The language of knowledge spaces in pre-topology 1

Abstract

The theory of knowledge spaces (KST) which is regarded as a mathematical framework for the assessment of knowledge and advices for further learning. Now the theory of knowledge spaces has many applications in education. From the topological point of view, we discuss the language of the theory of knowledge spaces by the axioms of separation and the accumulation points of pre-topology respectively, which establishes some relations between topological spaces and knowledge spaces; in particular, we show that the language of the regularity of pre-topology in knowledge spaces and give a characterization for knowledge spaces by inner fringe of knowledge states. Moreover, we study the relations of Alexandroff spaces and quasi ordinal spaces; then we give an application of the density of pre-topological spaces in primary items for knowledge spaces, which shows that one person in order to master an item, she or he must master some necessary items. In particular, we give a characterization of a skill multimap such that the delineated knowledge structure is a knowledge space, which gives an answer to a problem in [14] or [18] whenever each item with finitely many competencies; further, we give an algorithm to find the set of atom primary items for any finite knowledge space.

Keywords

Knowledge space knowledge structure,learning space pre-topological space skill multimap quasi ordinal space Alexandroff space separation of axiom primary item Primary 54A05 secondary 54A25 54B05 54B10 54D05 54D70

1 Introduction and preliminaries

Doignon and Falmagne (1999) introduced the theory of knowledge spaces (KST) which is regarded as a mathematical framework for the assessment of knowledge and advices for further learning [8, 14]. Indeed, KST makes a dynamic evaluation process; of course, the accurate dynamic evaluation is based on individuals’ responses to items and the quasi order on domain Q [8]. In particular, KST has many applications in education. In particular, a practical application is the ALEKS System, that is, assessment and learning in knowledge spaces; ALEKS is a artificial intelligence evaluation and learning system, which is known as the most effective learning system. ALEKS is based on the knowledge space theory and uses adaptive questioning to quickly and accurately locate what students have learned and what they have not learned, see [8, 14].

A field of knowledge is a non-empty set of items or questions, denoted by Q. A subset H of Q is called a knowledge state if an individual is capable of solving it under the ideal conditions. A family $H$ of knowledge states is called knowledge structure if ${\emptyset, Q} \subseteq H$ , which is denoted by $(Q, H)$ . Sometimes we simply say that $H$ is the knowledge structure if the domain can be omitted without ambiguity. There are two important types of knowledge structures, namely, knowledge spaces and learning spaces. If $H^{'} \subseteq H$ implies $⋃ H^{'} \in H$ , then the knowledge structure $H$ is called a knowledge space. Moreover, a special knowledge structure $H$ is called a learning space if $H$ satisfy learning smoothness and learning consistency, see [6, 14]. Moreover, each learning space must be finite. Next we give some notations and concepts.

Let $F \subseteq 2^{Q}$ , and let $F_{t} = {K \in F : t \in Q}$ for each t ∈ Q, where Q is a field of knowledge. For each t ∈ Q, put $t^{*} = {r \in Q | H_{r} = H_{t}};$ then t^∗ is called a notion. Therefore, it follows that $t^{*} = {r \in Q | H_{r} = H_{t}} .$ Moreover, if t^∗ is a single item for each t ∈ Q, then $(Q, H)$ is said to be discriminative. In [18], a knowledge structure $(Q, H)$ is called bi-discriminative if $H_{t} ⊈ H_{r}$ and $H_{r} ⊈ H_{t}$ for any distinct t, r ∈ Q. For each $L \in H$ , put $L^{*} = {t^{*} : t \in L},$ $H^{*} = {L^{*} : L \in H},$ and $Q^{*} = {t^{*} : t \in Q} .$ Then the knowledge structure $(Q^{*}, H^{*})$ is discriminative which is produced from $(Q, H)$ . We say that $(Q^{*}, H^{*})$ is the discriminative reduction of $(Q, H)$ . A knowledge space is called a quasi ordinal space if it is closed under intersection. A quasi ordinal space is called an ordinal space if it is discriminative. For a detailed description of KST, the reader may refer to Falmagne and Doignon [8–16].

This paper is stimulated by a recent paper Danilov (2009), where Danilov discussed the knowledge spaces based on the topological point of view. Indeed, the concept of a knowledge space is a generalization of the concept of topological space [7]. It is well known that Császár (2002) in [1] introduced the notions of generalized topological spaces. A generalized topology on a set Z is a subfamily $T$ of 2^Z such that $T$ is closed under arbitrary unions. Clearly, $\emptyset \in T$ . Short introductions to the theory of generalized topology are contained in [1–4].

The theory of topology has many practical applications in real life. By applying the theory of topology, we can better discover some deeper laws inherent in it; in particular, these promote the research in some disciplines. In order to better understand the theory of knowledge space and promote its research, it need to describe the language of the theory of knowledge spaces in pre-topology, which is main motivation of the this paper. We also find that some concepts of knowledge spaces can be well explained in topology. We are interested in a special generalized topological spaces, that is, pre-topological spaces. Indeed, J. Li first discuss the pre-topology (that is, the subbase for the topology) with the applications in rough sets, see [19–21]. Then D. Liu in [23, 24] discuss some properties of pre-topology. However, the theoretical framework of pre-topology is relatively fragmented, and the study of the relations of pre-topology and knowledge spaces are not sufficient. We will seek some deeper connections between topological properties and the properties of knowledge spaces; in particular, we give some characterizations of knowledge spaces from the axioms of separation and the density of topology, which help to understand the theory of knowledge spaces from topological point of view. We systematically study the theory of pre-topology in two aspects. On one hand, we unify the terms of the Császár’s and Li’s such that the theory of pre-topology is well applied in knowledge spaces. On the other hand, in order to give the integrity of the whole theoretical framework of the theory of pre-topology, we systematically list and give some propositions and theorems without proofs except for some part of new results. The main content of this paper is Section 3 in [22].

This paper is organized as follows. Section 1 presents some relevant background about KST. Some concepts of pre-topological spaces are listed in Section 2. The applications of the theory of pre-topological spaces in knowledge spaces are studied in section 3. Section 4 summarizes the main results of this paper.

Denote the sets of real number, rational number, positive integers, the closed unit interval and all non-negative integers by $ℝ$ , $ℚ$ , $ℕ$ , I and ω, respectively. For any sets A and B, we denote (A ∪ B) \ (A ∩ B) by A △ B. Moreover, readers may refer to [14, 17, 22] for terminology and notations not explicitly given here.

2 Some concepts of pre-topological spaces

In this section, we mainly list some concepts of pre-topological spaces which are introduced in [22, Section 2].

Definition 1. [1, 7, 19] A pre-topology on a set Z is a subfamily $T$ of 2^Z such that $⋃ T = Z$ and $⋃ T^{'} \in T$ for any $T^{'} \subseteq T$ . Each element of $T$ is called an open set of the pre-topology.

Clearly, each topological space is a pre-topological space. Now we give some examples of pre-topological spaces which are not topological spaces.

Example 1. (1) Let X be a set with the cardinality |X|≥2. Put $T = {\emptyset} \cup {U : U \subseteq X, | U | \geq 2} .$ Then $(X, T)$ is a pre-topological space, which is not a topological space.

(2) Let X be a set with the cardinality |X| = ω. Put $T = {\emptyset} \cup {U : U \subseteq X, | U | = ω} .$ Then $(X, T)$ is a pre-topological space, which is not a topological space.

(3) Let X = {a, b, c, d}. Put $T = {\emptyset, {a, b}, {a, c}, {a, b, c}, X}$ . Then $(X, T)$ is a pre-topological space which is not a topological space.

Definition 2. [22] A subset D of a pre-topological space Z is closed provided Z \ D is open in Z.

Take an arbitrary subset F of a pre-topological space (Z, τ); then it follows from Proposition [22, Proposition 7] that $⋂ {C : F \subseteq C, Z \ C τ}$ is closed in Z, which is called the closure of F and denoted by $\bar{F}$ .

Definition 3. [22] Let $B$ be a family of non-empty subsets of the set Z. Then $B$ is called a pre-base on Z if $⋃ B = Z$ .

Definition 4. [22] Let $P$ be a pre-base of a pre-topological space Z. Then we say that $P$ is an atom pre-base on the set Z if for each z ∈ Z and $z \in B \in P$ there does not exist $P \in P \ {B}$ such that z ∈ P ⊆ B.

Definition 5. [22] Let Z be a pre-topological space and z ∈ Z. The family $V_{z} \subseteq U_{z}$ is an open neighborhood pre-base at z if for each $U \in U_{z}$ there exists $V \in V_{z}$ with V ⊆ U.

