The language of knowledge spaces in pre-topology 1

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 ∈ H , we say that the set H I = { t ∈ H : H \ { t } ∈ H } is the inner fringe of H, and that the set H O = { t ∈ Q \ H : H ∪ { t } ∈ H } is the outer fringe of H. Moreover, the set H F = H I ∪ 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 } ∈ 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 ∈ H and t ∈ H, it follows that t ∈ H I iff q ∉ Q \ ( H \ { t } ) ¯ for any q ∈ H \ {t}.

Proof. Necessity. Assume that t ∈ H I , then H \ { t } ∈ H , hence (H\ {t}) ∩ (Q \ (H \ {t})) = ∅, that is, q ∉ Q \ ( H \ { t } ) ¯ for any q ∈ H \ {t}.

Sufficiency. Assume that q ∉ Q \ ( H \ { t } ) ¯ for any q ∈ H \ {t}, then for each q ∈ H \ {t} there exists G ( q ) ∈ H such that G (q)∩ (Q \ (H \ {t})) = ∅, hence G (q) ⊆ H \ {t}. Therefore, H \ { t } = ⋃ q ∈ H \ { t } G ( q ) ∈ 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 ∈ H and t ∈ Q \ H. Then t ∈ H O iff t ∉ (Q \ H)  ^d in ( Q , τ H ) .

Proof. Necessity. Assume t ∈ H O , then H ∪ { t } ∈ H , hence ( H ∪ { t } ) ∩ ( ( Q \ H ) \ { t } ) = ∅ , thus t ∉ (Q \ H)  ^d in ( Q , τ H ) .

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

By Propositions 1 and 2, the following theorem holds.

Theorem 4. Let ( Q , H ) be a knowledge space, H ∈ H and t ∈ Q. Then t ∈ H F iff either ( H \ { t } ) ∩ Q \ ( H \ { t } ) ¯ = ∅ 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 = { ∅ } ∪ { O : | Q \ O | ≤ 2 , O ⊆ 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 ∈ H \ { ∅ } , 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 ∈ H p and L ∈ 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 = { ∅ } ∪ { U : | Q \ U | ≤ 3 , U ⊂ 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 ∈ Q | there exists R g ∈ μ ( g ) with R g ⊆ R } . Then, the mapping p is said to be the problem function induced by (Q, S, μ). Put H = { p ( R ) | R ∈ 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 ∈ H iff there is P H ⊆ ⋃ t ∈ Q μ M ( t ) such that H = ⋃ D ∈ 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 ′ ∈ H . Indeed, for each H ∈ H ′ , there exists a subset P H ⊆ ⋃ t ∈ Q μ M ( t ) such that ⋃ D ∈ P H p ( D ) = H . Put P = ⋃ H ∈ H ′ P H . Clearly, we have ⋃ H ′ = ⋃ H ∈ H ′ ( ⋃ D ∈ P H p ( D ) ) = ⋃ D ∈ P p ( D ) . From our assumption, it follows that ⋃ H ′ ∈ H .

Necessity. Let ( Q , H ) be the knowledge space which is delineated by (Q, S, μ). Take any H ⊆ Q. Assume that H ∈ 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 ⊆ ⋃ t ∈ Q μ M ( t ) such that H = ⋃ D ∈ P H p ( D ) . Since ( Q , H ) is a knowledge space and each p ( D ) ∈ H , it follows that ⋃ D ∈ P H p ( D ) ∈ H , hence H ∈ 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 ⊆ ⋃ t ∈ Q ′ μ ( t ) , if C\ D ≠ ∅ for any C ∈ μ _M  (g) and D ∈ M , then C \ ⋃ M ≠ ∅ for each C ∈ μ _M  (g).

Proof. By Theorem 5, we need to prove that for any H ⊆ Q, the set H ∈ H iff there exists M ⊆ ⋃ t ∈ Q μ M ( t ) such that H = ⋃ D ∈ M p ( D ) .

Suppose that H ∈ 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 ∈ M p ( D ) . Clearly, we have ⋃ D ∈ M p ( D ) ⊆ p ( ⋃ M ) . Assume that p ( ⋃ M ) \ ⋃ D ∈ M p ( D ) ≠ ∅ . Take any g ∈ p ( ⋃ M ) \ ⋃ D ∈ M p ( D ) . Hence C\ D ≠ ∅ for each C ∈ μ _M  (g) and D ∈ M , then from the condition (★) it follows that C \ ⋃ M ≠ ∅ for any C ∈ μ _M  (g), which leads to a contradiction with g ∈ p ( ⋃ M ) \ ⋃ D ∈ M p ( D ) .

