Ibn Al-Haitham Jour. for Pure & Appl. Sci. 34 (1) 2021 105 Topological Structure of Generalized Rough Graphs Department of Mathematics, College of Education for Pure Sciences (Ibn Al-Haitham), University of Baghdad, Baghdad, Iraq. Abstract The main purpose of this paperis to introduce a topological space(𝐷, 𝜏𝐷 ), which is induced by reflexive graph and tolerance graph 𝐷, such that 𝐷 may be infinite. Furthermore, we offer some properties of (𝐷, 𝜏𝐷 ) such as connectedness, compactness, LindelΓΆf and separate properties. We also study the concept of approximation spaces and get the sufficient and necessary condition that topological space is approximation spaces. Keywords. Reflexive graph, tolerance graph, transmitting expression, approximation spaces. 1. Introduction Graph theory [1] is a tool for optimization and solving practical application in all fields such as engineering study and representation of economic and social networks, complex general systems, information theory and others. In particular, graphs are one of the prime objects of study in mathematics. Rough set was offered by Pawlak [2] as a method for dealing with uncertainly of imprecise data, the equivalence relation is the cornerstone of Pawlak , s theory of rough set. Topology is a major mathematics branch with independent theoretic frame work and wide applications. Z. Li [3] offered the concept of transmitting expression of relation and produced several important results of rough sets topological properties. We can apply topological approaches to the theory of rough set and search the connection between rough set theory and topological theory. The topological properties of various rough operators have been debated in [4]. We built on some of the results in [5-10], [11-15] and [16]. Ibn Al Haitham Journal for Pure and Applied Science Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index Doi: 10.30526/34.1.2559 Article history: Received 2, February 2020, Accepted20, Febraury,2019, Published in January 2021 Samah Sarmad Yousif Yaqoub Yousif Samah Sarmad samahsarmad0@gmail.com yoyayousif@yahoo.com mailto:samahsarmad0@gmail.com mailto:samahsarmad0@gmail.com mailto:%20%20%20yoyayousif@yahoo.com 106 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 34 (1) 2021 2. Generalized Rough Graphs Generated By Graphs. We will remember several fundamental concepts of the theory of rough set. In this article,𝐷 = (𝑉(𝐷), 𝐸(𝐷)) is a graph where𝑉(𝐷) implies the universe which may be infinite, the power set of 𝑉(𝐷) symbolized by 𝑃(𝑉(𝐷)) and the closure of subgraph 𝑄 in 𝐷 symbolized by 𝑄 wherever 𝑉(𝐷) is a topological space. Let 𝐷 = (𝑉(𝐷), 𝐸(𝐷)) be a graph. For each subgraph 𝑄 of 𝐷, we will define operators π·βˆ’ and 𝐷+ from 𝑃(𝑉(𝐷))to itself as the following: π·βˆ’(𝑄) = {ɍ ∈ 𝑉(𝐷): if (ɍ, 𝑒) ∈ 𝐸(𝐷), then 𝑒 ∈ 𝑉(𝑄)}, 𝐷+(𝑄) = { ɍ ∈ 𝑉(𝐷): there exists 𝑒 ∈ 𝑉(𝑄) such that (ɍ, 𝑒) ∈ 𝐸(𝐷)}. π·βˆ’(𝑄) is named lower approximation of 𝑄 and 𝐷+(𝑄) is named upper approximation of 𝑉(𝑄). The pair (𝑉(𝐷), 𝐸(𝐷)) is named generalized rough graph or generalized approximation space. 