Definition 6. [22] Let Z be a pre-topological space (Z, τ), C ⊆ Z and z ∈ Z. We say that z is an accumulation point of C if U∩ (C \ {z}) ≠ ∅ for any U ∈ τ with z ∈ U. The derived set of C is the set of all accumulation points of C, and is denoted by C^d.

Definition 7. [22] If B is a subset of a pre-topological space (Z, τ), then the set $⋃ {W \subseteq B : W \in τ}$ is called the interior of B and is denoted by intB or B°.

Definition 8. [22] Let h : Y → Z be a mapping and y ∈ Y, where (Y, τ) and (Z, υ) are two pre-topological spaces. The mapping h is pre-continuous at z if h^-1 (W) ∈ τ with z ∈ h^-1 (W) for any W ∈ υ with h (z) ∈ W.

Definition 9. [22] Let p : Y → Z be a surjective mapping between two pre-topological spaces Y and Z. The mapping p is called a pre-quotient mapping provided W is open in Z iff p^-1 (W) is open in Y.

Definition 10. [22] A pre-topological space (Z, τ) is called a T₀-space if for any y, z ∈ Z with y ≠ z there exists W ∈ τ such that W ∩ {y, z} is an exact one-point set.

Definition 11. [22] A pre-topological space (Z, τ) is called a T₁-space if for any y, z ∈ Z with y ≠ z there are V, W ∈ τ such that V ∩ {y, z} = {y} and W ∩ {y, z} = {z}.

Definition 12. [22] A pre-topological space (Z, τ) is called a T₂-space, or a Hausdorff space, if for any y, z ∈ Z with y ≠ z there are V, W ∈ τ such that y ∈ V, z ∈ W and V∩ W = ∅.

Definition 13. [22] A subset D of a pre-topological space Z is called dense in Z provided $\bar{D} = Z$ .

Definition 14. [22] Let Z be a T₁ pre-topological space. We say that Z is a T₃ pre-topological space, or a regular space, if for every z ∈ Z and every closed set A of Z with z ∉ A there are open subsets V and O such that V∩ O = ∅, z ∈ V and A ⊆ O.

Definition 15. [22] Let Z be a T₁ pre-topological space. Then Z is a $T_{3 \frac{1}{2}}$ pre-topological space, or a completely regular pre-topological space, or a Tychonoff pre-topological space, provide for each z ∈ Z and each closed subset C ⊆ Z with z ∉ C there exists a pre-continuous mapping r : Z → I so that r (z) =0 and r (x) =1 for each x ∈ C.

Definition 16. [22] A T₁ pre-topological space Z is called a T₄ pre-topological space, or a normal pre-topological space, provided for any two disjoint closed subsets C, D ⊆ Z there exist disjoint two open sets U, W in Z so that C ⊆ U and D ⊆ W.

Table 1
The equivalent terminologies in pre-topological spaces and knowledge spaces

Pre-topological space Knowledge space Relation

1 Open set Knowledge state ⇔

2 T₀-space Discriminative ⇔

3 T₁-space Bi-discriminative ⇔

4 Atom pre-base Base ⇒

5 Subspace Projection ⇔

6 T₀-reduction Discriminative reduction ⇔

7 Alexandroff space Quasi ordinal space ⇔

8 T₀-Alexandroff space Ordinal space ⇔

9 Locally inner closed points Inner fringe ⇔

10 Locally outer closed points Outer fringe ⇔

11 Locally closed points Fringe ⇔

12 Tight 1-connected Well-graded ⇔

13 Pre-quotient pre-topology Discriminative reduction ⇔

	Pre-topological space	Knowledge space	Relation
1	Open set	Knowledge state	⇔
2	T₀-space	Discriminative	⇔
3	T₁-space	Bi-discriminative	⇔
4	Atom pre-base	Base	⇒
5	Subspace	Projection	⇔
6	T₀-reduction	Discriminative reduction	⇔
7	Alexandroff space	Quasi ordinal space	⇔
8	T₀-Alexandroff space	Ordinal space	⇔
9	Locally inner closed points	Inner fringe	⇔
10	Locally outer closed points	Outer fringe	⇔
11	Locally closed points	Fringe	⇔
12	Tight 1-connected	Well-graded	⇔
13	Pre-quotient pre-topology	Discriminative reduction	⇔

Definition 17. Let Z be a pre-topological space. A pair O, W of two disjoint nonempty open sets of Z is a separation of Z whenever Z = O ∪ W. The pre-topological space Z is called connected provided there is no separation of Z.

Definition 18. [22] Let (Z, τ) be a finite pre-topological space. We say that Z is n-connected provided for any distinct open sets U and W there are open subsets O₀, O₁, …, O_m such that O₀ = U, O₁, …, O_m = W and |O_i ▵ O_i+1| = n for any 0 ≤ i ≤ m - 1.

We say that Z is tight n-connected provided for any distinct open sets V and W there exist open subsets O₀, O₁, …, O_m such that O₀ = V, O₁, …, O_m = W and |O_i ▵ O_i+1| = n for any 0 ≤ i ≤ m - 1, where m = |V ▵ W|.

Definition 19. [22] Let C be a subset of a pre-topological space Z. The set C is chain connected in Z, if for every open covering $U$ in Z and any x, y ∈ C, there can find a finite subfamily {U₁, U₂, …, U_n} of $U$ , such that U_i∩ U_i+1 ≠ ∅ for any i = 1, 2, …, n - 1, x ∈ U₁ and y ∈ U_n. If C = Z, we say that Z is chain-connected.

Definition 20. [22] Let Z be a pre-topological space. If for any y, z ∈ Z there exists a pre-continuous function r : I → Z such that r (0) = y and r (1) = z, then Z is said to be pathwise connected, where I with the usual topology.

3 The language of pre-topology in knowledge spaces

In this section, we will discuss the language of knowledge spaces in the theory of pre-topology. From Section 2 in [22], we see that each knowledge space is just a pre-topological space discussed above in the present paper. First, we list some terminologies in the theory of pre-topological spaces and knowledge spaces respectively that are equivalent, see the following table.

We always say that ( $Q, τ_{H}$ ) is a pre-topological space with $τ_{H} = H$ for a knowledge space ( $Q, H$ ).

3.1 The applications of axioms of separation in knowledge spaces

In this subsection, we discuss some applications of axioms of separation in knowledge spaces. First, the following two theorems in [18] are provided.

Theorem 1. [18] Let ( $Q, H$ ) be a knowledge space. Then ( $Q, τ_{H}$ ) is T₀ iff ( $Q, H$ ) is discriminative.

Theorem 2. [18] Let ( $Q, H$ ) be a knowledge space. Then ( $Q, τ_{H}$ ) is T₁ iff ( $Q, H$ ) is bi-discriminative.

Definition 21. [14] Assume that $(Q, H)$ is a discriminative knowledge structure. For each $H \in H$ , we say that the set $H^{I} = {t \in H : H \ {t} \in H}$ is the inner fringe of H, and that the set $H^{O} = {t \in Q \ H : H \cup {t} \in H}$ is the outer fringe of H. Moreover, the set $H^{F} = H^{I} \cup H^{O}$ is said to be the fringe of H.

The following theorem shows that we can give a characterization for the bi-discriminative knowledge spaces by inner fringe of Q. Since the proof is easy, we left it to the reader.

Theorem 3. For a knowledge space ( $Q, H$ ), it is bi-discriminative iff $Q \ {t} \in H$ for each t ∈ Q, that is, $Q^{I} = Q$ .

Moreover, the following proposition shows that we can give a characterization for knowledge spaces by inner fringe of knowledge states.

Proposition 1. Let ( $Q, H$ ) be a knowledge space. For each $H \in H$ and t ∈ H, it follows that $t \in H^{I}$ iff $q \notin \bar{Q \ (H \ {t})}$ for any q ∈ H \ {t}.

Proof. Necessity. Assume that $t \in H^{I}$ , then $H \ {t} \in H$ , hence (H\ {t}) ∩ (Q \ (H \ {t})) = ∅, that is, $q \notin \bar{Q \ (H \ {t})}$ for any q ∈ H \ {t}.

Sufficiency. Assume that $q \notin \bar{Q \ (H \ {t})}$ for any q ∈ H \ {t}, then for each q ∈ H \ {t} there exists $G (q) \in H$ such that G (q)∩ (Q \ (H \ {t})) = ∅, hence G (q) ⊆ H \ {t}. Therefore, $H \ {t} = ⋃_{q \in H \ {t}} G (q) \in H .$ □

The following proposition shows that we can give a characterization of outer fringe of a state for knowledge spaces by the derived points.

