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The theorem states this:

The relation R on a set A is transitive if and only if $R^n \subseteq R$ for n = 1, 2, 3,...

What I'm reading is that the nth power of that set is transitive if the set is a subset of that original set.

I've been trying this out with zero-one matrices (in C++, with the boolean product of the matrices) but it's not working as I expect it to. Specifically, I'm kind of confused about n. Does n go to infinity, or does it represent the size of the set?

Could someone explain this theorem to me, or tell me what I'm misunderstanding?

Asaf Karagila
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Austin Moore
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    This might help. – angryavian Feb 28 '14 at 06:04
  • @angryavian Thank you that was helpful. Specifically this comment by Harald Hanche-Olsen: "Since you are looking at a a matrix representation of the relation, an easy way to check transitivity is to square the matrix. The relation is transitive if and only if the squared matrix has no nonzero entry where the original had a zero. (If you don't know this fact, it is a useful exercise to show it.)" – Austin Moore Feb 28 '14 at 06:11
  • Older related questions: http://math.stackexchange.com/questions/551906/let-r-be-a-relation-on-set-a-prove-that-r2-subseteq-r-r-is-transitive and http://math.stackexchange.com/questions/440790/relation-r-r-circ-r-subseteq-r-implies-r-is-transitive – Martin Sleziak Feb 28 '14 at 07:49
  • "What I'm reading is that the nth power of that set is transitive if the set is a subset of that original set." Well, that's not what it's saying. It's saying that if, for all $n$, the $n$th power of a relation is contained in the relation, then the relation is transitive. But you have to know what it means to compute a power of a relation. – Gerry Myerson Feb 28 '14 at 08:08

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Suppose that you've enumerated the elements of your finite set: $$ A = \{a_1, \ldots, a_d\}. $$ Then, the zero-one matrix $M = (m_{ij})$, representing the relation $R$ has entries $$ m_{i,j} = \begin{cases} 1, &(a_i, a_j) \in R \\ 0, &(a_i, a_j) \notin R. \end{cases}$$ Now, $R$ is transitive means that whenever $(a_i, a_j) \in R$ and $(a_j, a_k) \in R$, it follows that $(a_i, a_k) \in R$. Translated into the matrix language, $$ \text{if } m_{i,j} = 1 \text{ and } m_{j,k} = 1 \text{ for any } j, \text{ then } m_{i,k} = 1, $$ or using the Boolean operations of conjunction (and) and disjunction (or), $$ \left( m_{i,1} \wedge m_{1,k} \right) \vee \cdots \vee \left( m_{i,n} \wedge m_{d,k} \right) \le m_{i,k}, $$ or simply $$ M^2 \le M. $$ Then, $M^3 = M M^2 \le M M \le M$ and by induction, $$ M^n \le M \text{ for all } n. $$

Sammy Black
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