𝑄 is named generalized exact graph or definable graph if π·βˆ’(𝑄) = 𝐷+(𝑄).While 𝑄 is called undefinable graph if π·βˆ’(𝑄) β‰  𝐷+(𝑄). If 𝐷 is an equivalence graph, a generalized rough graph (𝑉(𝐷), 𝐸(𝐷))means the rough graph in the Pawlak , s sense. Definition 2.1. Let 𝐷 = (𝑉(𝐷), 𝐸(𝐷)) be a nonempty graph. We define 𝜏𝐷, for each 𝑄 βŠ† 𝐷 by 𝜏𝐷 = {𝑄 βŠ† 𝐷: π·βˆ’π‘„ = 𝑉(𝑄)}. If 𝐷 is a reflexive graph, then 𝜏𝐷 constitutes a topology on 𝑉(𝐷), 𝜏𝐷can be named the topology produced by 𝐷. Definition 2.2. If𝐷is a reflexive graph, then (𝐷, 𝜏𝐷 ) is named the topological space produced by 𝐷. Definition 2.3. Let𝐷 be a graph, if 𝐷 is both reflexive and symmetric graph then 𝐷 is called tolerance graph. Definition 2.4. Let 𝐷𝛼 and 𝐷𝛽 be two graphs on 𝑉(𝐷𝛼 ) = 𝑉(𝐷𝛽 ) = 𝑉(𝐷). 𝐷𝛽 is named transmitting expression of 𝐷𝛼 , if for everyɍ, 𝑒 ∈ 𝑉(𝐷),(ɍ, 𝑒) ∈ 𝐸(𝐷𝛽 )if and only if (ɍ, 𝑒) ∈ 𝐸(𝐷𝛼 ) or there exists {𝑣1, 𝑣2, 𝑣3, … , 𝑣𝑛 } βŠ† 𝑉(𝐷) where(ɍ, 𝑣1) ∈ 𝐸(𝐷𝛼 ), (𝑣1, 𝑣2) ∈ 𝐸(𝐷𝛼 ), … , (𝑣𝑛, 𝑒) ∈ 𝐸(𝐷𝛼 ). Proposition 2.5. Let 𝐷𝛼 be a graph and 𝐷𝛽 the transmitting expression of 𝐷𝛼 , then 𝐷𝛽 is a transitive graph, furthermore, (1) If 𝐷𝛼 is reflexive, then 𝐷𝛽 is also reflexive, (2) If 𝐷𝛼 is symmetric, then 𝐷𝛽 is also symmetric, (3) If 𝐷𝛼 is transitive, then 𝐷𝛽 = 𝐷𝛼 . Proof. (1) Let 𝐷𝛼 is reflexive graph, then for each ɍ ∈ 𝑉(𝐷),(ɍ, ɍ) ∈ 𝐸(𝐷𝛼 ), since 𝐷𝛽 is a transmitting expression of 𝐷𝛼 , then (ɍ, ɍ) ∈ 𝐸(𝐷𝛽 ), so 𝐷𝛽 is a reflexive graph. 107 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 34 (1) 2021 (2) Let (ɍ, 𝑒) ∈ 𝐷𝛽 , since 𝐷𝛽 the transmitting expression of 𝐷𝛼 , then (ɍ, 𝑒) ∈ 𝐷𝛼 or there exists {𝑣1, 𝑣2, 𝑣3, … , 𝑣𝑛 } βŠ† 𝑉(𝐷) where (ɍ, 𝑣1) ∈ 𝐷𝛼 , (𝑣1, 𝑣2) ∈ 𝐷𝛼 , … , (𝑣𝑛, 𝑒) ∈ 𝐷𝛼 if and only if (ɍ, 𝑒) ∈ 𝐷𝛽 , because 𝐷𝛼 is symmetric, so (𝑒, ɍ) ∈ 𝐷𝛼 or there exists {𝑣1, 𝑣2, 𝑣3, … , 𝑣𝑛 } βŠ† 𝑉(𝐷) where,(𝑒, 𝑣𝑛 ) ∈ 𝐷𝛼 , … , (𝑣2, 𝑣1) ∈ 𝐷𝛼 , (𝑣1, ɍ) ∈ 𝐷𝛼 if and only if (𝑒, ɍ) ∈ 𝐷𝛽 , which implies to 𝐷𝛽 is symmetric. (3) Let (ɍ, 𝑒) ∈ 𝐸(𝐷𝛽 ), we have to show that (ɍ, 𝑒) ∈ 𝐸(𝐷𝛼 ).Since 𝐷𝛽 the transmitting expression of 𝐷𝛼 , then(ɍ, 𝑒) ∈ 𝐸(𝐷𝛽 )if and only if (ɍ, 𝑒) ∈ 𝐷𝛼 or there exists {𝑣1, 𝑣2, 𝑣3, … , 𝑣𝑛 } βŠ† 𝑉(𝐷) where (ɍ, 𝑣1) ∈ 𝐸(𝐷𝛼 ), (𝑣1, 𝑣2) ∈ 𝐸(𝐷𝛼 ), (𝑣2, 𝑣3) ∈ 𝐸(𝐷𝛼 ) … , (𝑣𝑛, 𝑒) ∈ 𝐸(𝐷𝛼 ). (i) If (ɍ, 𝑒) ∈ 𝐸(𝐷𝛼 ) the prove is complete. (ii) If there exists {𝑣1, 𝑣2, 𝑣3, … , 𝑣𝑛 } βŠ† 𝑉(𝐷) where (ɍ, 𝑣1) ∈ 𝐸(𝐷𝛼 ), (𝑣1, 𝑣2) ∈ 𝐸(𝐷𝛼 ), (𝑣2, 𝑣3) ∈ 𝐸(𝐷𝛼 ) … , (𝑣𝑛 , 𝑒) ∈ 𝐸(𝐷𝛼 ), and we have that 𝐷𝛼 is transitive, so (ɍ, 𝑒) ∈ 𝐸(𝐷𝛼 ), which means 𝐷𝛽 = 𝐷𝛼 . Definition2.6. (1) Let (𝐷, 𝜏𝐷 ) be a topological space and ℬ𝐷 a base of (𝐷, 𝜏𝐷 ), where (𝐷, 𝜏𝐷 ) is induced by a reflexive graph 𝐷. Then B ∈ ℬ𝐷 is called maximal element of ℬ𝐷 if does not exist 𝐡𝚀 ∈ ℬ𝐷 \{𝐡} such that 𝐡 βŠ† 𝐡 𝚀. (2)The set of all maximal elements of ℬ𝐷 symbolized by ℬ𝐷 βˆ— . Because ⋃ℬ𝐷 = 𝑉(𝐷), ℬ𝐷 βˆ— is referred to as the