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Covering dimension and finite spaces
Applied Mathematics and Computation 218 (2011) 3122–3130
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Covering dimension and finite spaces q D.N. Georgiou ⇑, A.C. Megaritis University of Patras, Department of Mathematics, 265 04 Patras, Greece
a r t i c l e
i n f o
a b s t r a c t Finite topological spaces, that is spaces with a finite number of points, have a wide range of applications in many areas such as computer graphics and image analysis. In this paper we study the covering dimension of a finite topological space. In particular, we give an algorithm for computing the covering dimension of a finite topological space using matrix algebra. Ó 2011 Elsevier Inc. All rights reserved.
Keywords: Covering dimension Finite spaces Incidence matrix
1. Preliminaries The class of finite topological spaces was first studied by Alexandroff in 1937 in [1]. There is a strong relationship between finite spaces and finite simplicial complexes (see [12]). On the other hand, all compact Hausdorff spaces arise by approximation using finite T0-spaces (see [11]). Finally, finite spaces play an important role in dimension theory (see [15]). Together with the theory of continua, dimension theory is the oldest branch of topology. It is possible to define the dimension of a topological space X in three different ways, the small inductive dimension ind(X), the large inductive dimension Ind(X), and the covering dimension dim(X). The three dimension functions coincide in the class of separable metric spaces. In larger classes of spaces the dimensions ind(X), Ind(X), and dim(X) diverge. The small and large inductive dimensions for finite topological spaces are studied by some other authors (see, for example, [2,3,13]). A topological space X is finite if the set X is finite. In what follows we denote by X = {x1, . . . , xn} a finite space of n elements and by Ui the smallest open set of X containing the point xi, i = 1, . . . , n. Also, we denote by x the first infinite cardinal. The space X is T0 if and only if Ui = Uj implies xi = xj for every i, j (see [1]). Let X = {x1, . . . , xn} be a finite space of n elements. The n n matrix T = (tij), where
t ij ¼
1; if xi 2 Uj 0;
otherwise
is called the incidence matrix of X (see [3]). We observe that
Uj ¼ fxi : t ij ¼ 1g;
j ¼ 1; . . . ; n:
We denote by c1, . . . , cn the n columns of the matrix T and by 1 the n 1 matrix with all elements equal to one, that is,
0 1 1 B1C B C C 1¼B B .. C: @.A 1
q
Work supported by the Caratheodory Programme of the University of Patras.
⇑ Corresponding author.
E-mail addresses: [email protected] (D.N. Georgiou), [email protected] (A.C. Megaritis). 0096-3003/$ - see front matter Ó 2011 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2011.08.040
D.N. Georgiou, A.C. Megaritis / Applied Mathematics and Computation 218 (2011) 3122–3130
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Let
1 0 1 c1j c1i Bc C B c2j C B 2i C B C C B C ci ¼ B B .. C and cj ¼ B .. C @ . A @ . A cni cnj 0
are two n 1 matrices. Then, by maxci we denote the maximum
maxfc1i ; c2i ; . . . ; cni g and by ci + cj the n 1 matrix
1 c1i þ c1j B c2i þ c2j C C B C: ci þ cj ¼ B .. C B A @ . 0
cni þ cnj Also, we write ci 6 cj if and only if cki 6 ckj for each k = 1, . . . , n. For the following notions see for example [4]. Let X be a space. A cover of X is a non-empty set of subsets of X, whose union is X. A family r of subsets of X is said to be a refinement of a family c of subsets of X if each element of r is contained in an element of c. Define the order of a family r of subsets of a space X as follows: (a) ord(r) = 1 if and only if r consists of the empty set only. (b) ord(r) = k, where k 2 x, if and only if the intersection of any k + 2 distinct elements of r is empty and there exist k + 1 distinct elements of r, whose intersection is not empty. (c) ord(r) = 1, if and only if for every k 2 x there exist k distinct elements of r, whose intersection is not empty. We denote by dim the function, called covering dimension, with domain the class of all spaces and range the set
x [ {1, 1}, satisfying the following condition: dim(X) 6 k, where k 2 {1} [ x if and only if for every finite open cover c of the space X there exists a finite open cover r of X, refinement of c, such that ord(r) 6 k. Finite topological spaces and the notion of dimension play an important role in digital spaces, computer graphics, image synthesis and image analysis (see, [9,10,14]). Many researchers are currently working in these areas (see, for example, [5,7,8]). In this paper we study the covering dimension for finite topological spaces. In particular, we give an algorithm for computing the covering dimension of a finite space X using the notion of the incidence matrix of X. In Section 2 we present some propositions concerning the covering dimension of a finite topological space. In Section 3 we give the algorithm for computing the covering dimension. Finally, in Section 4 we give some remarks on the algorithm. 2. Finite spaces and covering dimension In this section we present some propositions concerning the covering dimension of a finite topological space. Proposition 2.1. Let X = {x1, . . . , xn} be a finite space. Then, dim(X) 6 k, where k 2 x if and only if there exists an open cover fUj1 ; . . . ; Ujm g of X such that ordðfUj1 ; . . . ; Ujm gÞ 6 k.
Proof. Let dim(X) 6 k, where k 2 x. We prove that there exists an open cover fUj1 ; . . . ; Ujm g of X such that ordðfUj1 ; . . . ; Ujm gÞ 6 k. Let m = min{m 2 x : there exist j1, . . . , jm 2 {1, . . . , n} such that fUj1 ; . . . ; Ujm g is an open cover of Xg and c ¼ fUj1 ; . . . ; Ujm g be an open cover of X. Since dim(X) 6 k, there exists an open cover r = {V1, . . . , Vl} of X, refinement of c, such that ord(r) 6 k. For the proof of the proposition it suffices to prove that c # r. We suppose that there exists a 2 {1, . . . , m} such that Uja R r. Since r is a cover of X, there exists b 2 {1, . . . , l} such that xja 2 V b . By the fact that Uja is the smallest open set of X containing the point xja we have that Uja # V b . Also, since Uja R r, we have Uja – V b . Therefore, Uja V b . Since r is a refinement of c, there exists c 2 {1, . . . , m} such that V b # Ujc . Hence,
Uja Ujc : We observe that the family c n fUja g is an open cover of X with m 1 elements, which is a contradiction by the choice of m. Thus, c # r.
