Research Article  Open Access
On the Cardinality of the Topologies on a Finite Set
Abstract
Let be the number of all labeled topologies having open sets that we can define on points, and let be the number of those which are nonhomeomorphic. In this paper, we compute these numbers for and arbitrary . The numbers of all unlabeled and nontopologies with open sets are also given for .
1. Introduction
Let be a finite set with elements. The total number of all labelled topologies one can define on is a long standing open question. No reasonable explicit or recursive counting formula for is known. Moreover, the number of all labelled topologies definable on is also an open problem. Recall that a topology is a topology satisfying the separate axiom: for all , there is an open set containing one but not the other.
A classical and well known result is that each topology on corresponds to a quasiorder on (reflexive and transitive relation) and each topology corresponds to a partial order (reflexive, transitive, and antisymmetric relation). This bijective correspondence is due, independently, to Alexandroff [1] and Birkhoff [2]. On the other hand, several authors (Comtet [3], Evans et al. [4], Renteln [5] and others) have observed the following connexion between the and the : , where the are the Stirling numbers of the second kind. This formula was later refined by ErnÃ© [6] to , where is any topological property which is invariant under lattice isomorphisms. Numerically, the values have been found for by several authors (see, e.g., [3]), for by Evans et al. [4], for by ErnÃ© [6], for by Das [7], and for by ErnÃ© and Stege [8]. Later, the calculations were pursued by Brinkmann and McKay [9, 10] until .
Another approach is the study of the topologies by their number of open sets. For this, let be the number of all labeled topologies with open sets that we can define on and let be the number of those which are . As for , there is no explicit formula giving the or the , while some values are known. When is large, Sharp [11] and Stephen [12] showed that there is no topology with open sets. In 1971, Stanley [13] enumerated the for and arbitrary , and in , the author [14] extended the range to . For small values of , ErnÃ© and Stege [8, 15] determined the and for all and . Independently, Benoumhani [16] computed by a direct method the for and arbitrary . Later, the range was extended in [17] until .
For the unlabelled case, let be the number of all unlabeled topologies having open sets, let be the number of those which are , and let be the number of those which are non. Unfortunately, all these numbers are missing until now. However, some values are obtained. For example, Stanley [13] computed the and the for and , respectively. In the opposite sense, the numbers are given for by ErnÃ© and Stege [15], see also the paper [17] for .
Our present paper is a continuation of the previous effort in the field. Here we give the values and until . For the , the calculations are given until . Our approach is based on the notion of level partitions (Definition 2). Let denote the number of all labeled topologies with open sets and levels; then, . In Section 2, we prove that if ; then, . The summation is then reduced to and this reduction permits us to enumerate, in Section 3, all the possibilities and find the desired values. The unlabeled case is studied in Section 4. In Lemma 13, we prove that any two topologies are homeomorphic if and only if their corresponding levels are also homeomorphic. This property enables us to calculate the desired numbers.
2. Basic Results
In this section, we prove that if is a topology with open sets, then has at most levels.
Definition 1. Let be a topology on . The nonempty open set is called a minimal open set if, for all in , we have
In the lattice terminology, a minimal open set is an atom. Thereby, for any topology , we define inductively the decreasing sequence of the subsets and the topologies on as follows: for , we set , , and we denote by the minimal opens sets of . We put , , , and we denote by the minimal open sets of . Again, , , , and so on. This process should stop after at most iterations. Let , and let be the increasing family defined as follows: We can see that for all , , , and then is a topology on . Now, we recall the notion of level.
Definition 2. Let be a topology on the finite set , and let For all , is called the th level of .
Lemma 3. There exists a finite index , such that , and is a Boolean algebra.
The smallest integer such that is the number of level of . The number is at most , and it is reached if and only if is the maximal chain. On the other hand, we observe that is a partition of , is a partition of , and and are both partitions of . Let us now give an example.
Example 4. Take and We have , , and . Then has three levels with , and . We see that , with and , , with and , and finally , with . The partially ordered set corresponding to is as follows: (5)
Since the elements of a minimal open set cannot be distinguishable, we have the following.
Lemma 5. If is , then every minimal open set of is a singleton. Moreover,â€‰ â€‰for all in , the topology is also .
A consequence of Lemma 5 is the following.
Theorem 6. Let be a topology with levels. Then is if and only if is for all .
It follows again that a topology is if and only if all its blocks are singletons and then . In Example 4, this equality is satisfied and of course is .
Lemma 7. Let be a topology with levels. Then, has at least open sets and at most open sets. In addition, these two bounds are optimal.
Proof. The th level is generated by , where and is arbitrary from the previous level . So, contains at least open sets, and then has at least open sets. On the other hand, if we take , then contains exactly open sets and the lower bound is optimal. Let us prove the upper bound. We see that each has the form , where each is a union of a certain number of the minimal open sets . So each generates at most open sets, and then contains at most open sets. All these numbers added give the desired upper bound; in addition, if we take , then contains exactly open sets and the upper bound is reached.
Lemma 8. Let be a topology with levels. Then has at most open sets, and this bound is optimal.
Proof. Since is , then , and then, for all , By injecting all these inequalities in the upper bound appearing in Lemma 7, we obtain the desired bound. The optimality is also a consequence of the optimality in Lemma 7.
It is trivial to see that two topologies having different number of levels are also different, so we have the following.
Theorem 9. For all integers and , we have
This last formula remains true in the non case. For large values of it reduces to the following.
Theorem 10. Let . If , then
Proof. As , then should be less or equal to .
This reduction enables us to enumerate easily all the possibilities and then to find the numbers . This is the content of the next section.
3. Computing until
As mentioned before, the values are known for . So, we have to compute them only for in the interval . Equivalently, we have to compute the numbers , for . Details are given only in the case , the others are similar. Thereby, let be a topology with two levels and let be the associated partition. The level is generated by , and then it contains open sets. The level is generated by , where and is arbitrary from . The question is then how to choose the in order to get a topology such that ? For this, we consider all the cases one by one and we use the notation if , and otherwise.
Case 1 (). In this case, has at least open sets which is outside the interval ; so there is no solution in this case.
Case 2 (). In this case, we prove that there are only 8 possibilities. For a given partition , of , we try to reconstruct all the topologies having this partition and such that is in the desired interval. For this, we have to remark first that:
if one of the is adjoined to the singleton , then , which is outside the desired interval;
if the are both adjoined to more than three elements, then , which is also outside the desired interval.
So, at least one of the should be adjoined to a set with two or three elements. Without loss of generality, suppose that is adjoined to . Here we enumerate 3 possibilities. The first one is when is adjoined to . The corresponding poset is (12) Obviously, the contribution of the open set to the topology is open sets, that of is open sets, and that of their union is also open sets. So, has open sets and the number of such topologies is
The second possibility is when is adjoined to , as follows: (13) Here the contribution ofâ€‰â€‰ is , that ofâ€‰â€‰ is , and that of their union is . So, the topology has open sets and the number of such topologies is
The third possibility is when is adjoined to , as follows: (14) The contribution ofâ€‰â€‰ is , that ofâ€‰â€‰ is , and that of their union is . So, the topology has open sets and the number of such topologies is
Suppose now that is adjoined to three elements . Remark that if is adjoined to more than four elements, then , which is excluded. So, should be adjoined to three or four elements. When is adjoined to three elements, we find four possibilities. The first one is when is adjoined to . The corresponding poset is (15) Here the contribution of is , that ofâ€‰â€‰ is , and that of their union is also . So, has open sets. The number of such topologies is .
The second possibility is when is adjoined to as follows: (16) Here the contribution ofâ€‰â€‰ is , that ofâ€‰â€‰ is , and that of their union is . So, has open sets. The number of such topologies is .
The third possibility is when is adjoined to as follows: (17) Here the contribution ofâ€‰â€‰ is , that ofâ€‰â€‰ is , and that of their union is . So, has open sets. The number of such topologies is .
The fourth possibility is when is adjoined to as follows: (18) Here the contribution ofâ€‰â€‰ is , that ofâ€‰â€‰ is , and that of their union is . So, has open sets. The number of such topologies is .
When is adjoined to four elements , we remark that if then , which is outside the desired interval; so the unique possibility is when is adjoined to as follows: (20) Here the contribution of is , that of is , and that of their union is also . So, has open sets and the number of such topologies is .
Then we have proved that there are only 8 possibilities. We enumerate them in Table 1.