minimal complete cover of (𝐷, 𝜏𝐷 ) according to the base ℬ𝐷. We will define a pseudo-metric map on graph 𝐷. Definition2.7. Let 𝐷 = (𝑉(𝐷), 𝐸(𝐷)) be a nonempty graph, then 𝑑: 𝑉(𝐷) Γ— 𝑉(𝐷) ⟢ [0, +∞) is called pseudo-metric map on 𝐷, if for all ɍ, 𝑣, 𝑒 ∈ 𝑉(𝐷), (a) 𝑑(ɍ, ɍ) = 0, (b) 𝑑(ɍ, 𝑣) = 𝑑(𝑣, ɍ), (c) 𝑑(ɍ, 𝑣) ≀ 𝑑(ɍ, 𝑒) + 𝑑(𝑒, 𝑣). For each ɍ ∈ 𝑉(𝐷), 𝑄 βŠ† 𝐷, πœ– > 0, 𝐡(ɍ, πœ–) = {𝑣 ∈ 𝑉(𝐷): 𝑑(ɍ, 𝑣) < πœ–}, 𝑑(ɍ, 𝑄) = 𝑖𝑛𝑓 {𝑑(ɍ, 𝑣): 𝑣 ∈ 𝑉(𝑄)}. If there exists pseudo-metric map𝑑on 𝐷where{𝐡(ɍ, πœ–): ɍ ∈ 𝑉(𝐷), πœ– > 0} configures a base of 𝐷, then a topological space (𝐷, 𝜏𝐷 ) is referred to as pseudo-metrizable space. Proposition 2.8. Let 𝐷 be pseudo-metrizable space. If𝑄 βŠ† 𝐷and 𝑑 is pseudo-metric map on 𝐷, then ɍ ∈ οΏ½Μ…οΏ½ if and only if 𝑑(ɍ, 𝑄) = 0. Proof. ɍ ∈ οΏ½Μ…οΏ½ if and only if for each πœ– > 0, 𝐡(ɍ, πœ–) ∩ 𝑉(𝑄) β‰  βˆ… if and only if for each πœ– > 0, there exists 𝑒 ∈ 𝑉(𝑄) such that 𝑑(ɍ, 𝑒) < πœ– if and only if 𝑖𝑛𝑓 {𝑑(ɍ, 𝑒): 𝑒 ∈ 𝑉(𝑄)} if and only if 𝑑(ɍ, 𝑄) = 0. 108 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 34 (1) 2021 Definition 2.9. Let (𝐷, 𝜏𝐷 ) be a topological space. 𝐷 is named a pseudo-discrete space if 𝑄 βŠ† 𝐷 is open in 𝐷 if and only if 𝑄 is closed in 𝐷. 3. The Properties of Topological Spaces Induced by a Reflexive Graph We will study through this part, the properties of the topological space (𝐷, 𝜏𝐷 ), where (𝐷, 𝜏𝐷 ) is produced by a reflexive graph 𝐷. Lemma3.1. Let 𝐷𝛼 be a reflexive graph and 𝐷𝛽 the transmitting expression of 𝐷𝛼 , for every ɍ ∈ 𝑉(𝐷), chose 𝐿ɍ = {𝑣 ∈ 𝑉(𝐷): (ɍ, 𝑣) ∈ 𝐸(𝐷𝛽 )}, then (1) 𝐿ɍ ∈ πœπ·π›Ό , (2) {𝐿ɍ} is an open neighborhood base of ɍ, (3) 𝐿ɍ is compact subset of (𝐷𝛼 , πœπ·π›Ό ), (4) ℬ𝐷𝛼 = {𝐿ɍ: ɍ ∈ 𝑉(𝐷)} is a base for (𝐷𝛼 , πœπ·π›Ό ). Proof. (1) It is sufficient to show that𝐿ɍ βŠ† 𝐼𝑛𝑑(𝐿ɍ). Let 𝑒 ∈ 𝐿ɍ,so (ɍ, 𝑒) ∈ 𝐸(𝐷𝛽 ), then there exists {𝑣1, 𝑣2, 𝑣3, … , 𝑣𝑛 } βŠ† 𝑉(𝐷) where (ɍ, 𝑣1) ∈ 𝐸(𝐷𝛼 ), (𝑣1, 𝑣2) ∈ 𝐸(𝐷𝛼 ), … , (𝑣𝑛 , 𝑒) ∈ 𝐸(𝐷𝛼 ), so 𝑒 ∈ [𝑣𝑛 ]𝐷𝛼 . For 𝑦 ∈ 𝑉(𝐷) such that 𝑦 ∈ 𝐿ɍ, (𝑣𝑛, 𝑦) ∈ 𝐸(𝐷𝛼 ), so(ɍ, 𝑣1) ∈ 𝐸(𝐷𝛼 ), (𝑣1, 𝑣2) ∈ 𝐸(𝐷𝛼 ), … , (𝑣𝑛 , 𝑦) ∈ 𝐸(𝐷𝛼 ), then (ɍ, 𝑦) ∈ 𝐸(𝐷𝛽 ), so 𝑦 ∈ [ɍ]𝐷𝛽 , then 𝑦 ∈ 𝐿ɍ, [𝑣𝑛 ]𝐷𝛼 βŠ† 𝐿ɍ, so𝑒 ∈ 𝐼𝑛𝑑(𝐿ɍ), which implies to 𝐿ɍ βŠ† 𝐼𝑛𝑑(𝐿ɍ). Hence, 𝐿ɍ ∈ πœπ·π›Ό . (2) Let 𝐡 ∈ 𝜏𝐷 such that ɍ ∈ 𝐡, we will show that 𝐿ɍ βŠ† 𝐡. Let 𝑒 ∈ 𝐿ɍ, then (ɍ, 𝑒) ∈ 𝐸(𝐷𝛼 ) or there exists {𝑣1, 𝑣2, 𝑣3, … , 𝑣𝑛 } βŠ† 𝑉(𝐷) where (ɍ, 𝑣1) ∈ 𝐸(𝐷𝛼 ) , (𝑣1, 𝑣2) ∈ 𝐸(𝐷𝛼 ) , … , (𝑣𝑛, 𝑒) ∈ 𝐸(𝐷𝛼 ). (i) If (ɍ, 𝑒) ∈ 𝐸(𝐷𝛼 ), then we claim 𝑒 ∈ 𝐡. For otherwise, 𝑒 ∈ 𝐡 𝑐, then, 𝐿ɍ βŠ† 𝐡 𝑐, but 𝐷𝛼 is reflexive, so ɍ ∈ 𝐿ɍ and 𝐿ɍ βŠ† 𝐡 𝑐 , then ɍ ∈ 𝐡𝑐, which is a contradiction. Hence, 𝑒 ∈ 𝐡. (ii) If there exists {𝑣1, 𝑣2, 𝑣3, … , 𝑣𝑛 } βŠ† 𝑉(𝐷) where (ɍ, 𝑣1) ∈ 𝐸(𝐷𝛼 ), (𝑣1, 𝑣2) ∈ 𝐸(𝐷𝛼 ), … , (𝑣𝑛, 𝑒) ∈ 𝐸(𝐷𝛼 ), then by (i) we get 𝑣1 ∈ 𝐡, 𝑣2 ∈ 𝐡, … , 𝑒 ∈ 𝐡. So 𝐿ɍ βŠ† 𝐡, Which implies to {𝐿ɍ}constitutes an open neighborhood base of ɍ. (3) Let {πΎπœ†|πœ† ∈ Ξ›} be an open cover of 𝐿ɍ then ɍ ∈ πΎπœ†π‘– for some πœ†π‘– ∈ Ξ›, then by (2) 𝐿ɍ βŠ† πΎπœ†π‘– . Therefore, 