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Conversely, we suppose that there exists an open cover fUj1 ; . . . ; Ujm g of X such that ordðfUj1 ; . . . ; Ujm gÞ 6 k and we prove that dim(X) 6 k. Let c be a finite open cover of the space X. It suffices to prove that the open cover fUj1 ; . . . ; Ujm g of X is a refinement of c. For every i 2 {1, . . . , n} there exists Vi 2 c such that xi 2 Ui # Vi. This means that the open cover {U1, . . . , Un} of X is a refinement of c. Since
fUj1 ; . . . ; Ujm g # fU1 ; . . . ; Un g; we have that the open cover fUj1 ; . . . ; Ujm g of X is a refinement of c. h Proposition 2.2. Let X = {x1, . . . , xn} be a finite space, where n > 1. Then,
dimðXÞ 6 n 2: T T Proof. We consider the open cover {U1, . . . , Un} of X. If nj¼1 Uj ¼ ;, then we have dim(X) 6 n 2. Now, let nj¼1 Uj – ; and Tn T x 2 j¼1 Uj for some j0 2 {1, . . . , n}. Since Uj0 is the smallest open set of X containing the point xj0 , we have Uj0 # nj¼1 Uj . Also, Tj0n Tn j¼1 Uj # Uj0 . Therefore, Uj0 ¼ j¼1 Uj . Hence the family fU1 ; . . . ; Uj0 1 ; Uj0 þ1 . . . ; Un g is an open cover of X with n 1 elements and, therefore,
ordðfU1 ; . . . ; Uj0 1 ; Uj0 þ1 . . . ; Un gÞ 6 n 2: Thus, by Proposition 2.1, dim(X) 6 n 2. h Remark 2.3. We note that the inequality dim(X) 6 n 2 in Proposition 2.2 cannot be replaced by the inequality dim(X) < n 2. Indeed, let X = {x1, . . . , xn}. We consider the family
B ¼ ffx1 g; fx1 ; x2 g; . . . ; fx1 ; xn gg: S
Since V2BV = X, {x1, xi} \ {x1, xj} = {x1} for all i, j 2 {2, . . . , n} and {x1} \ {x1, xk} = {x1} for all k 2 {2, . . . , n}, the family B is a base for a topology s on X. We observe that for the finite space (X, s) we have dim(X) = n 2. In the following propositions we suppose that X = {x1, . . . , xn} is a finite space with n elements, T = (tij), i = 1, . . . , n, j = 1, . . . , n, the incidence matrix of X, and c1, . . . , cn the n columns of the matrix T. Proposition 2.4. If cj = 1 for some j 2 {1, . . . , n}, then dim(X) = 0. Proof. Since cj = 1, we have tij = 1 for every i = 1, . . . , n and, therefore,
Uj ¼ fxi : t ij ¼ 1g ¼ fx1 ; . . . ; xn g ¼ X: Thus, the family {Uj} is an open cover of X. Since ord({Uj}) = 0, by Proposition 2.1, we have dim(X) = 0. h Proposition 2.5. Let cji ; i ¼ 1; . . . ; m, be m columns of the matrix T. Then, cj1 þ þ cjm P 1 if and only if the family fUj1 ; . . . ; Ujm g is an open cover of X. Proof. Let cj1 þ þ cjm P 1. We prove that X ¼ Uj1 [ [ Ujm . It suffices to prove that X # Uj1 [ [ Ujm . Let xi0 2 X. Then, by the definition of the matrix T and by the assumption cj1 þ þ cjm P 1, there exists j 2 {1, . . . , m} such that ti0 jj ¼ 1. Since Ujj ¼ fxi : t ijj ¼ 1g, we have xi0 2 Ujj . Thus, X # Uj1 [ [ Ujm . Conversely, we suppose that the family fUj1 ; . . . ; Ujm g is an open cover of X. Then, for every i 2 {1, . . . , n} there exists j(i) 2 {1, . . . , m} such that xi 2 UjjðiÞ . Therefore, by the definition of the matrix T; tijjðiÞ ¼ 1. Thus, every element of the n 1 matrix cj1 þ þ cjm is larger or equal to 1, which means that cj1 þ þ cjm P 1. h Proposition 2.6. Let cji , i = 1, . . . ,m, be m columns of the matrix T and k ¼ maxðcj1 þ þ cjm Þ, that is k is the maximum element of the n 1 matrix cj1 þ þ cjm . Then,
ordðfUj1 ; . . . ; Ujm gÞ ¼ k 1: Proof. We observe that k is a positive integer larger than or equal to 1. Since maxðcj1 þ þ cjm Þ ¼ k, by the definition of the matrix T, there exist i0 2 {1, . . . , n} and jqk 2 fj1 ; . . . ; jm g; k ¼ 1; . . . ; k, such that ti0 jq þ þ ti0 jq ¼ k and, therefore, t i0 jqk ¼ 1 for 1 k every k = 1, . . . , k. This means that xi0 2 Ujqk for every k = 1, . . . , k and, hence, Ujq \ \ Ujq – ;. Also, we observe that, since k 1 k is the maximum element of the n 1 matrix cj1 þ þ cjm , the intersection of any k + 1 distinct elements of fUj1 ; . . . ; Ujm g is empty. Thus, ordðfUj1 ; . . . ; Ujm gÞ ¼ k 1. h
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Proposition 2.7. Let fUi1 ; . . . ; Uil g be an open cover of X,m ¼ minfm 2 x : there exist j1 ; . . . ; jm 2 f1; . . . ; ng such that fUj1 ; . . . ; Ujm g is an open cover of Xgand fUj1 ; . . . ; Ujm g an open cover of X. Then,
fUj1 ; . . . ; Ujm g # fUi1 ; . . . ; Uil g: Proof. The proof is similar to that of Proposition 2.1.
h
Corollary 2.8. Letm ¼ minfm 2 x : there exist j1 ; . . . ; jm 2 f1; . . . ; ng such that fUj1 ; . . . ; Ujm g is an open cover of Xgand fUj1 ; . . . ; Ujm g be an open cover of X such that
ordðfUj1 ; . . . ; Ujm gÞ ¼ k P 0: Then, for every open cover fUr1 ; . . . ; Url g of X we have
ordðfUr1 ; . . . ; Url g P k: Proof. Let fUr1 ; . . . ; Url g be an open cover of X. By Proposition 2.7,
fUj1 ; . . . ; Ujm g # fUi1 ; . . . ; Uil g: Since ordðfUj1 ; . . . ; Ujm gÞ ¼ k, we have ordðfUr1 ; . . . ; Url g P k. h Proposition 2.9. Let cji , i = 1, . . . , m, be m columns of the matrix T such that cj1 þ þ cjm P 1. If cr1 þ þ crq j 1 for every q < m, then
dimðXÞ ¼ maxðcj1 þ þ cjm Þ 1: Proof. Since cj1 þ þ cjm P 1, by Proposition 2.5, the family fUj1 ; . . . ; Ujm g is an open cover of X. Let k ¼ maxðcj1 þ þ cjm Þ. Then, by Proposition 2.6,
ordðfUj1 ; . . . ; Ujm gÞ ¼ k 1 and, therefore, by Proposition 2.1, dim(X) 6 k 1. We prove that dim(X) = k 1. We suppose that dim(X) < k 1. Then, by Proposition 2.1, there exists an open cover fUr1 ; . . . ; Url g of X with ordðfUr1 ; . . . ; Url gÞ < k 1. Since cj1 þ þ cjm P 1 and cr1 þ þ crq j1 for every q < m, by Proposition 2.5, we havem ¼ minft 2 x : there exist j1 ; . . . ; jt 2 f1; . . . ; ng such that fUj1 ; . . . ; Ujt g is an open cover of Xg. Moreover, since