Proposition 2. Let ( $Q, H$ ) be a knowledge space, $H \in H$ and t ∈ Q \ H. Then $t \in H^{O}$ iff t ∉ (Q \ H) ^d in $(Q, τ_{H})$ .

Proof. Necessity. Assume $t \in H^{O}$ , then $H \cup {t} \in H$ , hence $(H \cup {t}) \cap ((Q \ H) \ {t}) = \emptyset,$ thus t ∉ (Q \ H) ^d in $(Q, τ_{H})$ .

Sufficiency. Suppose that t ∉ (Q \ H) ^d in $(Q, τ_{H})$ , then there exists $G \in H$ such that t ∈ G and G∩ ((Q \ H) \ {t}) = ∅. Hence L ⊆ H ∪ {t}, then $H \cup G = H \cup {t} \in H$ , which shows $t \in H^{O}$ . □

By Propositions 1 and 2, the following theorem holds.

Theorem 4. Let ( $Q, H$ ) be a knowledge space, $H \in H$ and t ∈ Q. Then $t \in H^{F}$ iff either $(H \ {t}) \cap \bar{Q \ (H \ {t})} = \emptyset$ and t ∈ H or t ∉ (Q \ H) ^d in $(Q, τ_{H})$ and t ∉ H.

In knowledge assessment, the device of the items is very important. Since the time, the manpower and the material resources are limited, ones hope that the process of the assessment is efficient, and that the items are not repeated as far as possible in the process of the assessment. Further, ones hope that the selection of items in the knowledge assessment should not only have depth but also have breadth. However, if the device of the items is not appropriate, then it is possible that the process of the knowledge assessment is invalid, see the following example.

Example 2. In a knowledge assessment, a teacher in order to check the level for a course of a student, this teacher designs these five items (that is, knowledge points), that is, let Q = {a₁, a₂, a₃, a₄, a₅}. Now assume ( $Q, H$ ) is a knowledge space on Q, where and $H = {\emptyset} \cup {O : | Q \ O | \leq 2, O \subseteq Q} .$ Then ( $Q, H$ ) is a bi-discriminative. Assume this student is capable of solving of items {a₁, a₂} (that is, she/his knowledge state indeed). Then, for each $K \in H \ {\emptyset}$ , the teacher devices an exercise E_K which contains all knowledge points of K, then it is obvious that the student can not complete any exercise since this student is only capable of solving of at most two knowledge points; then it is problem for the teacher to check the level for this course of this student.

In a knowledge assessment device, we always find that the methods are inadequate, hence we have to provide further deepening and enriching the methods in the knowledge assessments. Therefore, we have the following subclass of bi-discriminative knowledge spaces.

Definition 22. A knowledge space ( $Q, H$ ) is completely discriminative provided there exists $H \in H_{p}$ and $L \in H_{q}$ with H∩ L = ∅ for any distinct p, q ∈ Q.

The [18, Example 3] is a completely discriminative knowledge space. Obviously, Example 2 is a bi-discriminative knowledge space which is not completely discriminative. If we enrich the methods of knowledge assessment in Example 2 as follows $H = {\emptyset} \cup {U : | Q \ U | \leq 3, U \subset Q},$ then we will know the level for the course of this student.

Definition 23. [11] Let μ : Q → 2^{2^S\{∅}} \ {∅} be a mapping, where Q and S are non-empty sets of items and skills respectively. Then the triple (Q, S, μ) is called a skill multimap. The elements of μ (t) are said to be competencies for every t ∈ Q. Moreover, if the elements of each μ (t) are pairwise incomparable, then (Q, S, μ) is called a skill function.

Definition 24. [11] For a skill multimap (Q, S, μ), we define a mapping p : 2^S → 2^Q as follows: for each R ∈ 2^S, put $p (R) = {g \in Q | there exists R_{g} \in μ (g) with R_{g} \subseteq R} .$ Then, the mapping p is said to be the problem function induced by (Q, S, μ). Put $H = {p (R) | R \in 2^{S}} .$ We say that the knowledge structure $(Q, H)$ is delineated by (Q, S, μ).

In [14, p117, Problem 6], the authors posed the following problem.

Problem 1. Under which condition on a skill multimap is the delineated structure a knowledge space?

Indeed, in [18] the authors also asked which kind of skill multimaps delineates knowledge spaces. The following theorem gives an answer to this problem when each item with finitely many competencies. For a skill multimap (Q, S, μ) and each t ∈ Q, we denote μ_M (t) by the set of all minimum elements in μ (t).

Theorem 5. Let (Q, S, μ) be a skill multimap, where each μ (t) is a finite set. Then the delineated knowledge structure $(Q, H)$ is a knowledge space iff, for any H ⊆ Q, $H \in H$ iff there is $P_{H} \subseteq ⋃_{t \in Q} μ_{M} (t)$ such that $H = ⋃_{D \in P_{H}} p (D)$ .

Proof. Sufficiency. Let p be the problem function induced by μ, and let $(Q, H)$ be the knowledge structure which is delineated by (Q, S, μ). Take any a subfamily $H^{'}$ of $H$ . We claim that $⋃ H^{'} \in H$ . Indeed, for each $H \in H^{'}$ , there exists a subset $P_{H} \subseteq ⋃_{t \in Q} μ_{M} (t)$ such that $⋃_{D \in P_{H}} p (D) = H$ . Put $P = ⋃_{H \in H^{'}} P_{H}$ . Clearly, we have $⋃ H^{'} = ⋃_{H \in H^{'}} (⋃_{D \in P_{H}} p (D)) = ⋃_{D \in P} p (D) .$ From our assumption, it follows that $⋃ H^{'} \in H$ .

Necessity. Let $(Q, H)$ be the knowledge space which is delineated by (Q, S, μ). Take any H ⊆ Q. Assume that $H \in H$ . Hence we can take a subset R ⊆ S such that p (R) = H. For each t ∈ H, there exists a C_t ∈ μ_M (t) such that C_t ⊆ R. We claim that H = ⋃ _t∈Hp (C_t). Clearly, H ⊆ ⋃ _t∈Hp (C_t); moreover, since each C_t ⊆ R and p (R) = H, it follows that ⋃_t∈Hp (C_t) ⊆ H. Therefore, H = ⋃ _t∈Hp (C_t). Now assume that there exists $P_{H} \subseteq ⋃_{t \in Q} μ_{M} (t)$ such that $H = ⋃_{D \in P_{H}} p (D)$ . Since $(Q, H)$ is a knowledge space and each $p (D) \in H$ , it follows that $⋃_{D \in P_{H}} p (D) \in H$ , hence $H \in H$ . □

Corollary 1. Let (Q, S, μ) be a skill multimap such that each μ (t) is a finite set. If, for any subset Q′ ⊆ Q and g ∈ Q, the following condition (★) holds, then $(Q, H)$ is a knowledge space.

(★) For any $M \subseteq ⋃_{t \in Q^{'}} μ (t)$ , if C\ D ≠ ∅ for any C ∈ μ_M (g) and $D \in M$ , then $C \ ⋃ M \neq \emptyset$ for each C ∈ μ_M (g).

Proof. By Theorem 5, we need to prove that for any H ⊆ Q, the set $H \in H$ iff there exists $M \subseteq ⋃_{t \in Q} μ_{M} (t)$ such that $H = ⋃_{D \in M} p (D)$ .

Suppose that $H \in H$ , then there exists a subfamily $M$ of ⋃_t∈H μ_M (t) such that $H = p (⋃ M)$ . We claim that $H = p (⋃ M) = ⋃_{D \in M} p (D)$ . Clearly, we have $⋃_{D \in M} p (D) \subseteq p (⋃ M)$ . Assume that $p (⋃ M) \ ⋃_{D \in M} p (D) \neq \emptyset$ . Take any $g \in p (⋃ M) \ ⋃_{D \in M} p (D)$ . Hence C\ D ≠ ∅ for each C ∈ μ_M (g) and $D \in M$ , then from the condition (★) it follows that $C \ ⋃ M \neq \emptyset$ for any C ∈ μ_M (g), which leads to a contradiction with $g \in p (⋃ M) \ ⋃_{D \in M} p (D)$ .

Pick any H ⊆ Q, and assume that there is $M \subseteq ⋃_{t \in Q} μ_{M} (t)$ such that $H = ⋃_{D \in M} p (D)$ . Then $H \in H$ since $H = p (⋃ M)$ by the condition (★). □

Corollary 2. Let (Q, S, μ) be a skill multimap. If, for each t ∈ Q and C ∈ μ (t), there exists s_C ∈ C such that {s_C} ∈ μ (t), then the delineated knowledge structure is a knowledge space.