𝐿ɍ is a compact subset of (𝐷, 𝜏𝐷 ). (4) It is obvious by (2) Remark 3.2. (1) Let 𝐷 = (𝑉(𝐷), 𝐸(𝐷)) be a graph, for each ɍ, 𝑒 ∈ 𝑉(𝐷), if (ɍ, 𝑒) ∈ 𝐸(𝐷)and(𝑒, ɍ) ∈ 𝐸(𝐷), then 𝐿ɍ = 𝐿𝑒. (2) For all 𝐡 ∈ ℬ𝐷, 𝐡 cannot be represented as the union of some elements of ℬ𝐷 \{𝐡}. Otherwise, there exists π’ͺ𝐷 βŠ† ℬ𝐷 \{𝐡} such that 𝐡 = βˆͺ π’ͺ𝐷. By 𝐡 ∈ ℬ𝐷 , there exists ɍ ∈ 𝑉(𝐷) where 𝐡 = 𝐿ɍ. Because ɍ ∈ 𝐡, there exists 𝑄 ∈ π’ͺ𝐷 such that ɍ ∈ 𝑄 βŠ† 𝐡. By Lemma 3.1, 𝐿ɍ βŠ† 𝑄. Then 𝐡 = 𝑄, so we obtain a contradiction. Hence, 𝐻 cannot be represented as the union of some elements of ℬ𝐷 \{𝐡}. 109 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 34 (1) 2021 (3) Let ℋ𝐷 form a base for (𝐷, 𝜏𝐷 ). Then ℬ𝐷 βŠ† ℋ𝐷 . Otherwise, there exists 𝐡 ∈ ℬ𝐷 but 𝐡 βˆ‰ ℋ𝐷 . Notice that 𝐡 ∈ ℬ𝐷, there exists ɍ ∈ 𝐡 where 𝐡 = 𝐿ɍ. Because ℋ𝐷 is a base for (𝐷, 𝜏𝐷 ), there exists ℋ𝐷 𝚀 βŠ† ℋ𝐷 such that 𝐡 = βˆͺ ℋ𝐷 𝚀 . Thus ɍ ∈ 𝐻 βŠ† 𝐡 for some 𝐻 ∈ ℋ𝐷 𝚀 . By using Lemma 3.1, 𝐡 βŠ† 𝐻. So 𝐡 = 𝐻 ∈ ℋ𝐷 and that means a contradiction. Therefore, ℬ𝐷 βŠ† ℋ𝐷 . Theorem 3.3.Let (𝐷, 𝜏𝐷 ) be a topological space generated by a reflexive graph 𝐷, then (1) (𝐷, 𝜏𝐷 ) is a first countable space, (2) (𝐷, 𝜏𝐷 ) is a locally compact space, (3) If 𝐷 is countable, then (𝐷, 𝜏𝐷 ) is second countable space. Proof. (1) By lemma 3.1(2) {𝐿ɍ} is an open neighborhood base of ɍ, then (𝐷, 𝜏𝐷 )is a first countable space. (2) By lemma 3.1(3), we have for each ɍ ∈ 𝑉(𝐷), ɍ has compact neighborhood. Hence (𝐷, 𝜏𝐷 ) is locally compact space. (3) By lemma 3.1(4),ℬ𝐷 = {𝐿ɍ: ɍ ∈ 𝑉(𝐷)} is a base for (𝐷, 𝜏𝐷 ), which implies to there exists a countable base for 𝜏𝐷, so (𝐷, 𝜏𝐷 ) is a second countable space. Theorem 3.4. If 𝐷 = (𝑉(𝐷), 𝐸(𝐷)) is a reflexive graph and 𝐷𝛽 the transmitting expression of 𝐷, then (𝐷, 𝜏𝐷 ) = (𝐷, πœπ·π›½ ). Proof. According to Lemma 3.1(4), ℬ𝐷 = {𝐿ɍ: ɍ ∈ 𝑉(𝐷)} is a base for (𝐷, 𝜏𝐷 ), we will prove that ℬ𝐷 = {𝐿ɍ: ɍ ∈ 𝑉(𝐷)} is also a base for (𝐷, πœπ·π›½ ). By definition 𝐿ɍ = {𝑒 𝑉(𝐷): (ɍ, 𝑒) ∈ 𝐷𝛽 } ∈ πœπ·π›½ . Let ɍ ∈ 𝐾 ∈ πœπ·π›½ , for any 𝑒 ∈ 𝐿ɍ, then (ɍ, 𝑒) ∈ 𝐷𝛽 , since ɍ ∈ 𝐾, by Lemma 3.1(2) ɍ ∈ 𝐿ɍ βŠ† 𝐾. So ℬ𝐷 is also base for πœπ·π›½ , hence (𝐷, 𝜏𝐷 ) = (𝐷, πœπ·π›½ ). Lemma 3.5. Let (𝐷, 𝜏𝐷 )be atopological space generated by a reflexive graph 𝐷, if ℬ𝐷 βˆ— the minimal complete cover of (𝐷, 𝜏𝐷 ) according to the base ℬ𝐷, then for all 𝐹 ∈ ℬ𝐷 βˆ— , ⋃(ℬ𝐷 \ {𝐹}) β‰  𝑉(𝐷) and ⋃(ℬ𝐷 βˆ— \{𝐹}) β‰  𝑉(𝐷). Proof. Suppose that ⋃(ℬ𝐷\{𝐹}) = 𝑉(𝐷), then there exists ℬ𝐷 πœ„ βŠ† ℬ𝐷 \{𝐹} such that 𝐹 βŠ† βˆͺ ℬ𝐷 πœ„ , Since 𝐹 ∈ ℬ𝐷 βˆ— βŠ† ℬ𝐷, there exists ɍ ∈ 𝑉(𝐷) such that 𝐹 = 𝐿ɍ, so ɍ ∈ 𝐹 πœ„ for some πΉπœ„ ∈ ℬ𝐷 πœ„ . By Lemma 3.1(2), 𝐹 = 𝐿ɍ βŠ† 𝐹 πœ„. Consequently 𝐹 is not a maximal element of ℋ𝐺 which implies a contradiction. So, ⋃(ℬ𝐷 \{𝐹}) β‰  𝑉(𝐷). Since ⋃(ℬ𝐷 \{𝐹}) β‰  𝑉(𝐷), ⋃(ℬ𝐷 βˆ— \{𝐹}) β‰  𝑉(𝐷). Lemma 3.6.Let (𝐷, 𝜏𝐷 ) be a topological space generated by a reflexive graph 𝐷, If ℬ𝐷 βˆ— the minimal complete cover of (𝐷, 𝜏𝐷 ) according to the base ℬ𝐷 and ℋ𝐷 an open