ordðfUj1 ; . . . ; Ujm gÞ ¼ k 1; by Corollary 2.8, we have
ordðfUr1 ; . . . ; Url gÞ P k 1 which is a contradiction. Thus, dim(X) = k 1. h 3. An algorithm for computing the covering dimension In this section we give an algorithm for computing the covering dimension of finite topological spaces using the Propositions 2.4 and 2.9. Let X = {x1, . . . , xn} be a finite space of n elements and T = (tij) the n n incidence matrix. Then our intended algorithm contains n steps: Step 1. Read the n columns c1, . . . , cn of the matrix T. If some column is equal to 1, then print
dimðXÞ ¼ 0: Otherwise go to the Step 2. 0
0
Step 2. Find the sums cj1 þ cj2 for each j1, j2 2 {1, . . . , n}. If there exist j1 ; j2 2 f1; . . . ; ng such that cj0 þ cj0 P 1, then print 1
2
dimðXÞ ¼ maxðcj0 þ cj0 Þ 1: 1
2
Otherwise go to the Step 3. 0
0
0
Step 3. Find the sums cj1 þ cj2 þ cj3 for each j1, j2, j3 2 {1, . . . , n}. If there exist j1 ; j2 ; j3 2 f1; . . . ; ng such that cj0 þ cj0 þ cj0 P 1, 1 2 3 then print
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dimðXÞ ¼ max cj0 þ cj0 þ cj0 1: 1
0
3
Otherwise go to the Step 4. ... Step n 1. Find the sums
cj1 þ cj2 þ þ cjn1 ; 0
0
0
for each j1, j2, . . . , jn1 2 {1, . . . , n}. If there exist j1 ; j2 ; . . . ; jn1 2 f1; . . . ; ng such that cj0 þ cj0 þ þ cj0 1
2
dimðXÞ ¼ maxðcj0 þ cj0 þ þ cj0 Þ 1: 1
0
n1
Otherwise go to the Step n. Step n. Find the sum c1 + c2 + + cn and print
dimðXÞ ¼ maxðc1 þ c2 þ þ cn Þ 1: Example 3.1 (1) Let X = {x1, x2, x3} with the topology
s ¼ f;; fx2 g; fx1 ; x2 g; fx2 ; x3 g; Xg: We observe that U1 = {x1, x2}, U2 = {x2}, U3 = {x2, x3}. Therefore,
0
1 0
0
1
B C T ¼ @ 1 1 1 A; 0
0 1 1 B C c1 ¼ @ 1 A;
0 1
0 1 0 B C c2 ¼ @ 1 A;
0
0 1 0 B C and c3 ¼ @ 1 A:
0
1
Moreover,
0 1 1 B C c1 þ c2 ¼ @ 2 A j 1;
0 1 1 B C c1 þ c3 ¼ @ 2 A > 1;
0
0 1 0 B C c2 þ c3 ¼ @ 2 A j 1
1
1
and
maxðc1 þ c3 Þ ¼ 2: Thus, dim(X) = max(c1 + c3) 1 = 1. (2) Let X = {x1, x2, x3, x4} with the topology
s ¼ f;; fx1 g; fx1 ; x2 g; fx1 ; x3 g; fx1 ; x4 g; fx1 ; x2 ; x3 g; fx1 ; x2 ; x4 g; fx1 ; x3 ; x4 g; Xg: We observe that
U1 ¼ fx1 g;
U2 ¼ fx1 ; x2 g;
U3 ¼ fx1 ; x3 g;
U4 ¼ fx1 ; x4 g:
Therefore,
0
1 1 1 1
1
C B B0 1 0 0C C T¼B B 0 0 1 0 C; A @ 0 0 0 1
0 1 1 B C B0C C c1 ¼ B B 0 C; @ A 0
0 1 1 B C B1C C c2 ¼ B B 0 C; @ A 0
0 1 1 B C B0C C c3 ¼ B B 1 C; @ A 0
0 1 1 B C B0C C and c4 ¼ B B 0 C: @ A 1
Moreover,
0 1 2 B C B1C C c1 þ c2 ¼ B B 0 C j 1; @ A 0
0 1 2 B C B0C C c1 þ c3 ¼ B B 1 C j 1; @ A 0
0 1 2 B C B0C C c1 þ c4 ¼ B B 0 C j 1; @ A 1
n1
P 1, then print
D.N. Georgiou, A.C. Megaritis / Applied Mathematics and Computation 218 (2011) 3122–3130
0 1 2 B1C B C c2 þ c3 ¼ B C j 1; @1A
0 1 2 B1C B C c2 þ c4 ¼ B C j 1; @0A
0
0 1 2 B0C B C c3 þ c4 ¼ B C j 1; @1A
1
0 1 3 B1C B C c2 þ c3 þ c4 ¼ B C > 1; @1A
and
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1
maxðc2 þ c3 þ c4 Þ ¼ 3:
1 Thus, dim(X) = max(c2 + c3 + c4) 1 = 2. (3) Let X = {x1, x2, x3, x4, x5} with the topologys = {;, {x2}, {x5}, {x2, x4}, {x2, x5}, {x1, x2, x3}, {x2, x4, x5}, {x1, x2, x3, x4}, {x1, x2, x3, x5}, X}. We observe that
U1 ¼ fx1 ; x2 ; x3 g;
U2 ¼ fx2 g;
U3 ¼ fx1 ; x2 ; x3 g;
U4 ¼ fx2 ; x4 g;
U5 ¼ fx5 g:
Therefore,
0
1 0
B B1 1 B T¼B B1 0 B @0 0 0 0 0 1 1 B C B1C B C C c1 ¼ B B 1 C; B C @0A 0
1 0
0
1
C 1 1 0C C 1 0 0C C; C 0 1 0A 0 0 1 0 1 0 B C B1C B C C c2 ¼ B B 0 C; B C @0A 0
0 1 1 B C B1C B C C c3 ¼ B B 1 C; B C @0A 0
0 1 0 B C B1C B C C c4 ¼ B B 0 C; B C @1A 0
0 1 0 B C B0C B C C and c5 ¼ B B 0 C: B C @0A 1
Moreover,
0 1 1 B C B2C B C C c1 þ c2 ¼ B B 1 C j 1; B C @0A 0
0 1 2 B C B2C B C C c1 þ c3 ¼ B B 2 C j 1; B C @0A 0
0 1 0 B C B2C B C C c1 þ c4 ¼ B B 1 C j 1; B C @1A 0
0 1 1 B C B1C B C C c1 þ c5 ¼ B B 1 C j 1; B C @0A 1
0 1 1 B C B2C B C C c2 þ c3 ¼ B B 1 C j 1; B C @0A 0
0 1 0 B C B2C B C C c2 þ c4 ¼ B B 1 C j 1; B C @1A 0
0 1 0 B C B1C B C C c2 þ c5 ¼ B B 0 C j 1; B C 0 @ A 1
0 1 1 B C B2C B C C c3 þ c4 ¼ B B 1 C j 1; B C 1 @ A 0
0 1 1 B C B1C B C C c3 þ c5 ¼ B B 1 C j 1; B C 0 @ A 1
0 1 0 B C B1C B C C c4 þ c5 ¼ B B 0 C j 1; B C @1A 1
0 1 1 B C B2C B C C c1 þ c4 þ c 5 ¼ B B 1 C > 1; B C @1A 1
Thus, dim(X) = max(c1 + c4 + c5) 1 = 1.
and
maxðc1 þ c4 þ c5 Þ ¼ 2:
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D.N. Georgiou, A.C. Megaritis / Applied Mathematics and Computation 218 (2011) 3122–3130
4. Remarks on the algorithm for computing the covering dimension of finite topological spaces Remark 4.1. Let A = (aij) be a n n matrix and B = (bij) a m m matrix. The Kronecker product of A and B (see, for example, [6]) is the mn mn block matrix
0
B AB¼B @
a11 B . . . a1n B .. .
..
.. .
.
an1 B . . . ann B
1
C C: A
More explicitly, the Kronecker product of A and B is the matrix
0
a11 b11 Ba b B 11 21
B .. B B . B Ba b B 11 m1 B B .. B . B B B an1 b11 B B an1 b 21 B B .. B @ .
an1 bm1
a11 b12 . . . a11 b1m . . . a1n b11 a11 b22 . . . a11 b2m . . . a1n b21 .. .
..
.. .
.
..
.. .
.
.. .
..
.. .
.
..
.. .
.
.. .
an1 b12 . . . an1 b1m . . . ann b11 an1 b22 . . . an1 b2m . . . ann b21
ann b12 ann b22
.. . an1 bm2
.. . ann bm2
..
.. . . . . . an1 bmm
..
.. . . . . . ann bm1
C C C C . . . a1n bmm C C C C .. .. C: . . C C . . . ann b1m C C . . . ann b2m C C C .. .. C A . . . . . ann bmm ..
a11 bm2 . . . a11 bmm . . . a1n bm1 a1n bm2 .. .
1
a1n b12 . . . a1n b1m a1n b22 . . . a1n b2m C C .. .
.