We use the same reasoning for all the rest of the cases . Note that we obtain 13 possibilities for the case , also 13 for , 7 for , 6 for , and 5 for all the cases . Since the machinery is the same as above, we have omitted the tedious details for the readerâ€™s convenience. The total number of topologies with two levels and open sets is then given by Table 2.

Using the same approach for three levels, we obtain two possibilities for the case , nine for , one for , eight for , one for , two for , two for all the cases , and finally one for all the cases . Omitting again the tedious details, the numbers with are given by Table 3.

For four levels, we obtain only one possibility for the case . The numbers with are given by Table 4.
Adding all these computations, we arrive at the following.
Theorem 11. Let . The number of all labeled topologies on points and having open sets is given in Table 5.
All the other topologies have either or open sets.

Corollary 12. According to ErnÃ©â€™s formula , the total number of nontopologies on arbitrary points and having open sets may be known until .
4. Computing Unlabeled Topologies
As mentioned in the Introduction, the numbers and are computed by Stanley [13], respectively, for and . In this section, we extend the range until for the topologies and until for the nontopologies. Then we rediscover Stanleyâ€™s results. The following observation is the key of our result.
Lemma 13. Let and be two topologies on points. Then and are homeomorphic if and only if their corresponding levels are, respectively, homeomorphic.
Proof. It is trivial to see that if the corresponding levels are, respectively, homeomorphic, then the two topologies are homeomorphic. Let us prove the nontrivial sense. Suppose that is an homeomorphism from into , and let be, respectively, their corresponding partitions. Since the first blocks are open and singletons, then and so . But is also an homeomorphism from into and for the same reason, we obtain . We repeat this process to obtain that , and for all . On the other hand, let , and , . As the above, we have , and then for all . Since and are open with respect to and , respectively, is also an homeomorphism from the subspace into the subspace . But and ; so for , we obtain that is homeomorphic to . For , we have which is homeomorphic to and then which is homeomorphic to . We repeat this process to obtain that is homeomorphic to for all , and the proof is done.
Corollary 14. Lemma 13 holds if and are non.
It follows that the number of unlabelled topologies is exactly the number of the possibilities given in the labelled case (see the above section). So, omitting the easy computations for , we obtain the following.
Theorem 15. Let . The number of all unlabeled topologies, with open sets, is detailed in Table 6, totaling unlabeled topologies. All the other unlabelled topologies have open sets.
