} ∈ 𝒩(x) such that W_{mn} ⊆ [m, n], denote ${W}_{1}=\underset{m\in {M}_{1}}{\cap}\underset{n\in {N}_{1}}{\cap}$ W_{mn}, then W_{1} ∈ 𝒩(x) and ${W}_{1}\underset{m\in {M}_{1}}{\cap}\underset{n\in {N}_{1}}{\cap}[m,n]\subseteq [y,z]$ because M_{1} and N_{1} are both finite. We denote A_{x} = {a ∈ P : a ≪_{α} x} and B_{x} = {b ∈ P : b ≪_{α} x}. Let ${D}_{x}=\{\underset{a\in {A}_{0}}{\cap}\underset{b\in {B}_{0}}{\cap}[a,b]:\phantom{\rule{thinmathspace}{0ex}}{A}_{0}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\subseteq}_{f\phantom{\rule{thinmathspace}{0ex}}in}\phantom{\rule{thinmathspace}{0ex}}{A}_{x}$ and B_{0} ⊆_{fin} B_{x}}, 𝒟 = {r, U) ∈ (∪ D_{x}) × D_{x} : r ∈ U}. Define “≤” on 𝒟 by: (r_{1}, U_{1}) ≤ (r_{2}, U_{2}) ⇔ U_{2} ⊆ U_{1}, then “≤” is a preorder on 𝒟 and 𝒟 is directed. Let x_{(r,U)} = r for (r, U) ∈ 𝒟, we will prove (x_{(r,U) ∈ 𝒟} O_{2}-convergence to x in the next paragraph.Since P is an α-doubly continuous poset, we know sup A_{x} = inf B_{x} = x, for each a ∈ A_{x} and b ∈ B_{x}, [a, b] ∈ 𝒟, when (r, U) ≥ (x, [a, b]) i.e. U ⊆[a, b], x_{(r,U)} = r ∈ U ⊆[a, b], so a ≤ (x_{(r,U)})_{(r,U)}∈𝒟 ≤ b holds eventually, by the definition of O_{2}-convergence we know that (x_{(r,U)})_{(r,U)}∈𝒟 O_{2}-convergence to x.For poset P, the O_{2}-convergence is topological, then (x_{(r,U)})_{(r,U)}∈𝒟 converges to x with respect to the topology 𝒯, hence x_{(r,U)} ∈ W_{1} holds eventually. There exist finite subsets A_{1} ⊆_{fin} A_{x} and B_{1} ⊆_{fin} B_{x} such that when $(r,U)\ge ({r}_{0},\underset{a\in {A}_{1}}{\cap}\underset{b\in {B}_{1}}{\cap}[a,b])$, we have x_{(r, U)} ∈ W_{1}. In particular for each $t\in \underset{a\in {A}_{1}}{\cap}\underset{b\in {B}_{1}}{\cap}[a,b],(t,\underset{a\in {A}_{1}}{\cap}\underset{b\in {B}_{1}}{\cap}[a,b]\ge ({r}_{0},\underset{a\in {A}_{1}}{\cap}\underset{b\in {B}_{1}}{\cap}[a,b])$, then $x(t,\underset{a\in {A}_{1}}{{\displaystyle \cap}}\underset{b\in {B}_{1}}{{\displaystyle \cap}}[a,b])=t\in {W}_{1}$, we conclude that $\underset{a\in {A}_{1}}{\cap}\underset{b\in {B}_{1}}{\cap}[a,b]\subseteq {W}_{1}\subseteq [y,z]$. To prove P satisfies the condition (2) in Definition 3.2, it is sufficient to show y ≪_{α} c for each $c\in \underset{a\in {A}_{1}}{\cap}\underset{b\in {B}_{1}}{\cap}[a,b]$. If there exists a net (x_{j})_{j∈J} O_{2}-convergence to c, then (x_{j})_{j∈J} converges to c with respect to the topology 𝒯; this means x_{j} is eventually in W_{1} since W_{1} ∈ 𝒯 and $c\in \underset{a\in {A}_{1}}{\cap}\underset{b\in {B}_{1}}{\cap}[a,b]\subseteq {W}_{1}$, therefore ${x}_{j}\in {W}_{1}\underset{m\in {M}_{1}}{\cap}\underset{n\in {N}_{1}}{\cap}[m,n]\subseteq [y,z]$ hold eventually, we conclude that y ≪_{α} c and z ⊳_{α} c.*