Pick any H ⊆ Q, and assume that there is M ⊆ ⋃ t ∈ Q μ M ( t ) such that H = ⋃ D ∈ M p ( D ) . Then H ∈ 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 = { ∅ , 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 ⊆ ⋃ t ∈ Q ′ μ ( t ) , if C\ D ≠ ∅ for any C ∈ μ (g) and D ∈ M , then C \ ⋃ M ≠ ∅ 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 ∈ O there exists W ∈ 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 ∈ H iff there exists P H ⊆ ⋃ t ∈ Q μ M ( t ) such that H = ⋃ D ∈ P H p ( D ) ;

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 ∈ H t , we have Q \ H ∈ H or H is not an atom at t.

Proof. Take an arbitrary t ∈ Q, and any H ∈ H t . Assume that Q \ H ∉ H , then H is not closed in ( Q , τ H ) . Since ( Q , τ H ) is regular and t ∉ Q \ H, there are L ∈ H and K ∈ H with t ∈ L, Q \ H ⊆ K and L∩ K = ∅. Because Q \ K ∉ H , it follows that K∩ H ≠ ∅; moreover, it is obvious that L ⊆ Q \ K ⊆ H, then L ≠ H since Q \ H ∉ 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 ∈ H and t ∈ Q, if t ∈ K O , then Q \ ( K ∪ { t } ) ∈ 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 ) ⊆ F 1 ( q ) ⇔ F 2 ( p ) ⊆ F 2 ( q ) .

Proof. The necessity is obvious. In order to prove the converse implication, suppose that O ∈ F 1 . Hence O ⊆ ⋃ p ∈ O ( ⋂ F 2 ( p ) ) ∈ F 2 . Put O ′ = ⋃ p ∈ O ( ⋂ F 2 ( p ) ) . We conclude that O = O′. Indeed, take any u ∈ O′. Then there exists v ∈ O such that u ∈ ⋂ F 2 ( v ) . Hence F 2 ( v ) ⊆ F 2 ( u ) , then F 1 ( v ) ⊆ F 1 ( u ) . We obtain u ∈ ⋂ F 1 ( v ) , which implies that u ∈ O. This gives O′ ⊆ O, hence O = O′. We conclude that F 1 ⊆ 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;

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 ∈ H . If H ∩ L is empty, then H ∩ L ∈ H . Assume that H∩ L ≠ ∅. For each p ∈ H ∩ L, since ⋂ H p ∈ H , we have ⋂ H p ⊂ H and ⋂ H p ⊂ L , hence ⋂ H p ⊂ H ∩ L . Therefore, H ∩ L = ⋃ p ∈ H ∩ L ⋂ H p , which shows that H ∩ L ∈ 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 } ∈ 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 ) ∩ O LC ≠ ∅ 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 ∈ ( O ▵ W ) ∩ 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 ∈ ( O ▵ W ) ∩ 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 ∈ 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 H = { ∅ , { 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 } . 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;

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 ∈ G ∈ H } and M ( t ) = ⋂ M ( t ) . Since ( Q , H ) is a quasi ordinal space, it follows that M ( t ) ∈ 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 ∈ H . If H =∅, then H ∈ 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 ∈ H M ( t ) ∈ 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 } ∈ H for each t ∈ Q. Indeed, take any t ∈ Q. Obviously, ( Q , H ) is a T₁ Alexandroff space. Therefore, it follows that { t } = ⋂ { H ∈ H : t ∈ H } ∈ 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 ) ∈ 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 | ≤ | 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 | ≤ κ and ⋃ 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 ′ ⊂ O . Hence | O ′ | ≤ c ( X ) = κ , and Z = ⋃ O ′ ¯ by the maximality of O ′ . For each O ∈ O ′ , fix a W O ∈ V such that O ⊂ W _O . Put W = { W O : O ∈ O ′ } . Then ⋃ 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 = 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 ⊆ Z , 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 | ≤ 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 H ∩ D ¯ = H ¯ for each H ∈ H , then |Q|≤2^2|D|.

Proof. For every t ∈ Q, the family N ( t ) = { U ∩ A : U ∈ H t } of subsets of D. Since ( Q , τ H ) is Hausdorff and L ∩ D ¯ = L ¯ for any L ∈ H , we conclude that the intersection of the closures of all members of N ( t ) equals {t}; thus N ( t ) ≠ 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|≤2²³ and |D|=3 < 4. However, { z 2 } = { z 2 } ¯ = { z 2 , z 4 } ∩ { z 1 , z 2 , z 3 } ¯ ≠ { z 2 , z 4 } = { z 2 , z 4 } ¯ . Hence the condition “ H ∩ D ¯ = H ¯ for each H ∈ 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 ∈ Q } . For any distinct r, t ∈ Q, if | H r ′ | = | H t ′ | = m , then ⋂ H r ′ = ⋂ H t ′ or ( ⋂ H r ′ ) ∩ ( ⋂ H t ′ ) = ∅ .