Let X = {x1, . . . , xn} be a finite space of n elements and Y = {y1, . . . , ym} a finite space of m elements. It is known that if TX is the incidence matrix of X and TY is the incidence matrix of Y, then the incidence matrix of X Y is the Kronecker product TX TY of TX and TY (see, for example, [16]). Example 4.2. Let X = {x1, x2, x3} with the topology
sX ¼ f;; fx2 g; fx1 ; x2 g; fx2 ; x3 g; Xg and Y = {y1, y2, y3, y4} with the topology
sY ¼ f;; fy1 g; fy1 ; y2 g; fy1 ; y3 g; fy1 ; y4 g; fy1 ; y2 ; y3 g; fy1 ; y2 ; y4 g; fy1 ; y3 ; y4 g; Yg: The incidence matrix TX of X is
0
1 0 0
1
B C TX ¼ @ 1 1 1 A 0
0 1
and the incidence matrix TY of Y is
0
1 1 1 1 1 B0 1 0 0C B C TY ¼ B C: @0 0 1 0A 0
0
0 1
Therefore, the incidence matrix TXY of the product space X Y is
0
T XY
1 1 1 1 0
B B0 B B0 B B B0 B B1 B B B0 ¼ TX TY ¼ B B0 B B B0 B B0 B B B0 B B0 @ 0
1 0
0
0
0 0
0
0
0
0
0
0
0 0
0
0
0 1 0 0 0 1
0 0
0 0
0 0
0 0 0 0
0 0
0 0
1 1 1 1 1 1 1 1 1 1 1 0
0
0 1 0
0 0
1 0
0 1
0
0
0 1
0 0
0
1
0
0 1
0
0
0 1 0
0
0
0
0
0
0
0
0
0 1 1 1
0
0
0
0
0
0
0 0
1 0
0
0
0
0
0
0
0 0
0
1
0
0
0
0
0
0
0 0
0
0
0
1
C 0C C 0C C C 0C C 1C C C 0C C: 0C C C 1C C 1C C C 0C C 0C A 1
D.N. Georgiou, A.C. Megaritis / Applied Mathematics and Computation 218 (2011) 3122–3130
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We observe that
cr1 þ . . . þ crq j 1 for every q < 6;
c2 þ c3 þ c4 þ c10 þ c11 þ c12
0 1 3 B C B1C B C B1C B C B C B1C B C B6C B C B C B2C C ¼B B2C > 1 B C B C B2C B C B3C B C B C B1C B C B1C @ A 1
and max(c2 + c3 + c4 + c10 + c11 + c12) = 6. Thus,
dimðX YÞ ¼ maxðc2 þ c3 þ c4 þ c10 þ c11 þ c12 Þ 1 ¼ 5: Also, we observe that dim(X) = 1 and dim(Y) = 2. Remark 4.3. By Example 4.2 arises that the inequality
dimðX YÞ 6 dimðXÞ þ dimðYÞ for two finite spaces X and Y in general, is not true. Remark 4.4. Let X = {x1, . . . , xn} be a finite T0-space. Then, there exists a finite space X1 homeomorphic to X such that the incidence matrix T X 1 of X1 is an upper triangular matrix (see Theorem 7 of [16]). So in order to calculate the dim(X) it suffices to calculate dim(X1). Example 4.5. Let X = {x1, x2, x3} with the topology
s ¼ f;; fx2 g; fx1 ; x2 g; fx2 ; x3 g; Xg: We consider the space X1 = {x1, x2, x3} with the topology
s1 ¼ f;; fx1 g; fx1 ; x2 g; fx1 ; x3 g; X 1 g: We observe that the spaces X and X1 are homeomorphic. The incidence matrix T X 1 of X1 is
0
T X1
1 1 1 1 B C ¼ @ 0 1 0 A: 0 0 1
Since
0 1 0 1 0 1 2 1 1 B C B C B C c2 þ c3 ¼ @ 1 A þ @ 0 A ¼ @ 1 A > 1 1 1 0 and max(c2 + c3) = 2, we have
dimðXÞ ¼ maxðc2 þ c3 Þ 1 ¼ 1: Proposition 4.6. An upper bound on the number of iterations of the algorithm for computation of the covering dimension of a finite space X of n elements performs is the number 2n 1. Proof. We observe that the number of iterations the algorithm performs in Steps 1, 2, . . . , n 1 and n is n n n n ; ;...; ; and , respectively. Thus, the number of iterations the algorithm performs is 1 2 n1 n
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D.N. Georgiou, A.C. Megaritis / Applied Mathematics and Computation 218 (2011) 3122–3130
n 1
þ
n 2
þ þ
n
þ
n1
n n
¼
n X n k nk ¼ ð1 þ 1Þn 1 ¼ 2n 1: 11 k k¼1
Proposition 4.7. Let X be a finite space with n elements. If the incidence matrix TX of X is an upper triangular matrix, then an upper bound on the number of iterations the algorithm performs is the number 2n1. Proof. Since the incidence matrix TX is an upper triangular matrix, in order to find dim(X) we follow the next steps: Step 0. Read the column cn. If cn P 1, then
dimðXÞ ¼ 0: Otherwise go to the Step 1. Step 1. If there exists i 2 {1, . . . , n 1} such that ci + cn P 1, then
dimðXÞ ¼ maxðci þ cn Þ 1: Otherwise go to the Step 2. Step 2. If there exists i1, i2 2 {1, . . . , n 1} such that ci1 þ ci2 þ cn P 1, then
dimðXÞ ¼ maxðci1 þ ci2 þ cn Þ 1: Otherwise go to the Step 3.. . . Step n2. If there exists i1, . . . , in2 2 {1, . . . , n 1} such that
ci1 þ þ cin2 þ cn P 1; then
dimðXÞ ¼ maxðci1 þ þ cin2 þ cn Þ 1: Otherwise go to the Step n 1. Step n 1. If c1 + + cn1 + cn P 1, then
dimðXÞ ¼ maxðc1 þ þ cn1 þ cn Þ 1: By the above we have that the number of iterations the algorithm performs is
n1 0
þ
n1 1
þ
n1 2
þ þ
n1 n1
¼
n1 X n 1 k n1k ¼ ð1 þ 1Þn1 ¼ 2n1 : 11 k k¼0
Acknowledgement We are grateful to the referee for a number of helpful suggestions for improvement in the article. References [1] P. Alexandroff, Diskrete Räume, Mat. Sb. (N.S.) 2 (1937) 501–518. [2] D.W. Bass, Dimension and finite spaces, J. London Math. Soc. 44 (1969) 159–162. [3] A. El-Fattah El-Atik, M.E. Abd El-Monsef, E.I. Lashin, On finite T0 topological spaces, Proceedings of the Ninth Prague Topological Symposium (2001) 75– 90. [4] R. Engelking, Theory of Dimensions, Finite and Infinite, Heldermann Verlag, Berlin, 1995. [5] A.V. Evako, R. Kopperman, Y.V. Mukhin, Dimensional properties of graphs and digital spaces, J. Math. Imaging Vision 6 (2–3) (1996) 109–119. [6] H. Eves, Elementary Matrix Theory, Dover Publications, 1980. [7] A.S. Farrag, S.E. Abbas, Computer programing for construction and enumeration of all regular topologies and equivalence relations on finite sets, Appl. Math. Comput. 165 (1) (2005) 177–184. [8] El Zohny Habiba, On zero dimensional space and its retraction, Appl. Math. Comput. 155 (2) (2004) 515–519. [9] G.T. Herman, Geometry of Digital Spaces, Birkhauser, Boston, 1998. [10] E.D. Khalimsky, R. Kopperman, P.R. Meyer, Computer graphics and connected topologies on finite ordered sets, Topol. Appl. 36 (1990) 1–17. [11] Ralph Kopperman, On storage of topological information, Discret. Appl. Math. 147 (2005) 287–300. [12] M.C. McCord, Singular homology groups and homotopy groups of finite topological spaces, Duke Math. J. 33 (1966) 465474. [13] W. Mielants, On the number of topologies with given small inductive dimension on a finite set, Matematiche (Catania) 26 (1971) 27–34 (1972). [14] A. Rosenfeld, Digital topology, Amer. Math. Monthly 86 (1979) 621–630. [15] J.B. Sancho and M.T. Sancho, Dimension of Dense Subalgebras of C(X), Proceedings of the American Mathematical Society, vol. 105, No. 2 (Feb., 1989), pp. 491–499. [16] Mitsunobu Shiraki, On finite topological spaces, Rep. Fac. Sci. Kagoshima Univ. 1 (1968) 1–8.
