Equivalence of Definitions of Transitive Closure (Relation Theory)

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Theorem

The following definitions of the concept of Transitive Closure in the context of Relation Theory are equivalent:

Smallest Transitive Superset

Let $\RR$ be a relation on a set $S$.

The transitive closure of $\RR$ is defined as the smallest transitive relation on $S$ which contains $\RR$ as a subset.

Intersection of Transitive Supersets

Let $\RR$ be a relation on a set $S$.

The transitive closure of $\RR$ is defined as the intersection of all transitive relations on $S$ which contain $\RR$.

Finite Chain

Let $\RR$ be a relation on a set or class $S$.

The transitive closure of $\RR$ is the relation $\RR^+$ defined as follows:

For $x, y \in S$, $x \mathrel {\RR^+} y$ if and only if for some $n \in \N_{>0}$ there exist $s_0, s_1, \dots, s_n \in S$ such that $s_0 = x$, $s_n = y$, and:

\(\ds s_0\)

\(\RR\)

\(\ds s_1\)

\(\ds s_1\)

\(\RR\)

\(\ds s_2\)

\(\ds \)

\(\vdots\)

\(\ds \)

\(\ds s_{n - 1}\)

\(\RR\)

\(\ds s_n\)

Union of Compositions

Let $\RR$ be a relation on a set $S$.

Let:

$\RR^n := \begin{cases}

\RR & : n = 1 \\ \RR^{n-1} \circ \RR & : n > 1 \end{cases}$

where $\circ$ denotes composition of relations.

Finally, let:

$\ds \RR^+ = \bigcup_{i \mathop = 1}^\infty \RR^i$

Then $\RR^+$ is called the transitive closure of $\RR$.

Proof

Let $\RR$ be a relation on a set $S$.

First note that by Smallest Element is Unique, once it has been shown that some relation, $\QQ$, is the smallest transitive superset of $\RR$, it is the only such.

Thus we need only prove that each of the other definitions lead to relations with this property.

First we have:

Intersection of Transitive Supersets is Smallest Transitive Superset

Let $\RR$ be a relation on a set $S$.

Then the intersection of all transitive relations on $S$ that contain $\RR$ is the smallest transitive relation on $S$ that contains $\RR$.

The Finite Chain Definition is Equivalent to the Union of Compositions Definition

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Follows from the definition of composition of relations.

$\blacksquare$

Union of Compositions is Smallest Transitive Superset

$\RR^+$ is Transitive

By Relation contains Composite with Self iff Transitive, we can prove that $\RR^+$ is transitive by proving the following:

$\RR^+ \circ \RR^+ \subseteq \RR^+$

Let $\tuple {a, c} \in \RR^+ \circ \RR^+$.

Then:

$\exists b \in S: \tuple {a, b} \in \RR^+, \tuple {b, c} \in \RR^+$

Thus:

$\exists n \in \N: \tuple {a, b} \in \RR^n$

$\exists m \in \N: \tuple {b, c} \in \RR^m$

From Composition of Relations is Associative:

$\RR^{n + m} = \RR^n \circ \RR^m$

so:

$\tuple {a, c} \in \RR^{n + m} \subseteq \RR^+$

Since this holds for all $\tuple {a, c} \in \RR^+ \circ \RR^+$:

$\RR^+ \circ \RR^+ \subseteq \RR^+$

Thus $\RR^+$ is transitive.

$\Box$

$\RR^+$ contains $\RR$

$\RR \subseteq \RR^+$ by Set is Subset of Union of Family.

$\RR^+$ is Smallest

Let $\RR'$ be a transitive relation on $S$ such that $\RR \subseteq \RR'$.

We must show that $\RR^+ \subseteq \RR'$.

Let $\tuple {a, b} \in \RR^+$.

That is:

$a \mathrel \RR b$

Then:

$\exists n \in \N: \tuple {a, b} \in \RR^n$

Thus by the definition of composition of relations, there exists $x_{n - 1} \in S$ such that:

$a \mathrel {\RR^{n - 1} } x_{n - 1} \land x_{n - 1} \mathrel \RR b$

Likewise there exists $x_{n - 2} \in S$ such that:

$a \mathrel {\RR^{n - 2} } x_{n - 2} \land x_{n - 2} \mathrel \RR x_{n - 1}$

And so forth there exist elements $x_0, \dots, x_n \in S$ such that:

$x_0 = a$

$x_n = b$

$\forall k \in \N_n: x_k \mathrel \RR x_{k + 1}$

Since $\RR \subseteq \RR'$:

$\forall k \in \N_n: x_k \mathrel {\RR'} x_{k + 1}$

Since $\RR'$ is transitive:

$a \mathrel {\RR'} b$

That is:

$\tuple {a, b} \in \RR'$

Since this holds for all $\tuple {a, b} \in \RR^+$:

$\RR^+ \subseteq \RR'$

Since this holds for all transitive relations $\RR'$ that contain $\RR$:

$\RR^+$ is the smallest transitive relation containing $\RR$.

$\blacksquare$

Finite Chain Definition gives Smallest Transitive Superset

Let $S$ be a set or class.

Let $\RR$ be a relation on $S$.

Let $\RR^+$ be the transitive closure of $\RR$ by the finite chain definition.

That is, for $x, y \in S$ let $x \mathrel {\RR^+} y$ if and only if for some natural number $n > 0$ there exist $s_0, s_1, \dots, s_n \in S$ such that $s_0 = x$, $s_n = y$, and:

$\forall k \in \N_n: s_k \mathrel \RR s_{k+1}$

Then $\RR^+$ is transitive and if $\QQ$ is a transitive relation on $S$ such that $\RR \subseteq \QQ$ then $\RR \subseteq \QQ$.

Also see

• Recursive Construction of Transitive Closure