cover of (𝐷, 𝜏𝐷 ). Then for all 𝐹 ∈ ℬ𝐷 βˆ— , there exists 𝐻 ∈ ℋ𝐷 where 𝐹 βŠ† 𝐻. Proof. Since ℋ𝐷 an open cover of (𝐷, 𝜏𝐷 ), for any 𝐹 ∈ ℬ𝐷 βˆ— , there exists ℋ𝐷 𝚀 βŠ† ℋ𝐷 such that 𝐹 βŠ† ⋃ ℋ𝐷 𝚀 . Because 𝐹 ∈ ℬ𝐷 βˆ— βŠ† ℬ𝐷, then 𝐹 = 𝐿ɍ for some ɍ ∈ 𝐹, so there exists 𝐻 ∈ ℋ𝐷 𝚀 βŠ† ℋ𝐷 such that ɍ ∈ 𝐻. By Lemma 3.1, 𝐹 βŠ† 𝐻. 110 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 34 (1) 2021 Lemma 3.7.Let (𝐷, 𝜏𝐷 ) be a topological space generated by a reflexive graph 𝐷. If ℬ𝐷 βˆ— is the minimal complete cover of (𝐷, 𝜏𝐷 ) according to the base ℬ𝐷 and ℋ𝐷 an open cover of (𝐷, 𝜏𝐷 ), which is made up of some elements of ℬ𝐷, then ℬ𝐷 βˆ— βŠ† π’ͺ𝐷 . Proof. For each 𝐡 ∈ ℬ𝐷 βˆ— , we claim that 𝐡 ∈ π’ͺ𝐷 . If not, 𝐡 βˆ‰ π’ͺ𝐷 . Since ⋃ π’ͺ𝐷 = 𝑉(𝐷), ⋃(π’ͺ𝐷 \ {𝐡}) = 𝑉(𝐷), So ⋃(ℬ𝐷 \{𝐡}) = 𝑉(𝐷). By using Lemma3.5,⋃(ℬ𝐷 \{𝐡}) β‰  𝑉(𝐷), which implies a contradiction. So ℬ𝐷 βˆ— βŠ† π’ͺ𝐷 . Theorem 3.8.Let (𝐷, 𝜏𝐷 )be a topological space generated by a reflexive graph 𝐷, ℬ𝐷 βˆ— the minimal complete cover of (𝐷, 𝜏𝐷 )according to the base ℬ𝐷. Thenℋ𝐺 βˆ— is a finite set if and only if(𝐷, 𝜏𝐷 ) is compact space. Proof. The only if part clear by Lemma (4.6).Conversely, suppose that (𝐷, 𝜏𝐷 ) is compact, asℋ𝐺 is an open cover of (𝐷, 𝜏𝐷 ) then ℋ𝐺 has a finite subcover ℋ𝐺 πœ„ . By using Lemma 3.7, ℬ𝐷 βˆ— βŠ† ℬ𝐷 πœ„ , thus |ℬ𝐷 βˆ— | ≀ |ℬ𝐷 πœ„ |. Hence ℬ𝐷 βˆ— is a finite set. 4. The Properties of Topological Spaces generated by a Tolerance Graph Through this part, we will achieve the properties of (𝐷, 𝜏𝐷 ), where (𝐷, 𝜏𝐷 ) is a topological space induced by tolerance graph 𝐷. Lemma 4.1If (𝐷, 𝜏𝐷 ) is a topological space generated by a tolerance graph 𝐷, then for all 𝑄 βŠ† 𝐷, 𝑄 is open if and only if 𝑄 is closed. Proof. 𝑄 is open ⟺ 𝑄 = 𝐼𝑛𝑑(𝑄) ⟺ 𝑄𝑐 = 𝐼𝑛𝑑(𝑄𝑐 ) ⟺ 𝑄𝑐 is open graph ⟺ 𝑄 is closed. Theorem 4.2. If(𝐷, 𝜏𝐷 )is a topological space generated by a tolerance graph 𝐷. Then (𝐷, 𝜏𝐷 ) is discrete if and only if (𝐷, 𝜏𝐷 ) is 𝑇0 βˆ’ π‘ π‘π‘Žπ‘π‘’. Proof.The only if part is clear. We are going to prove the if part. Let (𝐷, 𝜏𝐷 ) be 𝑇0 βˆ’ π‘ π‘π‘Žπ‘π‘’. Depending on the Lemma 3.1(4), we have if 𝐷 is reflexive, then {𝐿ɍ: ɍ ∈ 𝑉(𝐷)} is a base for (𝐷, 𝜏𝐷 ). We claim that 𝐿ɍ = {ɍ} for any ɍ ∈ 𝑉(𝐷). Suppose that 𝐿ɍ β‰  {ɍ} for some ɍ ∈ 𝑉(𝐷). By Proposition 2.5, 𝐷𝛽 is an equivalent graph on 𝑉(𝐷), so 𝐿ɍ = [ɍ]𝐷𝛽 . Chose 𝑒 ∈ [ɍ]𝐷𝛽 such that 𝑒 β‰  ɍ. Since (𝐷, 𝜏𝐷 ) is 𝑇0 βˆ’ π‘ π‘π‘Žπ‘π‘’, there exists an open subgraph 𝑂 where ɍ ∈ 𝑉(𝑂) and 𝑒 βˆ‰ 𝑉(𝑂), or there exists an open subgraph π‘ˆ where 𝑒 ∈ 𝑉(π‘ˆ) and ɍ βˆ‰ 𝑉(π‘ˆ). If there exists an open subgraph 𝑉(𝑂) where ɍ ∈ 𝑉(𝑂) and ɍ βˆ‰ 𝑉(π‘ˆ), then ɍ ∈ 𝐿𝑣 βŠ† 𝑉(𝑂) for some 𝑣 ∈ 𝑉(𝐷)depending on the Lemma 3.1(4). It follows 𝑒 βˆ‰ 𝐿𝑣 . As 𝐷𝛽 is an equivalence graph on 𝑉(𝐷), [ɍ]𝐷𝛽 = [𝑣]𝐷𝛽 = 𝐿𝑣. Thus 𝑒 ∈ [ɍ]𝐷𝛽 = 𝐿𝑣 means a contradiction. Similarly if there exists an open subgraph π‘ˆwhere𝑒 ∈ 𝑉(π‘ˆ) and ɍ βˆ‰ 𝑉(π‘ˆ). Hence, {ɍ} is open for all ɍ ∈ 𝑉(𝐷). Therefore, all subgraphs of 𝑉(𝐷) are open