# Tricky transitive relations

I have a set $A = \{1, 2, 3\}$.

Relation $S = \{(1, 1), (1, 2), (3, 1) \}$

Relation $T = \{(1, 1), (3, 2), (3, 1) \}$

$S$ is not transitive, but $T$ is transitive. Why is that?

A relation $R$ transitive if $(a,b),(b,c)\in R\Rightarrow (a,c)\in R$.

In $S$, we have $(1, 1), (1, 2)$, and we also have $(1, 2)$. And we have $(3, 1), (1, 1)$, and also $(3, 1)$.

In $T$, we have $(3, 1), (1, 1)$, and we also have $(3, 1)$.

It seems like we have the same kind of situations in both $S$ and $T$, except $S$ has another $a, b, c$ triplet. What makes T transitive but S not?

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Oops, don't know why I didn't see $(3,1)$ and $(1,2)$, guess I was thinking too hard. –  badjr Dec 6 '12 at 2:06

$S$ is not transitive as @Conan noted. $T$ is transitive,because you can easily check that for $(1,1)$ the only pair which satisfy the definition is $(3,1)$: $$(1,1),(3,1)\in T\Rightarrow (3,1)\in T$$ For other pairs the antecedent of the definition is always wrong so the whole definition is satisfied for $T$.

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deezy, in S, we have (3,1) and (1,2), but we do not have (3,2)

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