Proof. Assume that ⋂ H r ′ ≠ ⋂ H t ′ . Then, without loss of generality, we assume that there is H ∈ H t ′ with r ∉ H. Now assume that ( ⋂ H r ′ ) ∩ ( ⋂ H t ′ ) ≠ ∅ . Then there is q ∈ ( ⋂ H r ′ ) ∩ ( ⋂ H t ′ ) , hence q ∈ H and q ∈ L for any L ∈ H r ′ , which implies that H q ′ = m + 1 , this is a contradiction. Therefore, ( ⋂ H r ′ ) ∩ ( ⋂ H t ′ ) = ∅ . □

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 ∈ Q } . Put A₁ = {p₁} . If B \ B ( 1 ) p 1 = ∅ , then we stop our induction, and put A = A₁. If B \ B ( 1 ) p 1 ≠ ∅ , 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 ∈ 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 ∈ Q } for any 1 < i ≤ k, where B ( i ) = B \ ⋃ j < i B ( j ) p j . If B \ ⋃ i ≤ k B ( i ) p i = ∅ , then we stop our induction, and put A = A _k ; otherwise, let B ( k + 1 ) = B \ ⋃ i ≤ 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 ∈ Q } . Then put A_k+1 = A _k  ∪ {p_k+1} . Since Q is finite, there exists a minimum n ∈ ℕ such that B \ ⋃ i ≤ n B ( i ) p i = ∅ . Moreover, from Lemma 3, it follows that, for any j ≤ n, if | B ( j ) | = 1 we may assume that p j ∉ ⋃ ( ⋃ 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 ∈ B ( 1 ) p 1 , then put B₁ = A \ {p₁}; otherwise, put B₁ = A. If p₁ ∈ B₁ and B∩ (B₁ \ {p₂}) ≠ ∅ for any B ∈ B ( 2 ) p 2 , then put B₂ = B₁ \ {p₂}; if p₁ ∈ B₁ and B∩ (B₁ \ {p₂}) = ∅ for some B ∈ B ( 2 ) p 2 , then put B₂ = B₁; if p₁ ∉ B₁, B∩ (B₁ \ {p₂}) ≠ ∅ for any B ∈ B ( 2 ) p 2 , and O∩ (B₁ \ {p₂}) ≠ ∅ whenever p 2 ∈ O ∈ 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 C ¯ ≠ 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 C ¯ = ∅ . 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 C ¯ ≠ Q for any subset C of Q with |C| < |D| when | B | ≤ k . Now assume that | B | = k + 1 . From our construction, we conclude that |A ∩ B|=1 for each B ∈ B ( n ) p n . We divide the rest proof into the following two cases.

Case 1: | B ( n ) p n | ≥ 2 .

Let B ′ = ⋃ B ( n ) p n , and put B ′ = ( B \ B ( n ) p n ) ∪ { 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 ′ | ≤ k , from our assumption it follows that F ¯ τ F ≠ Q for each subset F of Q with |F| < |D|, hence F ¯ τ H ≠ 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 | ≥ 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 ) ∪ { B ′ ′ } . Then τ L is coarser than τ H , hence D is dense in ( Q , τ L ) . Since | B ′ ′ | ≤ k , from our assumption it follows that F ¯ τ L ≠ Q for each subset F of Q with |F| < |D|, hence F ¯ τ H ≠ 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.

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

Since ∑ j = 1 5 T 3 j > ∑ 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.

Since ∑ j = 4 5 T 2 j ( 1 ) = ∑ j = 4 5 T 3 j ( 1 ) = 1 and ∑ j = 4 5 T 4 j ( 1 ) = ∑ 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.

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.

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 ∩ H ¯ = { ∅ , X } , where H ¯ = { K : Q \ K ∈ H } .

A knowledge space ( Q , H ) is called granular if for each state L ∈ 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 ) → ( 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 ∗ ∈ H ∗ , we have h^-1 (K^∗) = K. Moreover, let h^-1 (A) is open ( Q , H ) . Then there is a subfamily H ⊆ H with h - 1 ( A ) = ⋃ H , hence A = ⋃ H ∈ H h ( H ) = ⋃ H ∈ H H ∗ is open in ( Q ∗ , H ∗ ) . Therefore, h a pre-quotient mapping. □