which means that (𝐷, 𝜏𝐷 ) is discrete. Theorem 4.3.Let (𝐷, 𝜏𝐷 ) be a topological space generated by a tolerance graph 𝐷. Then, the statements are equivalent: (1) 𝑉(𝐷) βˆ• 𝐸(𝐷𝛽 ) is countable, (2) (𝐷, 𝜏𝐷 ) is a second countable space, (3) (𝐷, 𝜏𝐷 ) is a separable space, 111 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 34 (1) 2021 (4) (𝐷, 𝜏𝐷 ) is a lindelΓΆf space. Proof. (1)⟹(2). Since 𝐷𝛽 is an equivalence graph on 𝑉(𝐷), {𝐿ɍ: ɍ ∈ 𝑉(𝐷)} = 𝑉(𝐷) βˆ• 𝐸(𝐷𝛽 ). By Lemma 3.1(4), (𝐷, 𝜏𝐷 ) is second countable space. (2)⟹(1) Suppose that ℬ is a countable base for (𝐷, 𝜏𝐷 ), then for ɍ ∈ 𝑉(𝐷), there exists 𝐡ɍ ∈ ℬ such that ɍ ∈ 𝐡ɍ βŠ† 𝐿ɍ. By Lemma 3.1(4), ɍ ∈ 𝐿𝑒 βŠ† 𝐡ɍ for some 𝑒 ∈ 𝑉(𝐷). Since 𝐿ɍ = [ɍ]𝐷𝛽 = [𝑒]𝐷𝛽 = 𝐿𝑒, 𝐡ɍ = [ɍ]𝐷𝛽 , we define 𝑓: 𝑉(𝐷) βˆ• 𝐸(𝐷𝛽 ) ⟢ ℬ by 𝑓 ([ɍ]𝐷𝛽 ) = 𝐡ɍ, then 𝑓 is injective. So|𝑉(𝐷) βˆ• 𝐸(𝐷𝛽 )| ≀ |ℬ|. Hence 𝑉(𝐷) βˆ• 𝐸(𝐷𝛽 ) is countable. (2) ⟹ (3) and (2) ⟹ (4) are clear. (3) ⟹ (2). Suppose that 𝐢 is a countable dense subgraph of (𝐷, 𝜏𝐷 ). Put πœ† = {𝐿ɍ: ɍ ∈ 𝑉(𝐢)}, then πœ† is countable. By Lemma 3.1(4), for all ɍ ∈ 𝑉(𝐷) and open subgraph 𝑂 with ɍ ∈ 𝑉(𝑂), we have ɍ ∈ 𝐿𝑒 βŠ† 𝑉(𝑂) for some 𝑒 ∈ 𝑉(𝐷). Since 𝐢 is dense, 𝐿𝑒 ∩ 𝑉(𝐢) β‰  βˆ…, Chose 𝑣 ∈ 𝐿𝑒 ∩ 𝑉(𝐢), then 𝐿𝑣 ∈ πœ†. Since 𝐷𝛽 is an equivalence graph on 𝑉(𝐷), 𝐿𝑣 = [𝑣]𝐷𝛽 = [𝑒]𝐷𝛽 = 𝐿𝑒. It follows ɍ ∈ 𝐿𝑣 βŠ† 𝑉(𝑂). Therefore, πœ† is a base for (𝐷, 𝜏𝐷 ). Hence (𝐷, 𝜏𝐷 ) is a second countable space. (4) ⟹ (2). Suppose that 𝑉(𝐷) βˆ• 𝐸(𝐷𝛽 ) is not countable. Since 𝐷𝛽 is an equivalence graph on 𝑉(𝐷), {𝐿ɍ: ɍ ∈ 𝑉(𝐷)} = 𝑉(𝐷) βˆ• 𝐸(𝐷𝛽 ). It is obvious that {𝐿ɍ: ɍ ∈ 𝑉(𝐷)} is an open cover of (𝐷, 𝜏𝐷 ) but {𝐿ɍ: ɍ ∈ 𝑉(𝐷)} does not have any countable subcover Hence we get a contradiction. Theorem 5.4 Let (𝐷, 𝜏𝐷 ) be a topological space generated by a tolerance graph 𝐷. Then (𝐷, 𝜏𝐷 ) is a connected space if and only if 𝐸(𝐷𝛽 ) = 𝑉(𝐷) Γ— 𝑉(𝐷). Proof. Suppose that (𝐷, 𝜏𝐷 ) is connected, If 𝐸(𝐷𝛽 ) β‰  𝑉(𝐷) Γ— 𝑉(𝐷), then 𝑉(𝐷) Γ— 𝑉(𝐷) βˆ• 𝐸(𝐷𝛽 ) β‰  βˆ…. Chose (ɍ, 𝑒) ∈ (𝑉(𝐷) Γ— 𝑉(𝐷))\𝐸(𝐷𝛽 ), then 𝑒 βˆ‰ [ɍ]𝐷𝛽 = 𝐿ɍ. So 𝐿ɍ β‰  𝑉(𝐷) and 𝐿ɍ β‰  βˆ…. By Lemma 4.1, 𝐿ɍ is both open and closed, so we obtain a contradiction. Conversely, Suppose that 𝐸(𝐷𝛽 ) = 𝑉( 𝐷) Γ— 𝑉(𝐷), then 𝑉(𝐷) βˆ• 𝐸(𝐷𝛽 ) = [𝑉(𝐷)]. So 𝜏𝐷 = {𝑉(𝐷), βˆ…}, thus (𝐷, 𝜏𝐷 ) is connected. Theorem 4.6. Let (𝐷, 𝜏𝐷 ) be a topological space generated by a tolerance graph 𝐷. Then (1) (𝐷, 𝜏𝐷 ) is a locally connected space (2) (𝐷, 𝜏𝐷 ) is a locally separable space, (3) (𝐷, 𝜏𝐷 ) is a regular space, (4) (𝐷, 𝜏𝐷 ) is a normal space, (5) (𝐷, 𝜏𝐷 )is a pseudo-metrizable space. Proof.