Contents lists available at SciVerse ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Covering dimension and finite spaces q D.N. Georgiou ⇑, A.C. Megaritis University of Patras, Department of Mathematics, 265 04 Patras, Greece
a r t i c l e
i n f o
a b s t r a c t Finite topological spaces, that is spaces with a finite number of points, have a wide range of applications in many areas such as computer graphics and image analysis. In this paper we study the covering dimension of a finite topological space. In particular, we give an algorithm for computing the covering dimension of a finite topological space using matrix algebra. Ó 2011 Elsevier Inc. All rights reserved.
Keywords: Covering dimension Finite spaces Incidence matrix
1. Preliminaries The class of finite topological spaces was first studied by Alexandroff in 1937 in [1]. There is a strong relationship between finite spaces and finite simplicial complexes (see [12]). On the other hand, all compact Hausdorff spaces arise by approximation using finite T0-spaces (see [11]). Finally, finite spaces play an important role in dimension theory (see [15]). Together with the theory of continua, dimension theory is the oldest branch of topology. It is possible to define the dimension of a topological space X in three different ways, the small inductive dimension ind(X), the large inductive dimension Ind(X), and the covering dimension dim(X). The three dimension functions coincide in the class of separable metric spaces. In larger classes of spaces the dimensions ind(X), Ind(X), and dim(X) diverge. The small and large inductive dimensions for finite topological spaces are studied by some other authors (see, for example, [2,3,13]). A topological space X is finite if the set X is finite. In what follows we denote by X = {x1, . . . , xn} a finite space of n elements and by Ui the smallest open set of X containing the point xi, i = 1, . . . , n. Also, we denote by x the first infinite cardinal. The space X is T0 if and only if Ui = Uj implies xi = xj for every i, j (see [1]). Let X = {x1, . . . , xn} be a finite space of n elements. The n n matrix T = (tij), where
t ij ¼
1; if xi 2 Uj 0;
otherwise
is called the incidence matrix of X (see [3]). We observe that
Uj ¼ fxi : t ij ¼ 1g;
j ¼ 1; . . . ; n:
We denote by c1, . . . , cn the n columns of the matrix T and by 1 the n 1 matrix with all elements equal to one, that is,
0 1 1 B1C B C C 1¼B B .. C: @.A 1
q
Work supported by the Caratheodory Programme of the University of Patras.
⇑ Corresponding author.
E-mail addresses: [email protected] (D.N. Georgiou), [email protected] (A.C. Megaritis). 0096-3003/$ - see front matter Ó 2011 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2011.08.040
D.N. Georgiou, A.C. Megaritis / Applied Mathematics and Computation 218 (2011) 3122–3130
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Let
1 0 1 c1j c1i Bc C B c2j C B 2i C B C C B C ci ¼ B B .. C and cj ¼ B .. C @ . A @ . A cni cnj 0
are two n 1 matrices. Then, by maxci we denote the maximum
maxfc1i ; c2i ; . . . ; cni g and by ci + cj the n 1 matrix
1 c1i þ c1j B c2i þ c2j C C B C: ci þ cj ¼ B .. C B A @ . 0
cni þ cnj Also, we write ci 6 cj if and only if cki 6 ckj for each k = 1, . . . , n. For the following notions see for example [4]. Let X be a space. A cover of X is a non-empty set of subsets of X, whose union is X. A family r of subsets of X is said to be a refinement of a family c of subsets of X if each element of r is contained in an element of c. Define the order of a family r of subsets of a space X as follows: (a) ord(r) = 1 if and only if r consists of the empty set only. (b) ord(r) = k, where k 2 x, if and only if the intersection of any k + 2 distinct elements of r is empty and there exist k + 1 distinct elements of r, whose intersection is not empty. (c) ord(r) = 1, if and only if for every k 2 x there exist k distinct elements of r, whose intersection is not empty. We denote by dim the function, called covering dimension, with domain the class of all spaces and range the set
x [ {1, 1}, satisfying the following condition: dim(X) 6 k, where k 2 {1} [ x if and only if for every finite open cover c of the space X there exists a finite open cover r of X, refinement of c, such that ord(r) 6 k. Finite topological spaces and the notion of dimension play an important role in digital spaces, computer graphics, image synthesis and image analysis (see, [9,10,14]). Many researchers are currently working in these areas (see, for example, [5,7,8]). In this paper we study the covering dimension for finite topological spaces. In particular, we give an algorithm for computing the covering dimension of a finite space X using the notion of the incidence matrix of X. In Section 2 we present some propositions concerning the covering dimension of a finite topological space. In Section 3 we give the algorithm for computing the covering dimension. Finally, in Section 4 we give some remarks on the algorithm. 2. Finite spaces and covering dimension In this section we present some propositions concerning the covering dimension of a finite topological space. Proposition 2.1. Let X = {x1, . . . , xn} be a finite space. Then, dim(X) 6 k, where k 2 x if and only if there exists an open cover fUj1 ; . . . ; Ujm g of X such that ordðfUj1 ; . . . ; Ujm gÞ 6 k.
Proof. Let dim(X) 6 k, where k 2 x. We prove that there exists an open cover fUj1 ; . . . ; Ujm g of X such that ordðfUj1 ; . . . ; Ujm gÞ 6 k. Let m = min{m 2 x : there exist j1, . . . , jm 2 {1, . . . , n} such that fUj1 ; . . . ; Ujm g is an open cover of Xg and c ¼ fUj1 ; . . . ; Ujm g be an open cover of X. Since dim(X) 6 k, there exists an open cover r = {V1, . . . , Vl} of X, refinement of c, such that ord(r) 6 k. For the proof of the proposition it suffices to prove that c # r. We suppose that there exists a 2 {1, . . . , m} such that Uja R r. Since r is a cover of X, there exists b 2 {1, . . . , l} such that xja 2 V b . By the fact that Uja is the smallest open set of X containing the point xja we have that Uja # V b . Also, since Uja R r, we have Uja – V b . Therefore, Uja V b . Since r is a refinement of c, there exists c 2 {1, . . . , m} such that V b # Ujc . Hence,
Uja Ujc : We observe that the family c n fUja g is an open cover of X with m 1 elements, which is a contradiction by the choice of m. Thus, c # r.