Proof. Indeed, it is easily verified that the condition (★) holds in Corollary 1; moreover, each μ_M (q) consists of elements of singleton. Therefore, the delineated knowledge structure is a knowledge space by Corollary 1. □

It follows from the following example that the condition in Corollary 2 is a sufficient and non-necessary condition.

Example 3. Assume that (Q, S, μ) is a skill multimap such that μ (t) = {S} for each t ∈ Q. It is obvious that the delineated knowledge structure $H = {\emptyset, Q}$ is a knowledge space. However, the condition in Corollary 2 does not hold.

By Corollary 1, the following corollary holds.

Corollary 3. Let (Q, S, μ) be a skill function. If, for any subset Q′ ⊆ Q and g ∈ Q, the following condition (∗) holds, then $(Q, H)$ is a knowledge space.

(★) For each $M \subseteq ⋃_{t \in Q^{'}} μ (t)$ , if C\ D ≠ ∅ for any C ∈ μ (g) and $D \in M$ , then $C \ ⋃ M \neq \emptyset$ for each C ∈ μ (g).

Definition 25. [18] Suppose that Z is a non-empty set, and suppose that $O$ and $W$ are two family of subsets on Z. If for any $O \in O$ there exists $W \in W$ with W ⊆ O, then we say that $O$ is refined by $W$ which is denoted by $O ⋐ W$ . Otherwise, $O$ is not refined by $W$ which is denoted by $O ⋐ W$ . Further, if $O = {O}$ , then we say that O is refined by $W$ , which is denoted by $O ⋐ W$ ; otherwise, we say that O is not refined by $W$ , which is denoted by $O ⋐ W$ .

Theorem 6. For a skill multimap (Q, S, μ) with each μ (r) being finite, then the delineated knowledge structure $(Q, H)$ is completely discriminative iff for any distinct h, q in Q, there exist C_h ∈ μ_M (h) and C_q ∈ μ_M (q) such that, for any g ∈ Q, at most one of C_h⋐ μ_M (g) and C_q⋐ μ_M (g) holds.

Proof. Necessity. Assume that $(Q, H)$ is completely discriminative. Then for any distinct h, q in Q, there exist C_h ∈ μ_M (h) and C_q ∈ μ_M (q) such that p (C_h)∩ p (C_q) = ∅. For any g ∈ Q, without loss of generality, suppose that C_h⋐ μ_M (g), then g ∈ p (C_h). Since p (C_h)∩ p (C_q) = ∅, it follows that g ∉ p (C_q), then C\ C_q ≠ ∅ for any C ∈ μ_M (g). Hence C_q $⋐$ μ_M (g).

Sufficiency. For any distinct h, q in Q, it follows from the assumption that there exist C_h ∈ μ_M (h) and C_q ∈ μ_M (q) such that, for any g ∈ Q, at most one of C_h⋐ μ_M (g) and C_q⋐ μ_M (g) holds. Then it is easily verified that p (C_h)∩ p (C_q) = ∅. Therefore, $(Q, H)$ is completely discriminative. □

The following corollary is easily checked by Theorems 5 and 6.

Corollary 4. For a skill multimap (Q, S, μ) with each μ (r) being finite, then the delineated knowledge structure $(Q, H)$ is a completely discriminative knowledge space iff the following two conditions hold:

For any H ⊆ Q, the set $H \in H$ iff there exists $P_{H} \subseteq ⋃_{t \in Q} μ_{M} (t)$ such that $H = ⋃_{D \in P_{H}} p (D)$ ;

For any distinct h, q in Q, there exist C_h ∈ μ_M (h) and C_q ∈ μ_M (q) such that, for any g ∈ Q, at most one of C_h⋐ μ_M (g) and C_q⋐ μ_M (g) holds.

The following Theorems 7 and 8 show that the language of the regularity of pre-topology in knowledge spaces.

Theorem 7. Let ( $Q, H$ ) be a knowledge space, and let $(Q, τ_{H})$ be regular. For each t ∈ Q and $H \in H_{t}$ , we have $Q \ H \in H$ or H is not an atom at t.

Proof. Take an arbitrary t ∈ Q, and any $H \in H_{t}$ . Assume that $Q \ H \notin H$ , then H is not closed in $(Q, τ_{H})$ . Since $(Q, τ_{H})$ is regular and t ∉ Q \ H, there are $L \in H$ and $K \in H$ with t ∈ L, Q \ H ⊆ K and L∩ K = ∅. Because $Q \ K \notin H$ , it follows that K∩ H ≠ ∅; moreover, it is obvious that L ⊆ Q \ K ⊆ H, then L ≠ H since $Q \ H \notin H$ , hence H is not an atom at t. □

Corollary 5. Let ( $Q, H$ ) be a knowledge space with $(Q, τ_{H})$ being regular. If there exists an atom K at some t ∈ Q, then $(Q, τ_{H})$ is not a connected space. In particular, if Q is finite and $(Q, τ_{H})$ is regular, then $(Q, τ_{H})$ is not a connected space.

By a similar proof of Theorem 7, the following theorem holds.

Theorem 8. Let ( $Q, H$ ) be a knowledge space with $(Q, τ_{H})$ being regular. For each $K \in H$ and t ∈ Q, if $t \in K^{O}$ , then $Q \ (K \cup {t}) \in H$ or K ∪ {t} is not an atom at t.

Remark 1. It is an interesting topic to find the applications of axioms of separation of Tychonoff and normality in knowledge spaces.

3.2 Alexandroff spaces and quasi ordinal spaces

It is well known that a topological space is called an Alexandroff space provided the intersection of arbitrary a family of open subsets is open. From Theorem 1, we see that each ordinal space is equivalent to a T₀-Alexandroff space.

Theorem 9. The knowledge space ( $Q, H$ ) is an ordinal space iff ( $Q, τ_{H}$ ) is a T₀-Alexandroff space.

Proof. Since each Alexandroff space is equivalent to a quasi ordinal space, it follows from Theorem 1 that each ordinal space is equivalent to a T₀-Alexandroff space. □

The following theorem holds by [14, Theorem 3.8.3] and Theorem 9.

Theorem 10. There exists a bijection between the collection of all quasi orders $Q$ on Q and the collection of all Alexandroff spaces $T$ on Q.

If a relation on a set Z is reflexive and transitive, then we say this relation is a quasi order on Z, and if Z is equipped with a quasi order is called quasi ordered. Given a quasi-order Q; we can construct the Alexandroff space Q (Q) as the set Q with the topology generated by {] ← , q] : q ∈ Q}. Conversely, given an Alexandroff space (Q, τ); we can construct a quasi-order Q (Q) as the set with the order x ⪯ y iff x ∈ ⋂ τ (y).

Proposition 3. Let ( $Q, F_{1})$ and ( $Q, F_{2})$ be two pre-topologies on Q such that ( $Q, F_{1})$ and ( $Q, F_{2})$ are two Alexandroff spaces. Then $F_{1} = F_{2}$ iff, for any p, q ∈ Q, $F_{1} (p) \subseteq F_{1} (q) \Leftrightarrow F_{2} (p) \subseteq F_{2} (q)$ .

Proof. The necessity is obvious. In order to prove the converse implication, suppose that $O \in F_{1}$ . Hence $O \subseteq ⋃_{p \in O} (⋂ F_{2} (p)) \in F_{2} .$ Put $O^{'} = ⋃_{p \in O} (⋂ F_{2} (p))$ . We conclude that O = O′. Indeed, take any u ∈ O′. Then there exists v ∈ O such that $u \in ⋂ F_{2} (v)$ . Hence $F_{2} (v) \subseteq F_{2} (u)$ , then $F_{1} (v) \subseteq F_{1} (u)$ . We obtain $u \in ⋂ F_{1} (v)$ , which implies that u ∈ O. This gives O′ ⊆ O, hence O = O′. We conclude that $F_{1} \subseteq F_{2}$ , and by symmetry, $F_{1} = F_{2}$ . □

Proposition 4. If ( $Q, H$ ) is a knowledge structure, then the following statements are equivalent:

( $Q, H$ ) is a finite ordinal space;

( $Q, H$ ) is a learning space and $⋂ H_{p} \in H$ for each p ∈ Q;

( $Q, τ_{H}$ ) is a finite T₀-Alexandroff space.