(1) By lemma 3.1(2) every open neighborhood of ɍ contains 𝐿ɍ which is connected. (2)Since {𝐿ɍ} is an open neighborhood base of ɍ, we just need to show that 𝐿ɍ is a separable subset of (𝐷, 𝜏𝐷 ). Let {ɍ} Μ…Μ… Μ…Μ… be the closure of {ɍ} and suppose that there exists 𝑒 ∈ {ɍ}Μ…Μ… Μ…Μ… such that 𝑒 βˆ‰ 𝐿ɍ, so [ɍ]𝐷𝛽 ∩ [𝑒]𝐷𝛽 = βˆ…. For an open neighborhood 𝐿𝑒 of 𝑒, {ɍ} ∩ 𝐿𝑒 = βˆ… ,so 𝑒 βˆ‰ {ɍ} Μ…Μ… Μ…Μ… 112 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 34 (1) 2021 which is a contradiction, hence, 𝑒 ∈ 𝐿ɍ then {ɍ}Μ…Μ… Μ…Μ… βŠ† 𝐿ɍ. On the other hand, let 𝑒 ∈ 𝐿ɍ then 𝑒 ∈ [ɍ]𝐷𝛽 , then 𝐿ɍ = 𝐿𝑒. Suppose 𝑂 is an open neighborhood of 𝑒, so 𝐿𝑒 βŠ† 𝑉(𝑂) then 𝐿𝑒 ∩ 𝑉(𝑂) β‰  βˆ…, so 𝐿ɍ ∩ 𝑉(𝑂) β‰  βˆ…, then, {ɍ} ∩ 𝑉(𝑂) β‰  βˆ… then 𝑒 ∈ {ɍ}Μ…Μ… Μ…Μ… , so 𝐿ɍ βŠ† {ɍ}Μ…Μ… Μ…Μ… . Hence, 𝐿ɍ = {ɍ}Μ…Μ… Μ…Μ… , and we obtained that {ɍ} is countable dense subset of 𝐿ɍ which implies to 𝐿ɍ is separable subset of (𝐷, 𝜏𝐷 ). Hence, (𝐷, 𝜏𝐷 ) is locally separable space. (3) Let 𝑄 be closed subgraph of 𝐷 and ɍ ∈ 𝑉(𝑄)𝑐 , by Lemma 4.1 𝑄 is open if and only if 𝑄 is closed, so𝑄 and 𝑄𝑐 are two open disjoint subgraph of 𝐷 such that 𝑉(𝑄) βŠ† 𝑉(𝑄) and ɍ ∈ 𝑉(𝑄)𝑐. Hence (𝐷, 𝜏𝐷 ) is a regular space. (4) Let 𝑄, 𝑀 are two disjoint closed subgraphs of 𝐷, then by Lemma 4.1 they are also disjoint closed subgraphs of 𝑉(𝐷). But we have 𝑉(𝑄) βŠ† 𝑉(𝑄)and 𝑉(𝑀) βŠ† 𝑉(𝑀). Hence (𝐷, 𝜏𝐷 ) is a normal space. (5) Since there exists the trivial pseudo-metrizable map 𝑑 induced by the pseudo-metrizable space, where 𝑑: 𝑉(𝐷) Γ— 𝑉(𝐷) ⟢ [0, ∞), such that 𝑑 = { 1 𝑖𝑓 ɍ = 𝑒 0 𝑖𝑓 ɍ β‰  𝑒 For any ɍ ∈ 𝑉(𝐷) and πœ– > 0, 𝐡(ɍ, πœ–) = { {ɍ} π‘–π‘“πœ– < 1 𝑉(𝐷) 𝑖𝑓 πœ– β‰₯ 1 Then, {ɍ} ∈ 𝜏𝐷, so (𝐷, 𝜏𝐷 ) is pseudo-discrete, then 𝐡(ɍ, 1) = {𝑒 ∈ 𝑉(𝐷): 𝑑(ɍ, 𝑒) < 1} = {𝑒 ∈ 𝑉(𝐷): 𝑑(ɍ, 𝑒) = 0} = {ɍ}. Thus {𝐡(ɍ, πœ–): ɍ ∈ 𝑉(𝐷)π‘Žπ‘›π‘‘ πœ– > 0} forms a base for (𝐷, 𝜏𝐷 ). Hence (𝐷, 𝜏𝐷 ) is pseudo-metrizable. 5. Approximation spaces on digraph We will present the concept of approximation spaces in this part; furthermore, we will get their characterizations and properties. Definition 5.1. Let (𝐷, 𝜌) be a topological space, then (𝐷, 𝜌) is called an approximation space if there exists an equivalence graph 𝐷 = (𝑉(𝐷), 𝐸(𝐷))such that 𝜏𝐷 = 𝜌. According to Lemma 4.1, we get that approximating spaces are pseudo-discrete spaces. But the question, would pseudo-discrete spaces are approximating space? This problem is certainly answered by the following theorem. Theorem 5.2. 6.2 If (𝐷, 𝜌) is a topological space, we have the next equivalence: (1) (𝐷, 𝜌) is an approximating space, (2) (𝐷, 𝜌) is both pseudo-metrizable and pseudo-discrete, (3) (𝐷, 𝜌) is pseudo-discrete space. Proof. (1) ⟹ (2). It holds depending on Lemma 4.1 and Theorem 4.6. 113 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 34 (1) 2021 (2) ⟹ (1). Let(𝐷, 𝜌)be both pseudo-metrizable and pseudo-discrete, then there exists a pseudo-metric map 𝑑 on 𝑉(𝐷) where {𝐡(ɍ, πœ–): ɍ ∈ 𝑉(𝐷)π‘Žπ‘›π‘‘ πœ– > 0} is a base for (𝐷, 𝜌). We define a graph 𝐷 on 𝑉(𝐷) as thereinafter: For all ɍ, 𝑒 ∈ 𝑉(𝐷), (ɍ, 𝑒) ∈ 𝐸(𝐷) if and only if 𝑑(ɍ, 𝑒) = 0. Since 𝑑 is pseudo-metric on 𝑉(𝐷), so 𝐷 is an equivalence graph. We will prove that 𝜏𝐷 = 𝜌. Let 𝑄 ∈ 𝜌, by Proposition 2.8, οΏ½Μ…οΏ½ = {ɍ ∈ 𝑉(𝐷): 𝑑(ɍ, 𝑄) = 0}. Since (𝐷, 𝜌)is pseudo-discrete, 𝑄 is closed in (𝐷, 𝜌), so 𝑉(𝑄) = {ɍ ∈ 𝑉(𝐷): 𝑑(ɍ, 𝑄) = 0}. It is obvious that 𝑉(𝑄) βŠ† ⋃{[ɍ]𝐷 : ɍ ∈ 𝑉(𝑄)}. If 𝑒 ∈ [ɍ]𝐷 with