**Remark 4.2:** *The relations among* ${W}_{1},\underset{m\in {M}_{1}}{\cap}\underset{n\in {N}_{1}}{\cap}[m,n]$ *and* $\underset{a\in {A}_{1}}{\cap}\underset{b\in {B}_{1}}{\cap}[a,b]$ *are revealed in Fig 2*.We review the general relation between convergence and topology. If on any set *P* a class 𝓛 of pairs ((*x _{i}*)

**Fact 4.3:** *([11]). Given a poset P, the class 𝓛 is topological if and only if the following axioms are satisfied.(Constant) If (x_{i})_{i∈I} is a constant net with value x_{i} = x for every i ∈ I, then ((x_{i})_{i∈I}, x) ∈ 𝓛.(Subnets) If ((x_{i})_{i∈I}, x) ∈ 𝓛 and (y_{j})_{j∈J} is a subnet of (x_{i})_{i∈I}, then ((y_{j})_{j∈J} ((y_{j})_{j∈J}, x) ∈ 𝓛.(Divergence) If ((x_{i})_{i∈I},x) is not in 𝓛, then there exists a subnet (y_{j})_{j∈J} of (x_{i})_{i∈I} which has no subnet (z_{k})_{k∈K} so that ((z_{k})_{k∈K}, x) belongs to 𝓛.(Iterated Limits) If ((x_{i})_{i∈I}, x), ∈ 𝓛, and if ((x_{i, j})_{j∈J(i)}, x_{i}) ∈ 𝓛 for all i ∈ I, then ((x_{i, f(i)})_{i, f∈I×M},x) ∈ 𝓛, where $M=\prod _{{}_{i\in I}}J(i)$ is a product of directed sets.*

**Lemma 4.4:** *For an α-doubly continuous poset P, a net* (*x _{i}*)

**Proof:** *Necessary, if net ( x_{i})_{i∈I} O_{2}-converges to x ∈ P and y ≪_{α} x, z ⊳_{α} x, by the Definition of ≪_{α} and ⊳_{α} we know y ≤ x_{i} ≤ z holds eventually. Conversely, just let M = {y ∈ P : y ≪_{α} x} and N = {z ∈ P : z ⊳_{α} x}, then supM = infN = x. For each y ∈ M and z ∈ N, y ≤ x_{i} ≤ z holds eventually, we have (x_{i})_{i∈I} O_{2}-converges to x.*

**Lemma 4.5:** *For an O_{2}-doubly continuous poset P, a net* (

**Proof:** *Necessary, if net ( x_{i})_{i∈I} O_{2}-converges to x ∈ P and y ≪_{α} x, z ⊳_{α} x, by Definition 3.2 there exist finite A ⊆_{fin} {a ∈ P : a ≪_{α} x} and B ⊆_{fin} {b ∈ P : b ⊳_{α} x} such that y ≪_{α} c and z ⊳_{α} c for each c ∈ c ∈ $c\in \underset{m\in A}{\cap}\underset{n\in B}{\cap}\text{\hspace{0.17em}}[m,n]$ Since A and B are both finite, by Lemma 4.4, x_{i} ∈ ${x}_{i}\in \underset{m\in A}{\cap}\underset{n\in B}{\cap}\text{\hspace{0.17em}}[m,n]$ holds eventually, thus y ≪_{α} x_{i} and z ⊳_{α} x_{i} hold eventually. Conversely, since y ≪_{α} x_{i} and z ⊳_{α} x_{i} imply y ≤ x_{i} ≤ z, by Lemma 4.4 we know (x_{i})_{i∈I} O_{2}-converges to x ∈ P.*

**Proposition 4.6:** *For any poset P, the class 𝓛 satisfies axiom (Constant)*.

**Proof:** *Suppose x_{i} = x for every i ∈ I, just take M = N = {x}, sup M = inf N = x and x ≤ x_{i} = x ≤ x holds for all i ∈ I. Therefore ((x_{i})_{i∈I}, x) ∈ 𝓛.*

**Proposition 4.7:** *For any poset P, the class 𝓛 satisfies axiom (Subnets)*.

**Proof:** *Let (( x_{i})_{i∈I}, x) ∈ 𝓛. There exist subsets M and N such that sup M = inf N = x, and for every m ∈ M and every n ∈ N there exists k ∈ I such that m ≤ x_{i} ≤ n for all i ≥ k. For any subnet (y_{j})_{j∈J} of net (x_{i})_{i∈I}, there exists a mapping f : J ⟶ I such that y_{j} = x_{f(j)} for every j ∈ J, and for every i ∈ I there exists j_{i} ∈ J such that f(j) ≥ i for every j ≥ j_{i}. In particular, for k ∈ I, there exists j_{k} ∈ J such that f(j) ≥ k for all j ≥ J such that f(j) ≤ i for every j ≤ j_{i} . In particular, for k ∈ I, there exists j_{k} ∈ J such that f(j) ≥ k for all j ≥ j_{k}. Hence m ≤ y_{j} = x_{f(j)} ≤ n for all j ≥ j_{k}. Therefore ((y_{j})_{j∈J}, x) ∈ 𝓛.*