Proof. By Theorem 9, we have (1) ⇔ (3). It follows from [14, Theorem 3.8.7] that (1) ⇒ (2). Hence it suffices to prove (2) ⇒ (3). Since ( $Q, H$ ) is a learning space, it follows that ( $Q, H$ ) is finite and discriminative, hence ( $Q, τ_{H}$ ) is T₀. Suppose that $H, L \in H$ . If H ∩ L is empty, then $H \cap L \in H$ . Assume that H∩ L ≠ ∅. For each p ∈ H ∩ L, since $⋂ H_{p} \in H$ , we have $⋂ H_{p} \subset H$ and $⋂ H_{p} \subset L$ , hence $⋂ H_{p} \subset H \cap L$ . Therefore, $H \cap L = ⋃_{p \in H \cap L} ⋂ H_{p},$ which shows that $H \cap L \in H$ . Thus ( $Q, τ_{H}$ ) is a finite T₀-Alexandroff space. □

Let $H$ be a family of subsets of a finite set Q such that $H$ is closed under union. The family $H$ is called an antimatroid provided $Q = ⋃ H$ and $H$ satisfies the condition: for each non-empty element H of $H$ , there is t ∈ H such that $H \ {t} \in H$ .

Lemma 1. Any finite T₀-Alexandroff space X is antimatroid.

Proof. For each point x ∈ X, let V_x be a minimal open set containing the point x. Let U be any non-empty open set in X. Take an arbitrary point y₀ ∈ U. If U is a single set, then it is obviously true. Thus we assume that |U|≥2. If ⋃_y∈U\{y₀}V_y = U \ {y₀}, then U \ {y₀} is obviously open in X, hence the assertion of Lemma is true. Therefore, now we assume that ⋃_y∈U\{y₀}V_y ≠ U \ {y₀}, then there is y₁ ∈ U \ {y₀} with y₁ ∉ V_{y
₀} since X is T₀. If U \ V_{y
₀} = {y₁}, then the assertion of Lemma is true. Otherwise, |U \ V_{y
₀}|≥2. If V_{y
₀} ∪ ⋃ _{y∈U\(V_{y
₀}∪{y₁})}V_y = U \ {y₁}, then U \ {y₁} is obviously open in X, hence the assertion of Lemma is true. Therefore, we may assume that V_{y
₀} ∪ (⋃ _{y∈U\(V_{y
₀}∪{y₁})}V_y) ≠ U \ {y₁}, then there exists y₂ ∈ U \ (V_{y
₀} ∪ {y₁}) with y₂ ∉ V_{y
₁} ∪ V_{y
₀} since X is T₀. Since U is finite, by induction, it follows that there exists a point t ∈ U so that U \ {t} is open in X. □

Theorem 11. Any finite T₀-Alexandroff space X is tight 1-connected.

Proof. Since X is a finite T₀-Alexandroff space, we denote the minimal open neighborhood of x by W_x. By [22, Theorem 30], it only need to prove that $(O ▵ W) \cap O^{LC} \neq \emptyset$ for any open sets O and W with O ≠ W. We divide the rest proof into the following two cases.

Case 1: O\ W ≠ ∅.

It follow from Lemma 1 and [7, Lemma 6] that we can take a point z ∈ O \ W such that O \ {z} is open in X. Therefore, $z \in (O ▵ W) \cap O^{LC}$ .

Case 2: O\ W = ∅.

Indeed, we shall use an induction on the size of the set W \ O. If W \ O is a single set {x}, then $x \in (O ▵ W) \cap O^{LC}$ . Suppose that there is a point x ∈ W \ O such that O \ ∪ {x} is open if |W \ O| = n for each n ≥ 2. Now assume that |W \ O| = n + 1. Then it follow from Lemma l and [7, Lemma 6] that there exists a point z ∈ W \ O such that W (z) = W \ {z} is open in X. Then |W (z) \ O| = n. By our assumption, there exists y ∈ W (z) \ O such that O ∪ {y} is open in X. □

Let $F$ be a family of subsets of a finite set Z. We say that $F$ is well-graded provided, for any $K, H \in F$ with L ≠ H, there exists a finite sequence of states L = H₀, H₁, …, H_m = H so that d (H_i-1, H_i) =1 for 1 ≤ i ≤ m, where m = d (L, H).

Corollary 6. Each finite ordinal space $(Q, H)$ is well-graded.

Remark 2. There is a finite ordinal space $(Q, H)$ such that $(Q, τ_{H})$ is not connected. Indeed, let Q = {z₁, z₂, z₃, z₄, z₅, z₆} endowed with the following knowledge structure $\begin{matrix} H = {\emptyset, {z_{4}}, {z_{1}, z_{3}}, {z_{5}, z_{6}}, {z_{1}, z_{3}, z_{4}}, \\ {z_{1}, z_{2}, z_{3}}, {z_{4}, z_{5}, z_{6}}, {z_{1}, z_{3}, z_{5}, z_{6}}, \\ {z_{1}, z_{2}, z_{3}, z_{4}}, Q} . \end{matrix}$ Then $(Q, H)$ is a finite ordinal space, which is not connected since {z₁, z₂, z₃} is open and closed in $(Q, τ_{H}) .$

However, we have the following Theorem 12.

Theorem 12. If $(Q, H)$ is an ordinal space, then the following conditions are equivalent:

$(Q, τ_{H})$ is connected;

$(Q, τ_{H})$ is chain connected;

$(Q, τ_{H})$ is pathwise connected;

for any r, t ∈ Q, there exists a finite sequence q₀, …, q_n+1 ∈ Q such that r = q₀, q_n+1 = t and M (q_i)∩ M (q_j) ≠ ∅ if |i - j|≤1.

Proof. By [5, Theorem 2.7], the proof is easily verified. □

The following proposition gives a characterization of quasi ordinal knowledge spaces.

Proposition 5. Assume that $(Q, H)$ is a knowledge space. Then $(Q, H)$ is a quasi ordinal space iff each point in Q has a unique minimal state M (t) containing the point t.

Proof. Necessity. Let $(Q, H)$ be a quasi ordinal space with t ∈ Q. Put $M (t) = {G : t \in G \in H}$ and $M (t) = ⋂ M (t)$ . Since $(Q, H)$ is a quasi ordinal space, it follows that $M (t) \in H$ . From the definition of M (t), it is obvious that M (t) is a unique minimal state containing the point t.

Sufficiency. Assume each t ∈ Q has a unique minimal state M (t) containing the point t. Consider an arbitrary intersection of states, H = ⋂ _β∈AH_β, where every $H_{β} \in H$ . If H =∅, then $H \in H$ and we are done. If H≠ ∅, then take any t ∈ H and we have t ∈ H_β for each β ∈ A, hence M (t) ⊆ H_β for each β ∈ A since M (t) is the unique state containing the point t. Therefore, M (t) ⊆ H for each β ∈ A, thus $H = ⋃_{t \in H} M (t) \in H$ since $(Q, H)$ is a knowledge space. Therefore, $(Q, H)$ is a quasi ordinal space. □

Theorem 13. If $(Q, H)$ is a bi-discriminative quasi ordinal space, then $H = 2^{Q}$ .

Proof. It only need to prove that ${t} \in H$ for each t ∈ Q. Indeed, take any t ∈ Q. Obviously, $(Q, H)$ is a T₁ Alexandroff space. Therefore, it follows that ${t} = ⋂ {H \in H : t \in H} \in H .$ □

From [5, Theorem 2.9], the following two theorems hold.

Theorem 14. If $(Q, H)$ is a quasi ordinal space, then $Q \ M (t) \in H$ for each t ∈ Q iff the pre-topological space $(Q, τ_{H})$ is regular, where M (t) is the unique minimal state containing the point t.

Theorem 15. Let $(Q, H)$ be a quasi ordinal space. If $(Q, τ_{H})$ is a regular pre-topological space with a countable dense subset, then $(Q, τ_{H})$ is normal.

3.3 Primary items in knowledge spaces

In this section, we discuss the density of pre-topological spaces with the applications in knowledge spaces. By [22, Theorem 20], it is easily verified that the following theorem holds.

Theorem 16. The inequality $| Q | \leq | H |$ holds for each discriminative knowledge space $(Q, H)$ .

We say that a pairwise disjoint family consisting of non-empty open subsets of a pre-topological space (Z, τ) is called a cellular family. The cellularity of Z is defined as follows: $c (Z) = sup {| V | : V is a cellular family in Z} .$ Here, the cellularity of a pre-topological space maybe finite.

Lemma 2. Let Z be a pre-topological space such that c (Z) = κ. If $V$ is an open cover of Z, then there is a subcollection $W$ of $V$ with $| W | \leq κ$ and $\bar{⋃ W} = Z$ .

Proof. Let $O$ be the family of all non-empty open sets in Z that are subsets of some element of $V$ . It follows from Zorn’s lemma that we can take a maximal cellular family $O^{'} \subset O$ . Hence $| O^{'} | \leq c (X) = κ$ , and $Z = \bar{⋃ O^{'}}$ by the maximality of $O^{'}$ . For each $O \in O^{'}$ , fix a $W_{O} \in V$ such that O ⊂ W_O. Put $W = {W_{O} : O \in O^{'}}$ . Then $\bar{⋃ W} = Z$ . □

The following corollary holds by Lemma 2.

