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Let $A$ and $B$ be two finite sets with cardinalities $m$ and $n$, respectively. The total number of possible relations $R \subseteq A \times B$ between $A$ and $B$ is $2^{m \times n}$, since each relation is a subset of the Cartesian product $A \times B$.

However, I am interested in understanding these relations in terms of their structure rather than just counting all possible relations.

Example 1

Let $A = \{a_1, a_2\}$ and $B = \{b\}$. Hence, there are the following $2^{2 \times 1} = 4$ possible relations:

  1. $R_1 = \emptyset$.
  2. $R_2 = \{(a_1, b)\}$.
  3. $R_3 = \{(a_2, b)\}$.
  4. $R_4 = \{(a_1, b), (a_2, b)\}$.

However, in terms of structure, there are only 3 kinds of relations $S \subseteq A \times B$:

  1. $S_1$ which has the structure of $R_1$.
  2. $S_2$ which has the structure of $R_2$ and $R_3$.
  3. $S_3$ which has the structure of $R_4$.

This is because $R_2 = \{(a_1, b)\}$ and $R_3 = \{(a_2, b)\}$ are structurally the same, differing only in which element of $A$ is related to $b$.

Example 2

Now, let $B = {b_1, b_2}$ while keeping $A = {a_1, a_2}$. The total number of relations is $2^{2 \times 2} = 16$. The possible relations are:

  1. $R_1 = \emptyset$.

  2. $R_2 = \{(a_1, b_1)\}$.

  3. $R_3 = \{(a_1, b_2)\}$.

  4. $R_4 = \{(a_2, b_1)\}$.

  5. $R_5 = \{(a_2, b_2)\}$.

  6. $R_6 = \{(a_1, b_1), (a_1, b_2)\}$.

  7. $R_7 = \{(a_2, b_1), (a_2, b_2)\}$.

  8. $R_8 = \{(a_1, b_1), (a_2, b_1)\}$.

  9. $R_9 = \{(a_1, b_2), (a_2, b_2)\}$.

  10. $R_{10} = \{(a_1, b_2), (a_2, b_1)\}$.

  11. $R_{11} = \{(a_1, b_1), (a_2, b_2)\}$.

  12. $R_{12} = \{(a_1, b_1), (a_1, b_2), (a_2, b_1)\}$.

  13. $R_{13} = \{(a_1, b_1), (a_1, b_2), (a_2, b_2)\}$.

  14. $R_{15} = \{(a_1, b_1), (a_2, b_1), (a_2, b_2)\}$.

  15. $R_{14} = \{(a_1, b_2), (a_2, b_1), (a_2, b_2)\}$.

  16. $R_{16} = \{(a_1, b_1), (a_1, b_2), (a_2, b_1), (a_2, b_2)\}$.

In terms of structure, however, the distinct kinds of relations $S \subseteq A \times B$ are:

  1. $S_1$ which has the structure of $R_1$
  2. $S_2$ which has the structure of $R_2$-$R_5$.
  3. $S_3$ which has the structure of $R_6$-$R_7$.
  4. $S_4$ which has the structure of $R_8$-$R_9$.
  5. $S_5$ which has the structure of $R_{10}$-$R_{11}$.
  6. $S_6$ which has the structure of $R_{12}$-$R_{15}$
  7. $S_7$ which has the structure of $R_{16}$.

More generally

Here is the data I have gathered (by drawing case by case) on the number of relations of similar structure.

  • For $m=n=1$: 2 relations grouped in 2 kinds.
  • For $m=1$ and $n=2$: 4 relations grouped in 3 kinds.
  • For $m=1$ and $n=3$: 8 relations grouped in 4 kinds.
  • For $m=n=2$: 16 relations grouped in 7 kinds.
  • For $m=2$ and $n=3$: 64 relations grouped in 13 kinds.
  • For $m=n=3$: still checking.

Questions:

  • What is the general method for determining the number of such distinct kinds of relations $S \subseteq A \times B$ based on structure for arbitrary $m$ and $n$?
  • How does this classification extend to cases with more than two sets, such as relations $S \subseteq A_1 \times ... \times A_n$?
  • Is there any algorithm that can allow me to determine exactly the form of those relations?
  • Is there a specific mathematical term for this kind of classification, and in which field of mathematics is this topic studied in detail? (Any recommended textbooks or references would be greatly appreciated!)
Alex Kruckman
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elmo
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  • Can you provide a precise definition of structure? By your example, it could simply be the cardinality of the relation. – eti902 Nov 01 '24 at 18:20
  • @eti902 I used the term 'structure' just to mean, informally expressed, that there's a similarity of shape in the relations, as exemplified by $R_1$ and $R_2$, both of which have the structure or shape $S_2$. I'm not using these 'structure' or 'shape' in any technical sense here nor I have clear definitions in mind. My intention with this question is, in part, to find the mathematical concept and definition corresponding to what I'm trying to get at. – elmo Nov 01 '24 at 18:26
  • Can you provide an example where $B$ has more than one element? It is not clear (at least to me) how you distinguish two relations having the same cardinality – eti902 Nov 01 '24 at 18:31
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    Hi, @eti902. I just added the requested example. – elmo Nov 01 '24 at 19:33
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    You are trying to count the number of bipartite graphs (with parts of size $m$ and $n$) up to isomorphism. Your more general question asks for the number of $k$-partite $k$-hyper graphs up to isomorphism. The field of mathematics that studies this kind of question is called enumerative combinatorics. – Alex Kruckman Nov 01 '24 at 21:18

1 Answers1

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Given sets $A$ and $B$ you may say that two subsets $R$, $R^\prime\subset A\times B$ are structurally equivalent if there exists bijections $\phi:A\rightarrow A$ and $\psi:B\rightarrow B$ such that $$ \phi\times\psi(R)=R^\prime. $$ This is in fact an equivalence on the set $\mathrm{P}(A\times B)$ of subets of $A\times B$: just check the definition using the properties of bijections.

When $|A|=m<\infty$ and $|B|=n<\infty$ the number of equivalence classes is just the number of orbits of the natural action of $\mathcal{S}_m\times\mathcal{S}_n$ on $\mathrm{P}(A\times B)$. Here $\mathcal{S}_k$ denotes the group of permutations on $k$ objects, i.e. the bjiections of a set with $k$ elements.

A standard basic tool to compute the number of orbits for a finite group action is Burnside's Lemma but the general application in this case doesn't seem too feasible.

Andrea Mori
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