**Proposition 4.8:** *For an α-doubly continuous poset P, the class 𝓛 satisfies axiom (Divergence)*.

**Proof:** *Assume that (( x_{i})_{i∈I}, x) is not in 𝓛. By lemma 4. 4, there exist y_{0}, z_{0} ∈ P with y_{0} ≪_{α} x and z_{0} ⊳_{α} x such that y_{0} ≤ x_{i} ≤ z_{0} does not hold eventually. This means that for every i ∈ I we can choose a j_{i} ∈ I such that y_{0} ≰ x_{ji} or z_{0} ≱ x_{ji}. Let J = {j_{i} : i ∈ I} ⊆ I, then J is a cofinal subset of I. We now consider the subnet (x_{j})_{j∈J of (xi)i∈I, since y0 ≰ xj or z0 ≱ xj for every j ∈ J, by Lemma 4.4 again, (xj)j∈J has no subnet (zk)k∈K such that ((zk)k∈K, x}) belongs to 𝓛. Thus the class 𝓛 satisfies axiom (Divergence).*

**Proposition 4.9:** *For an O*_{2}-*doubly continuous poset P, the class 𝓛 satisfies axiom (Iterated Limits)*.

**Proof:** *Let (( x_{i})_{i∈I}, x) ∈ 𝓛 and ((x_{i, j})_{j∈Ji}, ∈ 𝓛 for every i ∈ I. For any y, z ∈ P with y ≪_{α} x and z ⊳_{α} x, by Lemma 4.5 we know y ≪_{α} x_{i} and z ⊳_{α} x_{i} hold eventually, i.e., there exists i_{0} ∈ I such that y ≪_{α} x_{i} and z ⊳_{α} x_{i} for all i ≥ i_{0}. Since ((x_{i, j})_{j∈Ji}, x_{i}) ∈ 𝓛, by Lemma 4.5 again, for every i ∈ I with i ≥ i_{0} there exists j_{i} ∈ J_{i} such that y ≪_{α} x_{i, j} and z ⊳_{α} x_{i, j} for every j ≥ j_{i}. Define ${f}_{0}\in M=\prod _{i\in I}J(i)$ as: f_{0}(i) = j_{i} for every i ≥ i_{0} and f_{0}(i) is an arbitrarily fixed element in J_{i} for every i ≱ i_{0}. Now we claim that ((x_{i, f(i)})_{(i, f)∈I×M}, x) ∈ 𝓛, because if (i, f) ∈ I × M and (i, f) ≥ (i_{0}, f_{0}) then y ≪_{α} x_{i, f(i)} and z ⊳_{α} x_{i, f(i)}, it means ((x_{i, f(i)})_{(i, f)∈I×M}, x) O_{2}-converges to x by Lemma 4.4. This completes the proof.From Propositions 4.6 –4.9 and Fact 4. 3, we have*

**Theorem 4.10:** *For any poset P, if P is O*_{2}-*doubly continuous, then O*_{2}-*convergence is topological*.Combining Theorem 4.1 with Theorem 4.1 0, we obtain the following theorem.

**Theorem 4.11:** *Let P be a poset. Then the O*_{2}-*convergence is topological if and only if P is O*_{2}-*doubly continuous*.By Proposition 3.4 and Theorem 4.1 0, we have the following corollary.

**Corollary 4.12:** *([5]). If P is an α*-doubly continuous poset, then O_{2}-convergence is topological.*

**Example 4.13:** *
*

*(1)**For the poset P*_{1}*showed in Example 3.5, the O*_{2}-*convergence is not topological*.*(2)**In Example 3.6, the O*_{2}-*convergence on the poset P*_{2}*is topological*.

This work is supported by the Natural Science Foundation of China (Grant No: 11371130) and (Grant No: 2014GXNSFBA118015).

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**Received**: 2016-01-06

**Accepted**: 2016-03-09

**Published Online**: 2016-04-23

**Published in Print**: 2016-01-01

**Citation Information: **Open Mathematics, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2016-0018. Export Citation

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