Corollary 7. Let $B$ be an atom pre-base of a finite pre-topological space Z. Then there is a maximally disjoint subfamily $B^{'}$ of $B$ such that $Z = \bar{B^{'}}$ .

Let $(Q, H)$ be a knowledge space. If D is a dense subset of $(Q, τ_{H})$ , we say that D is the primary items of Q. In other work, one person in order to master an item, she or he must master some items of D.

Definition 26. Let Z be a pre-topological space. The infimum of the set ${| D | : D \subseteq Z, \bar{D} = Z}$ is called the density of Z, which is denoted by d (Z).

For a knowledge space $(Q, H)$ , we say that A is an atom primary items of $(Q, H)$ if A is dense and |A| = d (Q). Moreover, the atom primary items of a knowledge space is not unique. Indeed, the sets {z₁, z₂, z₃}, {z₁, z₂, z₄} and {z₂, z₃, z₄} are all atom primary items of the pre-topological space (Z, τ) in [22, Example 4]. Clearly, we have the following proposition.

Proposition 6. For any knowledge space $(Q, H)$ , if A is an atom primary items of $(Q, H)$ , then $| A | \leq min {| Q |, w (Q, τ_{H})}$ .

From Lemma 2, the following proposition holds.

Proposition 7. For any knowledge space $(Q, H)$ , if A is an atom primary items of $(Q, H)$ , then |A| ≥ c (Q).

The following theorem gives a complement for [22, Example 6].

Theorem 17. Let $(Q, H)$ be a knowledge space, and let D be a dense subset of $(Q, τ_{H})$ . If $(Q, τ_{H})$ is Hausdorff and $\bar{H \cap D} = \bar{H}$ for each $H \in H$ , then |Q|≤2^{2^|D|}.

Proof. For every t ∈ Q, the family $N (t) = {U \cap A : U \in H_{t}}$ of subsets of D. Since $(Q, τ_{H})$ is Hausdorff and $\bar{L \cap D} = \bar{L}$ for any $L \in H$ , we conclude that the intersection of the closures of all members of $N (t)$ equals {t}; thus $N (t) \neq N (g)$ for g ≠ q. Clearly, the number of all distinct families $N (t)$ is not larger than 2^{2^|D|}, hence |Q|≤2^{2^|D|}. □

Remark 3. The Hausdorff pre-topological space (Z, τ) in [22, Example 4], which has an atom primary items D = {z₁, z₂, z₃} such that |Z|≤2^2³ and |D|=3 < 4. However, $\begin{matrix} {z_{2}} = \bar{{z_{2}}} = \bar{{z_{2}, z_{4}} \cap {z_{1}, z_{2}, z_{3}}} \neq {z_{2}, z_{4}} \\ = \bar{{z_{2}, z_{4}}} . \end{matrix}$ Hence the condition “ $\bar{H \cap D} = \bar{H}$ for each $H \in H$ ” is sufficient and not necessary.

Lemma 3. Let $(Q, H)$ be a finite knowledge space, and let $H^{'}$ be a subfamily of $H$ . Put $m = max {| H_{q}^{'} | : q \in Q}$ . For any distinct r, t ∈ Q, if $| H_{r}^{'} | = | H_{t}^{'} | = m$ , then $⋂ H_{r}^{'} = ⋂ H_{t}^{'}$ or $(⋂ H_{r}^{'}) \cap (⋂ H_{t}^{'}) = \emptyset$ .

Proof. Assume that $⋂ H_{r}^{'} \neq ⋂ H_{t}^{'}$ . Then, without loss of generality, we assume that there is $H \in H_{t}^{'}$ with r ∉ H. Now assume that $(⋂ H_{r}^{'}) \cap (⋂ H_{t}^{'}) \neq \emptyset$ . Then there is $q \in (⋂ H_{r}^{'}) \cap (⋂ H_{t}^{'})$ , hence q ∈ H and q ∈ L for any $L \in H_{r}^{'}$ , which implies that $H_{q}^{'} = m + 1$ , this is a contradiction. Therefore, $(⋂ H_{r}^{'}) \cap (⋂ H_{t}^{'}) = \emptyset$ . □

Now we give a method to construct a primary items for a finite pre-topological space as follows.

Let $B$ be an atom pre-base for a finite pre-topological space $(Q, H)$ . Now we define a dense subset A of $(Q, τ_{H})$ , by induction, as follows. First, let $B (1) = B$ , and take an arbitrary p₁ ∈ Q such that $| B (1)_{p_{1}} | = max {| B (1)_{q} | : q \in Q}$ . Put A₁ = {p₁} . If $B \ B (1)_{p_{1}} = \emptyset$ , then we stop our induction, and put A = A₁. If $B \ B (1)_{p_{1}} \neq \emptyset$ , then put $B (2) = B \ B (1)_{p_{1}}$ . Then take an arbitrary p₂ ∈ Q such that $| B (2)_{p_{2}} | = max {| B (2)_{q} | : q \in Q}$ . Put A₂ = A₁ ∪ {p₂}. Assume that we have define a subset A_k = {p_i : i ≤ k} (k ≥ 2) such that $| B (i)_{p_{i}} | = max {| B (i)_{q} | : q \in Q}$ for any 1 < i ≤ k, where $B (i) = B \ ⋃_{j < i} B (j)_{p_{j}}$ . If $B \ ⋃_{i \leq k} B (i)_{p_{i}} = \emptyset$ , then we stop our induction, and put A = A_k; otherwise, let $B (k + 1) = B \ ⋃_{i \leq k} B (i)_{p_{i}}$ . Then take an arbitrary p_k+1 ∈ Q such that $| B (k + 1)_{p_{k + 1}} | = max {| B (k)_{q} | : q \in Q}$ . Then put A_k+1 = A_k ∪ {p_k+1} . Since Q is finite, there exists a minimum $n \in ℕ$ such that $B \ ⋃_{i \leq n} B (i)_{p_{i}} = \emptyset$ . Moreover, from Lemma 3, it follows that, for any j ≤ n, if $| B (j) | = 1$ we may assume that $p_{j} \notin ⋃ (⋃_{i < j} B (i)_{p_{i}})$ . Then we put A = A_n. Clearly, A is dense in $(Q, H)$ .

Generally, A is not an atom primary items of $(Q, H)$ . We must delete some surplus points of A such that it is an atom primary items of $(Q, H)$ .

If B∩ (A \ {p₁}) ≠ ∅ for any $B \in B (1)_{p_{1}}$ , then put B₁ = A \ {p₁}; otherwise, put B₁ = A. If p₁ ∈ B₁ and B∩ (B₁ \ {p₂}) ≠ ∅ for any $B \in B (2)_{p_{2}}$ , then put B₂ = B₁ \ {p₂}; if p₁ ∈ B₁ and B∩ (B₁ \ {p₂}) = ∅ for some $B \in B (2)_{p_{2}}$ , then put B₂ = B₁; if p₁ ∉ B₁, B∩ (B₁ \ {p₂}) ≠ ∅ for any $B \in B (2)_{p_{2}}$ , and O∩ (B₁ \ {p₂}) ≠ ∅ whenever $p_{2} \in O \in B (1)_{p_{1}}$ , then put B₂ = B₁ \ {p₂}; otherwise, put B₂ = B₁. By induction, we can obtain the subsets B₁, …, B_n of A such that B₁ ⊃ … ⊃ B_n and B_n is dense in $(Q, H)$ . Clearly, we have p_n ∈ B_n. We claim that D = B_n is an atom primary items of $(Q, H)$ , see the following theorem.

Theorem 18. The set D is dense in $(Q, τ_{H})$ and |D| = d (Q), that is, D is an atom primary items of $(Q, H)$ .

Proof. From our construction of D above, it follows that D is dense in $(Q, τ_{H})$ . Hence it only need to prove that $\bar{C} \neq Q$ for ecah subset C of Q with |C| < |D|. Indeed, we shall prove by induction on the cardinality of atom pre-base $B$ .

Clearly, if $| B | = 1$ , then |D|=1, hence $\bar{C} = \emptyset$ . If $| B | = 2$ , it suffices to verify the case of D = {p₁, p₂}. Let C = {q₀}. By our construction, if C is dense in $(Q, τ_{H})$ , then $| B_{q_{0}} | > | B_{p_{1}} |$ , which is a contradiction. Assume that $\bar{C} \neq Q$ for any subset C of Q with |C| < |D| when $| B | \leq k$ . Now assume that $| B | = k + 1$ . From our construction, we conclude that |A ∩ B|=1 for each $B \in B (n)_{p_{n}}$ . We divide the rest proof into the following two cases.