ɍ ∈ 𝑉(𝑄), then, 𝑑(ɍ, 𝑄) ≀ 𝑑(𝑒, ɍ) = 0. So 𝑒 ∈ {ɍ ∈ 𝑉(𝐷): 𝑑(ɍ, 𝑄) = 0} = 𝑉(𝑄), that is mean𝑉(𝑄) βŠ‡ ⋃{[ɍ]𝐷 : ɍ ∈ 𝑉(𝑄)}. Thus 𝑉(𝑄) = ⋃{[ɍ]𝐷 : ɍ ∈ 𝑉(𝑄)}, it follows that 𝑉(𝑄) ∈ 𝜏𝐷, so𝜌 βŠ† 𝜏𝐷. On the other side, let ɍ ∈ 𝑉(𝐷), by Proposition 2.8, {ɍ}Μ…Μ… Μ…Μ… = {𝑒 ∈ 𝑉(𝐷): 𝑑(𝑒, ɍ) = 0}, then {ɍ}Μ…Μ… Μ…Μ… = {ɍ}𝐷. Now [ɍ]𝐷 is closed in (𝐷, 𝜌), since (𝐷, 𝜌)is pseudo-discrete,[ɍ]𝐷 ∈ 𝜌. Since {[ɍ]𝐷 : ɍ ∈ 𝑉(𝐷)} is a base for (𝐷, 𝜏𝐷 ), 𝜏𝐷 βŠ† 𝜌. Hence 𝜏𝐷 = 𝜌. This means that (𝐷, 𝜌) are an approximation space. (2) ⟹ (3). Clear. (3) ⟹ (2). Let(𝐷, 𝜌)be pseudo-discrete. For eachɍ ∈ 𝑉(𝐷), 𝐢(ɍ) denoted a connected component with ɍ ∈ 𝐢(ɍ), then 𝐢(ɍ) is closed in (𝐷, 𝜌). So {ɍ}Μ…Μ… Μ…Μ… βŠ† 𝐢(ɍ). Let 𝑒 ∈ 𝐢(ɍ), since 𝐢(ɍ) is a connected component with ɍ ∈ 𝐢(ɍ), there exists a connected subgraph 𝑄 of 𝐷whereɍ, 𝑒 ∈ 𝑉(𝑄). Since 𝐷 is pseudo-discrete, {ɍ}Μ…Μ… Μ…Μ… is both open and closed in (𝐷, 𝜌). Note that {ɍ}Μ…Μ… Μ…Μ… ∩ 𝑉(𝑄) is both open and closed in the subspace 𝑄 and 𝑄 is connected. Then {ɍ}Μ…Μ… Μ…Μ… ∩ 𝑉(𝑄) = 𝑉(𝑄), so 𝑒 ∈ {ɍ}Μ…Μ… Μ…Μ… . This indicates that 𝐢(ɍ) βŠ† {ɍ}Μ…Μ… Μ…Μ… . Thus 𝐢(ɍ) = {ɍ}Μ…Μ… Μ…Μ… . We define 𝑑: 𝑉(𝐷) Γ— 𝑉(𝐷) ⟢ [0, ∞) as follows: 𝑑(ɍ, 𝑒) = { 0 𝑖𝑓 𝐢(ɍ) = 𝐢(𝑒), 1 𝑖𝑓 𝐢(ɍ) β‰  𝐢(𝑒). The assumption that 𝑑 is pseudo-metric on 𝑉(𝐷)can be easily proved. For any ɍ ∈ 𝑉(𝐷) and πœ– > 0, 𝐡(ɍ, πœ–) = { {ɍ}Μ…Μ… Μ…Μ… 𝑖𝑓 πœ– ≀ 1, 𝑉(𝐷) 𝑖𝑓 πœ– > 1. Then 𝐡(ɍ, πœ–)will be closed in (𝐷, 𝜌). Because𝐷 is pseudo-discrete, 𝐡(ɍ, πœ–) ∈ 𝜌. Let ɍ ∈ 𝑉(𝐷) and 𝑉(𝑂) ∈ 𝜌 with ɍ ∈ 𝑉(𝑄). Since 𝐷is pseudo-discrete, 𝑉(𝑄)is closed in (𝐷, 𝜌). So{ɍ}Μ…Μ… Μ…Μ… βŠ† 𝑉(𝑄). By Proposition 2.8,{ɍ}Μ…Μ… Μ…Μ… = {𝑒 ∈ 𝑉(𝐷): 𝑑(ɍ, 𝑒) = 0}. Then . 𝑉(𝐷) π‘Žπ‘›π‘‘ πœ– > 0} is a base for (𝐷, 𝜌). Therefore(𝐷, 𝜌) is pseudo-metrizable. Corollary 5.3. Discrete spaces are approximating spaces. Theorem 5.4. Quotient maps preserve approximating spaces. 114 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 34 (1) 2021 Proof. Suppose that the image of an approximating space 𝐷under a quotient map 𝑓 is 𝐷′. We have to show that 𝐷′ is an approximating space. Since 𝑓is a quotient map, 𝑁 βŠ† 𝐷′ is open in 𝐷′ if and only if 𝑓 βˆ’1(𝑁) is open in 𝐷. By using Theorem, 5.2, 𝑓 βˆ’1(𝑁) is open in 𝐷 if and only if 𝑓 βˆ’1(𝑁) is closed in 𝐷. Since𝑓 is a quotient map, then, 𝑓 βˆ’1(𝑁) is closed if and only if 𝑁 βŠ† 𝐷′ is closed in 𝐷′. So𝑁 βŠ† 𝐷′is open in 𝐷′ if and only if 𝑁 is closed in𝐷′. According to Theorem 5.2,𝐷′ is an approximating space. Corollary5.5. Continuous maps do not preserve an approximating space. We will explicate Corollary 5.5. in the next example. Example 5.6.Suppose that𝑉(𝐷) is a real numbers set R given with the usual discrete topology and 𝑉(𝐷′) is a real numbers 𝑅given with the usual Euclidean topology, let 𝑓: 𝑉(𝐷) ⟢ 𝑉(𝐷′) be the identity map, it is obvious that 𝑓 is continuous map. According to the Corollary 5.3, 𝑉(𝐷) is an approximating space. But 𝑉(𝐷′) is not an approximating space. 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