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Conversely, we suppose that there exists an open cover fUj1 ; . . . ; Ujm g of X such that ordðfUj1 ; . . . ; Ujm gÞ 6 k and we prove that dim(X) 6 k. Let c be a finite open cover of the space X. It suffices to prove that the open cover fUj1 ; . . . ; Ujm g of X is a refinement of c. For every i 2 {1, . . . , n} there exists Vi 2 c such that xi 2 Ui # Vi. This means that the open cover {U1, . . . , Un} of X is a refinement of c. Since
fUj1 ; . . . ; Ujm g # fU1 ; . . . ; Un g; we have that the open cover fUj1 ; . . . ; Ujm g of X is a refinement of c. h Proposition 2.2. Let X = {x1, . . . , xn} be a finite space, where n > 1. Then,
dimðXÞ 6 n 2: T T Proof. We consider the open cover {U1, . . . , Un} of X. If nj¼1 Uj ¼ ;, then we have dim(X) 6 n 2. Now, let nj¼1 Uj – ; and Tn T x 2 j¼1 Uj for some j0 2 {1, . . . , n}. Since Uj0 is the smallest open set of X containing the point xj0 , we have Uj0 # nj¼1 Uj . Also, Tj0n Tn j¼1 Uj # Uj0 . Therefore, Uj0 ¼ j¼1 Uj . Hence the family fU1 ; . . . ; Uj0 1 ; Uj0 þ1 . . . ; Un g is an open cover of X with n 1 elements and, therefore,
ordðfU1 ; . . . ; Uj0 1 ; Uj0 þ1 . . . ; Un gÞ 6 n 2: Thus, by Proposition 2.1, dim(X) 6 n 2. h Remark 2.3. We note that the inequality dim(X) 6 n 2 in Proposition 2.2 cannot be replaced by the inequality dim(X) < n 2. Indeed, let X = {x1, . . . , xn}. We consider the family
B ¼ ffx1 g; fx1 ; x2 g; . . . ; fx1 ; xn gg: S
Since V2BV = X, {x1, xi} \ {x1, xj} = {x1} for all i, j 2 {2, . . . , n} and {x1} \ {x1, xk} = {x1} for all k 2 {2, . . . , n}, the family B is a base for a topology s on X. We observe that for the finite space (X, s) we have dim(X) = n 2. In the following propositions we suppose that X = {x1, . . . , xn} is a finite space with n elements, T = (tij), i = 1, . . . , n, j = 1, . . . , n, the incidence matrix of X, and c1, . . . , cn the n columns of the matrix T. Proposition 2.4. If cj = 1 for some j 2 {1, . . . , n}, then dim(X) = 0. Proof. Since cj = 1, we have tij = 1 for every i = 1, . . . , n and, therefore,
Uj ¼ fxi : t ij ¼ 1g ¼ fx1 ; . . . ; xn g ¼ X: Thus, the family {Uj} is an open cover of X. Since ord({Uj}) = 0, by Proposition 2.1, we have dim(X) = 0. h Proposition 2.5. Let cji ; i ¼ 1; . . . ; m, be m columns of the matrix T. Then, cj1 þ þ cjm P 1 if and only if the family fUj1 ; . . . ; Ujm g is an open cover of X. Proof. Let cj1 þ þ cjm P 1. We prove that X ¼ Uj1 [ [ Ujm . It suffices to prove that X # Uj1 [ [ Ujm . Let xi0 2 X. Then, by the definition of the matrix T and by the assumption cj1 þ þ cjm P 1, there exists j 2 {1, . . . , m} such that ti0 jj ¼ 1. Since Ujj ¼ fxi : t ijj ¼ 1g, we have xi0 2 Ujj . Thus, X # Uj1 [ [ Ujm . Conversely, we suppose that the family fUj1 ; . . . ; Ujm g is an open cover of X. Then, for every i 2 {1, . . . , n} there exists j(i) 2 {1, . . . , m} such that xi 2 UjjðiÞ . Therefore, by the definition of the matrix T; tijjðiÞ ¼ 1. Thus, every element of the n 1 matrix cj1 þ þ cjm is larger or equal to 1, which means that cj1 þ þ cjm P 1. h Proposition 2.6. Let cji , i = 1, . . . ,m, be m columns of the matrix T and k ¼ maxðcj1 þ þ cjm Þ, that is k is the maximum element of the n 1 matrix cj1 þ þ cjm . Then,
ordðfUj1 ; . . . ; Ujm gÞ ¼ k 1: Proof. We observe that k is a positive integer larger than or equal to 1. Since maxðcj1 þ þ cjm Þ ¼ k, by the definition of the matrix T, there exist i0 2 {1, . . . , n} and jqk 2 fj1 ; . . . ; jm g; k ¼ 1; . . . ; k, such that ti0 jq þ þ ti0 jq ¼ k and, therefore, t i0 jqk ¼ 1 for 1 k every k = 1, . . . , k. This means that xi0 2 Ujqk for every k = 1, . . . , k and, hence, Ujq \ \ Ujq – ;. Also, we observe that, since k 1 k is the maximum element of the n 1 matrix cj1 þ þ cjm , the intersection of any k + 1 distinct elements of fUj1 ; . . . ; Ujm g is empty. Thus, ordðfUj1 ; . . . ; Ujm gÞ ¼ k 1. h
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Proposition 2.7. Let fUi1 ; . . . ; Uil g be an open cover of X,m ¼ minfm 2 x : there exist j1 ; . . . ; jm 2 f1; . . . ; ng such that fUj1 ; . . . ; Ujm g is an open cover of Xgand fUj1 ; . . . ; Ujm g an open cover of X. Then,
fUj1 ; . . . ; Ujm g # fUi1 ; . . . ; Uil g: Proof. The proof is similar to that of Proposition 2.1.
h
Corollary 2.8. Letm ¼ minfm 2 x : there exist j1 ; . . . ; jm 2 f1; . . . ; ng such that fUj1 ; . . . ; Ujm g is an open cover of Xgand fUj1 ; . . . ; Ujm g be an open cover of X such that
ordðfUj1 ; . . . ; Ujm gÞ ¼ k P 0: Then, for every open cover fUr1 ; . . . ; Url g of X we have
ordðfUr1 ; . . . ; Url g P k: Proof. Let fUr1 ; . . . ; Url g be an open cover of X. By Proposition 2.7,
fUj1 ; . . . ; Ujm g # fUi1 ; . . . ; Uil g: Since ordðfUj1 ; . . . ; Ujm gÞ ¼ k, we have ordðfUr1 ; . . . ; Url g P k. h Proposition 2.9. Let cji , i = 1, . . . , m, be m columns of the matrix T such that cj1 þ þ cjm P 1. If cr1 þ þ crq j 1 for every q < m, then
dimðXÞ ¼ maxðcj1 þ þ cjm Þ 1: Proof. Since cj1 þ þ cjm P 1, by Proposition 2.5, the family fUj1 ; . . . ; Ujm g is an open cover of X. Let k ¼ maxðcj1 þ þ cjm Þ. Then, by Proposition 2.6,
ordðfUj1 ; . . . ; Ujm gÞ ¼ k 1 and, therefore, by Proposition 2.1, dim(X) 6 k 1. We prove that dim(X) = k 1. We suppose that dim(X) < k 1. Then, by Proposition 2.1, there exists an open cover fUr1 ; . . . ; Url g of X with ordðfUr1 ; . . . ; Url gÞ < k 1. Since cj1 þ þ cjm P 1 and cr1 þ þ crq j1 for every q < m, by Proposition 2.5, we havem ¼ minft 2 x : there exist j1 ; . . . ; jt 2 f1; . . . ; ng such that fUj1 ; . . . ; Ujt g is an open cover of Xg. Moreover, since