Case 1: $| B (n)_{p_{n}} | \geq 2$ .

Let $B^{'} = ⋃ B (n)_{p_{n}}$ , and put $B^{'} = (B \ B (n)_{p_{n}}) \cup {B^{'}}$ . Let $(Q, F)$ be the knowledge space generated by $B^{'}$ . Clearly, $τ_{F}$ is coarser than $τ_{H}$ , hence D is dense in $τ_{F}$ . Since $| B^{'} | \leq k$ , from our assumption it follows that ${\bar{F}}^{τ_{F}} \neq Q$ for each subset F of Q with |F| < |D|, hence ${\bar{F}}^{τ_{H}} \neq Q$ for each subset F of Q with |F| < |D|.

Case 2: $| B (n)_{p_{n}} | = 1$ .

If $| B (i)_{p_{i}} | = 1$ for each i ≤ n, then it is obvious. Otherwise, there exists a maximal m ≤ n such that $| B (i)_{p_{i}} | = 1$ for any i > m and $| B (m)_{p_{m}} | \geq 2$ . Then put $B^{''} = ⋃ B (m)_{p_{m}}$ . Let $(Q, L)$ be the knowledge space which is generated by the family $B^{''} = (B \ B (m)_{p_{m}}) \cup {B^{''}}$ . Then $τ_{L}$ is coarser than $τ_{H}$ , hence D is dense in $(Q, τ_{L})$ . Since $| B^{''} | \leq k$ , from our assumption it follows that ${\bar{F}}^{τ_{L}} \neq Q$ for each subset F of Q with |F| < |D|, hence ${\bar{F}}^{τ_{H}} \neq Q$ for each subset F of Q with |F| < |D|. □

Finally we give an algorithm for constructing the set of atom primary items for a finite knowledge spaces.

Sketch of Algorithm. Let Q = {q₁, …, q_m}, and list the atom pre-base $B$ as B₁, …, B_n. Form an m × n array T = (T_ij) with the rows and columns representing the atom pre-base $B$ and the elements of Q respectively; thus, both the rows and the columns are indexed from 1 to m and 1 to n respectively. Initially, set T_ij to 1 if q_i ∈ B_j; otherwise, set T_ij to 0.

First, take an arbitrary i₁ ∈ {1, …, m} such that $\sum_{j = 1}^{n} T_{i_{1} j} = max {\sum_{j = 1}^{n} T_{ij} : i \in {1, \dots, m}}$ , then swap the i₁-th row of the matrix with the first row, and swap all the columns N₁ = {j : T_i
₁
j = 1, j = 1, …, n} with the first n₁ columns, where n₁ = |N₁|. If n₁ = n, then we terminates this step, and put A = {p_i
₁}; otherwise, denote the new matrix by $(T_{ij}^{(1)})_{m \times n}$ .

For any k ≥ 2, take an arbitrary i_k ∈ {k, …, m} such that $\begin{matrix} \sum_{j = (\sum_{j = 1}^{k - 1} n_{j}) + 1}^{n} T_{i_{k} j}^{(k - 1)} = max {\sum_{j = (\sum_{j = 1}^{k - 1} n_{j}) + 1}^{n} \\ T_{ij}^{(k - 1)} : i \in {k, \dots, m}}, \end{matrix}$ then swap the i_k-th row of the matrix with the k-th row, and swap all the columns $N_{k} = {j : T_{i_{k} j} = 1, j = (\sum_{j = 1}^{k - 1} n_{j}) + 1, \dots, n}$ with the $j = (\sum_{j = 1}^{k - 1} n_{j}) + 1$ -th column to $j = (\sum_{j = 1}^{k} n_{j})$ -th column, where n_k = |N_k|. If n_k = 1, we may require the i_k satisfying that $T_{i_{k} j}^{(k)} = 0$ for any $j \leq \sum_{j = 1}^{k - 1} n_{j}$ . If $\sum_{j = 1}^{k} n_{j} = n$ , then we terminates this step, and put A = {p_i
₁, …, p_{i
_k}}; otherwise, denote the new matrix by $(T_{ij}^{(k)})_{m \times n}$ .

The set A = {p_i
₁, …, p_{i
_N}} and the sub-matrix $(T_{ij}^{(N)})_{N \times M}$ of $(T_{ij}^{(N)})_{m \times n}$ obtained after step N are the desired set and the desired matrix respectively, where $M = \sum_{i}^{N} n_{i}$ .

Then initialize D to A again.

For each l ≤ N, check whether, there exists $\sum_{k = 1}^{l - 1} n_{k} < j \leq \sum_{k = 1}^{l} n_{k}$ such that $T_{i_{l} j}^{(N)} = 0$ for any i_l > l; if the condition holds, the set D is invariant. If the condition does not hold, then we check whether there exists t < l such that p_{i
_t} ∉ D and $\sum_{j = a_{t - 1} + 1}^{a_{t}} T_{i_{l} j}^{(N)} = 1$ , where $a_{t} = \sum_{i = 1}^{t} n_{i}$ and n₀ = 0; otherwise, delete p_{i
_l} from D. (This terminates step N.)

The set D obtained after step N is the set of atom primary items for a finite knowledge spaces $(Q, H)$ .

Example 4. Let a knowledge space $(Q, H)$ have an atom pre-base ${{z_{1}}, {z_{2}}, {z_{1}, z_{3}}, {z_{2}, z_{3}, z_{4}}, {z_{1}, z_{3}, z_{4}, z_{5}}},$ where Q = {z₁, z₂, z₃, z₄, z₅}. Then

Table 2
The array T = (T_ij) of the atom pre-base $B$ and the elements of Q

{z₁} {z₂} {z₁, z₃} {z₂, z₃, z₄} {z₁, z₃, z₄, z₅}

z ₁ 1 0 1 0 1

z ₂ 0 1 0 1 0

z ₃ 0 0 1 1 1

z ₄ 0 0 0 1 1

z ₅ 0 0 0 0 1

	{z₁}	{z₂}	{z₁, z₃}	{z₂, z₃, z₄}	{z₁, z₃, z₄, z₅}
z ₁	1	0	1	0	1
z ₂	0	1	0	1	0
z ₃	0	0	1	1	1
z ₄	0	0	0	1	1
z ₅	0	0	0	0	1

Since $\sum_{j = 1}^{5} T_{3 j} > \sum_{j = 1}^{5} T_{ij}$ for any i ≠ 3, we swap the 3-th row of the matrix with the first row. Moreover, since T₃₃ = T₃₄ = T₃₅ = 1, it follows that n₁ = 3, then we swap the third column to fifth column with the first column to third column respectively. Then we have the following table.

Table 3

The array $T = (T_{ij}^{(1)})$ of the atom pre-base $B$ and the elements of Q

	{z₁, z₃}	{z₂, z₃, z₄}	{z₁, z₃, z₄, z₅}	{z₁}	{z₂}
z ₃	1	1	1	0	0
z ₁	1	0	1	1	0
z ₂	0	1	0	0	1
z ₄	0	1	1	0	0
z ₅	0	0	1	0	0

Since $\sum_{j = 4}^{5} T_{2 j}^{(1)} = \sum_{j = 4}^{5} T_{3 j}^{(1)} = 1$ and $\sum_{j = 4}^{5} T_{4 j}^{(1)} = \sum_{j = 4}^{5} T_{5 j}^{(1)} = 0$ , hence n₂ = 1 and n₃ = 1. Clearly, n₁ + n₂ + n₃ = 5, we terminates this step. Clearly, $(T_{ij}^{(2)}) = (T_{ij}^{(1)})$ , and $T = (T_{ij}^{(3)})$ as the following table.

Table 4

The array $T = (T_{ij}^{(3)})$ of the atom pre-base $B$ and the elements of Q

	{z₁, z₃}	{z₂, z₃, z₄}	{z₁, z₃, z₄, z₅}	{z₁}	{z₂}
z ₃	1	1	1	0	0
z ₁	1	0	1	1	0
z ₂	0	1	0	0	1

Since $T_{2 j}^{(3)} + T_{3 j}^{(3)} = 1$ for each j = 1, 2, 3, 4, 5, it follows that the set of atom primary items of $(Q, H)$ is D = {z₁, z₂}.

Remark 4. In Example 4, we see that the knowledge points must contain at least one knowledge point in D = {z₁, z₂} if a teacher in order to check the level for a course of a student.