ordðfUj1 ; . . . ; Ujm gÞ ¼ k 1; by Corollary 2.8, we have
ordðfUr1 ; . . . ; Url gÞ P k 1 which is a contradiction. Thus, dim(X) = k 1. h 3. An algorithm for computing the covering dimension In this section we give an algorithm for computing the covering dimension of finite topological spaces using the Propositions 2.4 and 2.9. Let X = {x1, . . . , xn} be a finite space of n elements and T = (tij) the n n incidence matrix. Then our intended algorithm contains n steps: Step 1. Read the n columns c1, . . . , cn of the matrix T. If some column is equal to 1, then print
dimðXÞ ¼ 0: Otherwise go to the Step 2. 0
0
Step 2. Find the sums cj1 þ cj2 for each j1, j2 2 {1, . . . , n}. If there exist j1 ; j2 2 f1; . . . ; ng such that cj0 þ cj0 P 1, then print 1
2
dimðXÞ ¼ maxðcj0 þ cj0 Þ 1: 1
2
Otherwise go to the Step 3. 0
0
0
Step 3. Find the sums cj1 þ cj2 þ cj3 for each j1, j2, j3 2 {1, . . . , n}. If there exist j1 ; j2 ; j3 2 f1; . . . ; ng such that cj0 þ cj0 þ cj0 P 1, 1 2 3 then print
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D.N. Georgiou, A.C. Megaritis / Applied Mathematics and Computation 218 (2011) 3122–3130
dimðXÞ ¼ max cj0 þ cj0 þ cj0 1: 1
0
3
Otherwise go to the Step 4. ... Step n 1. Find the sums
cj1 þ cj2 þ þ cjn1 ; 0
0
0
for each j1, j2, . . . , jn1 2 {1, . . . , n}. If there exist j1 ; j2 ; . . . ; jn1 2 f1; . . . ; ng such that cj0 þ cj0 þ þ cj0 1
2
dimðXÞ ¼ maxðcj0 þ cj0 þ þ cj0 Þ 1: 1
0
n1
Otherwise go to the Step n. Step n. Find the sum c1 + c2 + + cn and print
dimðXÞ ¼ maxðc1 þ c2 þ þ cn Þ 1: Example 3.1 (1) Let X = {x1, x2, x3} with the topology
s ¼ f;; fx2 g; fx1 ; x2 g; fx2 ; x3 g; Xg: We observe that U1 = {x1, x2}, U2 = {x2}, U3 = {x2, x3}. Therefore,
0
1 0
0
1
B C T ¼ @ 1 1 1 A; 0
0 1 1 B C c1 ¼ @ 1 A;
0 1
0 1 0 B C c2 ¼ @ 1 A;
0
0 1 0 B C and c3 ¼ @ 1 A:
0
1
Moreover,
0 1 1 B C c1 þ c2 ¼ @ 2 A j 1;
0 1 1 B C c1 þ c3 ¼ @ 2 A > 1;
0
0 1 0 B C c2 þ c3 ¼ @ 2 A j 1
1
1
and
maxðc1 þ c3 Þ ¼ 2: Thus, dim(X) = max(c1 + c3) 1 = 1. (2) Let X = {x1, x2, x3, x4} with the topology
s ¼ f;; fx1 g; fx1 ; x2 g; fx1 ; x3 g; fx1 ; x4 g; fx1 ; x2 ; x3 g; fx1 ; x2 ; x4 g; fx1 ; x3 ; x4 g; Xg: We observe that
U1 ¼ fx1 g;
U2 ¼ fx1 ; x2 g;
U3 ¼ fx1 ; x3 g;
U4 ¼ fx1 ; x4 g:
Therefore,
0
1 1 1 1
1
C B B0 1 0 0C C T¼B B 0 0 1 0 C; A @ 0 0 0 1
0 1 1 B C B0C C c1 ¼ B B 0 C; @ A 0
0 1 1 B C B1C C c2 ¼ B B 0 C; @ A 0
0 1 1 B C B0C C c3 ¼ B B 1 C; @ A 0
0 1 1 B C B0C C and c4 ¼ B B 0 C: @ A 1
Moreover,
0 1 2 B C B1C C c1 þ c2 ¼ B B 0 C j 1; @ A 0
0 1 2 B C B0C C c1 þ c3 ¼ B B 1 C j 1; @ A 0
0 1 2 B C B0C C c1 þ c4 ¼ B B 0 C j 1; @ A 1
n1
P 1, then print
D.N. Georgiou, A.C. Megaritis / Applied Mathematics and Computation 218 (2011) 3122–3130
0 1 2 B1C B C c2 þ c3 ¼ B C j 1; @1A
0 1 2 B1C B C c2 þ c4 ¼ B C j 1; @0A
0
0 1 2 B0C B C c3 þ c4 ¼ B C j 1; @1A
1
0 1 3 B1C B C c2 þ c3 þ c4 ¼ B C > 1; @1A
and
3127
1
maxðc2 þ c3 þ c4 Þ ¼ 3:
1 Thus, dim(X) = max(c2 + c3 + c4) 1 = 2. (3) Let X = {x1, x2, x3, x4, x5} with the topologys = {;, {x2}, {x5}, {x2, x4}, {x2, x5}, {x1, x2, x3}, {x2, x4, x5}, {x1, x2, x3, x4}, {x1, x2, x3, x5}, X}. We observe that
U1 ¼ fx1 ; x2 ; x3 g;
U2 ¼ fx2 g;
U3 ¼ fx1 ; x2 ; x3 g;
U4 ¼ fx2 ; x4 g;
U5 ¼ fx5 g:
Therefore,
0
1 0
B B1 1 B T¼B B1 0 B @0 0 0 0 0 1 1 B C B1C B C C c1 ¼ B B 1 C; B C @0A 0
1 0
0
1
C 1 1 0C C 1 0 0C C; C 0 1 0A 0 0 1 0 1 0 B C B1C B C C c2 ¼ B B 0 C; B C @0A 0
0 1 1 B C B1C B C C c3 ¼ B B 1 C; B C @0A 0
0 1 0 B C B1C B C C c4 ¼ B B 0 C; B C @1A 0
0 1 0 B C B0C B C C and c5 ¼ B B 0 C: B C @0A 1
Moreover,
0 1 1 B C B2C B C C c1 þ c2 ¼ B B 1 C j 1; B C @0A 0
0 1 2 B C B2C B C C c1 þ c3 ¼ B B 2 C j 1; B C @0A 0
0 1 0 B C B2C B C C c1 þ c4 ¼ B B 1 C j 1; B C @1A 0
0 1 1 B C B1C B C C c1 þ c5 ¼ B B 1 C j 1; B C @0A 1
0 1 1 B C B2C B C C c2 þ c3 ¼ B B 1 C j 1; B C @0A 0
0 1 0 B C B2C B C C c2 þ c4 ¼ B B 1 C j 1; B C @1A 0
0 1 0 B C B1C B C C c2 þ c5 ¼ B B 0 C j 1; B C 0 @ A 1
0 1 1 B C B2C B C C c3 þ c4 ¼ B B 1 C j 1; B C 1 @ A 0
0 1 1 B C B1C B C C c3 þ c5 ¼ B B 1 C j 1; B C 0 @ A 1
0 1 0 B C B1C B C C c4 þ c5 ¼ B B 0 C j 1; B C @1A 1
0 1 1 B C B2C B C C c1 þ c4 þ c 5 ¼ B B 1 C > 1; B C @1A 1
Thus, dim(X) = max(c1 + c4 + c5) 1 = 1.
and
maxðc1 þ c4 þ c5 Þ ¼ 2:
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4. Remarks on the algorithm for computing the covering dimension of finite topological spaces Remark 4.1. Let A = (aij) be a n n matrix and B = (bij) a m m matrix. The Kronecker product of A and B (see, for example, [6]) is the mn mn block matrix
0
B AB¼B @
a11 B . . . a1n B .. .
..
.. .
.
an1 B . . . ann B
1
C C: A
More explicitly, the Kronecker product of A and B is the matrix
0
a11 b11 Ba b B 11 21
B .. B B . B Ba b B 11 m1 B B .. B . B B B an1 b11 B B an1 b 21 B B .. B @ .
an1 bm1
a11 b12 . . . a11 b1m . . . a1n b11 a11 b22 . . . a11 b2m . . . a1n b21 .. .
..
.. .
.
..
.. .
.
.. .
..
.. .
.
..
.. .
.
.. .
an1 b12 . . . an1 b1m . . . ann b11 an1 b22 . . . an1 b2m . . . ann b21
ann b12 ann b22
.. . an1 bm2
.. . ann bm2
..
.. . . . . . an1 bmm
..
.. . . . . . ann bm1
C C C C . . . a1n bmm C C C C .. .. C: . . C C . . . ann b1m C C . . . ann b2m C C C .. .. C A . . . . . ann bmm ..
a11 bm2 . . . a11 bmm . . . a1n bm1 a1n bm2 .. .
1
a1n b12 . . . a1n b1m a1n b22 . . . a1n b2m C C .. .
.