3.4 The applications of connectedness and pre-quotient mapping in knowledge spaces

In this subsection, we discuss some applications of connectedness and pre-quotient mapping in knowledge spaces. First, we have the following proposition. The proof is easy, hence we omit it.

Proposition 8. If $(Q, H)$ is a knowledge space, then $(Q, τ_{H})$ is connected iff $H \cap \bar{H} = {\emptyset, X}$ , where $\bar{H} = {K : Q \ K \in H}$ .

A knowledge space $(Q, H)$ is called granular if for each state $L \in H_{t}$ and t ∈ Q, there exists an atom H at t with H ⊂ L. By Theorem 7, the following proposition holds.

Proposition 9. If $(Q, H)$ is a granular knowledge space and $(Q, τ_{H})$ is regular, then both H and Q \ H belong to $H$ for each atom H. Thus $(Q, τ_{H})$ is not connected.

Assume that σ is a mapping from a non-empty set Q into 2^{2^Q}. We say that σ is an attribution if σ (t)≠ ∅ for every t ∈ Q. The following theorem is obvious.

Theorem 19. Let $(Q, H)$ be a knowledge space which is derived from the attribution σ on Q. If there exist a proper subset Q′ of Q and C_t ∈ σ (t) for each t ∈ Q such that ⋃_t∈Q′C_t ⊂ Q′ and ⋃_t∈Q\Q′C_t ⊂ Q \ Q′, then $(Q, τ_{H})$ is not connected.

Theorem 20. Let $(Q, H)$ be a knowledge space. Then the mapping $h : (Q, H) \to (Q^{*}, H^{*})$ , which is defined by h (t) = t^∗ for each t ∈ Q, is a pre-quotient mapping, that is, the discriminative reduction $(Q^{*}, H^{*})$ is a pre-quotient pre-topology of $(Q, H)$ .

Proof. For each $K^{*} \in H^{*}$ , we have h^-1 (K^∗) = K. Moreover, let h^-1 (A) is open $(Q, H)$ . Then there is a subfamily $H \subseteq H$ with $h^{- 1} (A) = ⋃ H$ , hence $A = ⋃_{H \in H} h (H) = ⋃_{H \in H} H^{*}$ is open in $(Q^{*}, H^{*})$ . Therefore, h a pre-quotient mapping. □

4 Conclusions

The theory of knowledge spaces (KST) was originally regarded as a mathematical framework for knowledge training and assessment. In fact, the concept of a knowledge spaces is a generalization of topological spaces. Indeed, it is equivalent to the notion of pre-topological spaces in [22]. Therefore, the language of topology and its established generalizations are very useful for studying in the theory of knowledge spaces. For example, Falmagne’s discriminative resemble the axiom of separation T₀ for topologies. His notion of quasi ordinal space is just the well-known Alexandroff space in topology.

In order to study the theory of knowledge spaces, we systematically discuss the applications of axioms of separation for knowledge spaces. For a bi-discriminative knowledge space, we prove that the necessary and sufficient condition of an item belongs to some inner fringe or outer fringe of some knowledge state. Moreover, we give a characterization of a skill multimap such that answer to a problem in [14] or [18] whenever each item with finitely many competencies. Further, we discuss the relation of Alexandroff spaces and quasi ordinal spaces. Indeed, the concept of Alexandroff spaces in topology is equivalent to the quasi ordinal spaces in knowledge spaces. We prove that each bi-discriminative quasi ordinal space $(Q, H)$ has $H = 2^{Q}$ , and give a characterization of quasi ordinal spaces that are connected. Finally, we study the roles of the density of pre-topological spaces in primary items for knowledge spaces. We define the concept of primary items in knowledge spaces, that is, one person in order to master an item, she or he must master items. We give an algorithm to find the set of atom primary items for any finite knowledge space.

Our paper only takes a first step to study the language of pre-topology in the theory of knowledge spaces. There are many interesting topological properties need us to further discuss in the theory of knowledge spaces, such as chain-connectedness, path-connectedness, normality, etc. Moreover, how to combine the theory of knowledge spaces with the theory of topology and rough sets, which better reflects the interdisciplinary cross-integration. The theoretical framework introduced here are not sufficient; in particular, the method of studying the theory of knowledge spaces from the topological views are expected to prove useful for future work in this direction.

Footnotes

Declarations

References

Császár

Á.

, Generalized topology, generalizedcontinuity, Acta Math. Hungar. 96 (2002), 351–357.

Császár

Á.

, Separation axioms for generalizedtopologies, Acta Math. Hungar. 104 (2004), 63–69.

Császár

Á.

, On generalized neigghorhood systems, Acta Math. Hungar. 121 (2008), 395–400.

Császár

Á.

, Product of generalized topologies, Acta Math. Hungar. 123 (2009), 127–132.

Arenas

F.G.

, Alexandroff spaces, Acta. Math. Univ. Comenianae LXVIII(1) (1999), 17–25.

Cosyn

and Uzun

H.B.

, Note on two sufficient axioms for awell-graded knowledge space, J. Math. Psychol. 53(1) (2009), 40–42.

Danilov

V.I.

, Knowledge spaces from a topological point of view, J. Math. Psychol. 53 (2009), 510–517.

Doignon

J.P.

and Falmagne

J.C.

, Spaces for the assessment ofknowledge, Int. J. Man-Mach. Stud. 23(2) (1985), 175–196.

Doignon

J.P.

and Falmagne

J.C.

, Well-graded families of relations, Discrete Math. 173(1-3) (1997), 35–44.

10.

Doignon

J.P.

and Falmagne

J.C.

, Knowledge Spaces. Springer Berlin Heidelberg, 1999.

11.

Doignon

J.P.

and Falmagne

J.C.

. Learning Spaces. Springer-Verlag Berlin Heidelberg, 2011.

12.

Falmagne

J.C.

, A latent trait theory via a stochastic learningtheory for a knowledge space, Psychometrika 54(2) (1989), 283–303.

13.

Falmagne

J.C.

, Koppen

, Villano

Doignon

J.P.

and Johanessen

, Introduction to knowledge spaces: how to build, test andsearch them, Psychol. Rev. 97 (1990), 204–224.

14.

Falmagne

J.C.

and Doignon

J.P.

, Learning spaces: Interdisciplinaryapplied mathematics. Berlin, Heidelberg: Springer, 2011.

15.

Falmagne

J.C.

, Albert

, Doble

, Eppstein

and Hu

, Knowledge spaces: Applications in education. Springer Science and Business Media, 2013.

16.

Falmagne

J.C.

and Doignon

J.P.

, A Markovian procedure for assessingthe state of a system, J. Math. Psychol. 32(3) (1988b), 232–258.

17.

Engelking

, General Topology (revised and completed edition), Heldermann Verlag, Berlin, 1989.

18.

and Li

, A note on the separability of items in knowledgestructures delineated by skill multimaps, J. Math. Psychol. 98 (2020), 102427.

19.

, Topological methods on the theory of covering generalizedrough sets, PR & AI 17(1) (2004), 7–10.

20.

, The interior opeerators and the closure operators generated by a subbase, Adv. in Math. 35(4) (2006), 476–484.

21.

, The connectedness relative to a subbase for the topology, Adv. in Math. 36(4) (2007), 421–428.

22.

Lin

, Cao

, Li

The language of pre-topology inknowledge spaces, arXiv: 2111.14380.

23.

Liu

, The separateness relative to a subbase for the topology, College Math. 27(3) (2011), 59–65.

24.

Liu

, The regular space and relative regularity with regard to asubbase, Pure. Appl. Math. 29(6) (2013), 559–564.

The language of knowledge spaces in pre-topology 1

Abstract

Keywords

1 Introduction and preliminaries

2 Some concepts of pre-topological spaces

3.1 The applications of axioms of separation in knowledge spaces

3.2 Alexandroff spaces and quasi ordinal spaces

3.3 Primary items in knowledge spaces

Table 2 The array T = (T ij ) of the atom pre-base B and the elements of Q {z1} {z2} {z1, z3} {z2, z3, z4} {z1, z3, z4, z5} z 1 1 0 1 0 1 z 2 0 1 0 1 0 z 3 0 0 1 1 1 z 4 0 0 0 1 1 z 5 0 0 0 0 1

4 Conclusions

Footnotes

Declarations

References

Table 2
The array T = (T_ij) of the atom pre-base $B$ and the elements of Q

{z₁} {z₂} {z₁, z₃} {z₂, z₃, z₄} {z₁, z₃, z₄, z₅}

z ₁ 1 0 1 0 1

z ₂ 0 1 0 1 0

z ₃ 0 0 1 1 1

z ₄ 0 0 0 1 1

z ₅ 0 0 0 0 1