Let X = {x1, . . . , xn} be a finite space of n elements and Y = {y1, . . . , ym} a finite space of m elements. It is known that if TX is the incidence matrix of X and TY is the incidence matrix of Y, then the incidence matrix of X Y is the Kronecker product TX TY of TX and TY (see, for example, [16]). Example 4.2. Let X = {x1, x2, x3} with the topology
sX ¼ f;; fx2 g; fx1 ; x2 g; fx2 ; x3 g; Xg and Y = {y1, y2, y3, y4} with the topology
sY ¼ f;; fy1 g; fy1 ; y2 g; fy1 ; y3 g; fy1 ; y4 g; fy1 ; y2 ; y3 g; fy1 ; y2 ; y4 g; fy1 ; y3 ; y4 g; Yg: The incidence matrix TX of X is
0
1 0 0
1
B C TX ¼ @ 1 1 1 A 0
0 1
and the incidence matrix TY of Y is
0
1 1 1 1 1 B0 1 0 0C B C TY ¼ B C: @0 0 1 0A 0
0
0 1
Therefore, the incidence matrix TXY of the product space X Y is
0
T XY
1 1 1 1 0
B B0 B B0 B B B0 B B1 B B B0 ¼ TX TY ¼ B B0 B B B0 B B0 B B B0 B B0 @ 0
1 0
0
0
0 0
0
0
0
0
0
0
0 0
0
0
0 1 0 0 0 1
0 0
0 0
0 0
0 0 0 0
0 0
0 0
1 1 1 1 1 1 1 1 1 1 1 0
0
0 1 0
0 0
1 0
0 1
0
0
0 1
0 0
0
1
0
0 1
0
0
0 1 0
0
0
0
0
0
0
0
0
0 1 1 1
0
0
0
0
0
0
0 0
1 0
0
0
0
0
0
0
0 0
0
1
0
0
0
0
0
0
0 0
0
0
0
1
C 0C C 0C C C 0C C 1C C C 0C C: 0C C C 1C C 1C C C 0C C 0C A 1
D.N. Georgiou, A.C. Megaritis / Applied Mathematics and Computation 218 (2011) 3122–3130
3129
We observe that
cr1 þ . . . þ crq j 1 for every q < 6;
c2 þ c3 þ c4 þ c10 þ c11 þ c12
0 1 3 B C B1C B C B1C B C B C B1C B C B6C B C B C B2C C ¼B B2C > 1 B C B C B2C B C B3C B C B C B1C B C B1C @ A 1
and max(c2 + c3 + c4 + c10 + c11 + c12) = 6. Thus,
dimðX YÞ ¼ maxðc2 þ c3 þ c4 þ c10 þ c11 þ c12 Þ 1 ¼ 5: Also, we observe that dim(X) = 1 and dim(Y) = 2. Remark 4.3. By Example 4.2 arises that the inequality
dimðX YÞ 6 dimðXÞ þ dimðYÞ for two finite spaces X and Y in general, is not true. Remark 4.4. Let X = {x1, . . . , xn} be a finite T0-space. Then, there exists a finite space X1 homeomorphic to X such that the incidence matrix T X 1 of X1 is an upper triangular matrix (see Theorem 7 of [16]). So in order to calculate the dim(X) it suffices to calculate dim(X1). Example 4.5. Let X = {x1, x2, x3} with the topology
s ¼ f;; fx2 g; fx1 ; x2 g; fx2 ; x3 g; Xg: We consider the space X1 = {x1, x2, x3} with the topology
s1 ¼ f;; fx1 g; fx1 ; x2 g; fx1 ; x3 g; X 1 g: We observe that the spaces X and X1 are homeomorphic. The incidence matrix T X 1 of X1 is
0
T X1
1 1 1 1 B C ¼ @ 0 1 0 A: 0 0 1
Since
0 1 0 1 0 1 2 1 1 B C B C B C c2 þ c3 ¼ @ 1 A þ @ 0 A ¼ @ 1 A > 1 1 1 0 and max(c2 + c3) = 2, we have
dimðXÞ ¼ maxðc2 þ c3 Þ 1 ¼ 1: Proposition 4.6. An upper bound on the number of iterations of the algorithm for computation of the covering dimension of a finite space X of n elements performs is the number 2n 1. Proof. We observe that the number of iterations the algorithm performs in Steps 1, 2, . . . , n 1 and n is n n n n ; ;...; ; and , respectively. Thus, the number of iterations the algorithm performs is 1 2 n1 n
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D.N. Georgiou, A.C. Megaritis / Applied Mathematics and Computation 218 (2011) 3122–3130
n 1
þ
n 2
þ þ
n
þ
n1
n n
¼
n X n k nk ¼ ð1 þ 1Þn 1 ¼ 2n 1: 11 k k¼1
Proposition 4.7. Let X be a finite space with n elements. If the incidence matrix TX of X is an upper triangular matrix, then an upper bound on the number of iterations the algorithm performs is the number 2n1. Proof. Since the incidence matrix TX is an upper triangular matrix, in order to find dim(X) we follow the next steps: Step 0. Read the column cn. If cn P 1, then
dimðXÞ ¼ 0: Otherwise go to the Step 1. Step 1. If there exists i 2 {1, . . . , n 1} such that ci + cn P 1, then
dimðXÞ ¼ maxðci þ cn Þ 1: Otherwise go to the Step 2. Step 2. If there exists i1, i2 2 {1, . . . , n 1} such that ci1 þ ci2 þ cn P 1, then
dimðXÞ ¼ maxðci1 þ ci2 þ cn Þ 1: Otherwise go to the Step 3.. . . Step n2. If there exists i1, . . . , in2 2 {1, . . . , n 1} such that
ci1 þ þ cin2 þ cn P 1; then
dimðXÞ ¼ maxðci1 þ þ cin2 þ cn Þ 1: Otherwise go to the Step n 1. Step n 1. If c1 + + cn1 + cn P 1, then
dimðXÞ ¼ maxðc1 þ þ cn1 þ cn Þ 1: By the above we have that the number of iterations the algorithm performs is
n1 0
þ
n1 1
þ
n1 2
þ þ
n1 n1
¼
n1 X n 1 k n1k ¼ ð1 þ 1Þn1 ¼ 2n1 : 11 k k¼0
Acknowledgement We are grateful to the referee for a number of helpful suggestions for improvement in the article. References [1] P. Alexandroff, Diskrete Räume, Mat. Sb. (N.S.) 2 (1937) 501–518. [2] D.W. Bass, Dimension and finite spaces, J. London Math. Soc. 44 (1969) 159–162. [3] A. El-Fattah El-Atik, M.E. Abd El-Monsef, E.I. Lashin, On finite T0 topological spaces, Proceedings of the Ninth Prague Topological Symposium (2001) 75– 90. [4] R. Engelking, Theory of Dimensions, Finite and Infinite, Heldermann Verlag, Berlin, 1995. [5] A.V. Evako, R. Kopperman, Y.V. Mukhin, Dimensional properties of graphs and digital spaces, J. Math. Imaging Vision 6 (2–3) (1996) 109–119. [6] H. Eves, Elementary Matrix Theory, Dover Publications, 1980. [7] A.S. Farrag, S.E. Abbas, Computer programing for construction and enumeration of all regular topologies and equivalence relations on finite sets, Appl. Math. Comput. 165 (1) (2005) 177–184. [8] El Zohny Habiba, On zero dimensional space and its retraction, Appl. Math. Comput. 155 (2) (2004) 515–519. [9] G.T. Herman, Geometry of Digital Spaces, Birkhauser, Boston, 1998. [10] E.D. Khalimsky, R. Kopperman, P.R. Meyer, Computer graphics and connected topologies on finite ordered sets, Topol. Appl. 36 (1990) 1–17. [11] Ralph Kopperman, On storage of topological information, Discret. Appl. Math. 147 (2005) 287–300. [12] M.C. McCord, Singular homology groups and homotopy groups of finite topological spaces, Duke Math. J. 33 (1966) 465474. [13] W. Mielants, On the number of topologies with given small inductive dimension on a finite set, Matematiche (Catania) 26 (1971) 27–34 (1972). [14] A. Rosenfeld, Digital topology, Amer. Math. Monthly 86 (1979) 621–630. [15] J.B. Sancho and M.T. Sancho, Dimension of Dense Subalgebras of C(X), Proceedings of the American Mathematical Society, vol. 105, No. 2 (Feb., 1989), pp. 491–499. [16] Mitsunobu Shiraki, On finite topological spaces, Rep. Fac. Sci. Kagoshima Univ. 1 (1968) 1–8.