Equivalence of Definitions of Transitive Closure (Relation Theory)/Finite Chain is Smallest

Theorem
Let $S$ be a set or class.

Let $\mathcal R$ be a relation on $S$.

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

That is, for $x, y \in S$ let $x \mathrel{\mathcal R^+} y$ iff 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:


 * $\forall k \in \N_n: s_k \mathrel{\mathcal R} s_{k+1}$

Then $\mathcal R^+$ is transitive and if $\mathcal Q$ is a transitive relation on $S$ such that $\mathcal R \subseteq \mathcal Q$ then $\mathcal R \subseteq \mathcal Q$.

$\mathcal R^+$ is transitive
Let $x,y,z \in S$.

Let $x \mathrel{\mathcal R^+} y$ and $y \mathrel{\mathcal R^+} z$.

Then for some $m, n \in \N_{>0}$ there are $s_0, s_1, \dots, s_m$ and $t_0, t_1, \dots, t_n$ such that $s_0 = x$, $s_m = y$, $t_0 = y$, $t_n = z$, and the following hold:


 * $\forall k \in \N_m: s_k \mathrel{\mathcal R^+} s_{k+1}$
 * $\forall k \in \N_n: t_k \mathrel{\mathcal R^+} t_{k+1}$

Let $\left\langle {u_k} \right\rangle_{k \in \N_{m+n}}$ be defined thus:


 * $u_k = \cases {s_k & \text{if $k \le m$} \\

t_{k - m} & \text {if $k > m$}}$

Then clearly $u_k \mathrel{\mathcal R^+} u_{k+1}$ whenever $k < m$ and whenever $k > m$.

But $u_m = s_m = y = t_0 \mathrel{\mathcal R^+} t_1 = u_{m+1}$, so this holds also for $k = m$.

Furthermore, $u_0 = s_0 = x$ and $u_{m+n} = t_n = z$.

Therefore $x \mathrel{\mathcal R^+} z$.

As this holds for all such $x$ and $z$, $\mathcal R^+$ is transitive.

$\mathcal R^+$ is smallest
Let $\mathcal Q$ be any transitive relation on $S$ such that $\mathcal R \subseteq \mathcal Q$.

For any $x, y \in S$ such that $x \mathrel{\mathcal R^+} y$, let $d \left({x, y}\right)$ be the smallest natural number $n > 0$ such that there exist $s_0, s_1, \dots, s_n \in S$ such that $s_0 = x$, $s_n = y$, and:


 * $\forall k \in \N_n: s_k \mathrel{\mathcal R} s_{k+1}$

Such an $n$ always exists by the definition of $\mathcal R^+$ and the fact that $\N$ is well-ordered by $\le$.

We will show by induction on $n$ that for every $x, y$ such that $x \mathrel{\mathcal R^+}$ and $d \left({x, y}\right) = n$, $x \mathrel{\mathcal Q} y$. This will show that $\mathcal R^+ \subseteq \mathcal Q$.

If $d\left({x, y}\right) = 1$ then $x \mathrel{\mathcal R} y$, so $x \mathrel{\mathcal Q} y$.

Suppose that the result holds for $n$.

Let $d \left({x, y}\right) = n + 1$.

Then there exist $s_0, s_1, \dots, s_{n+1}$ such that $s_0 = x$, $s_{n+1} = y$, and:


 * $\forall k \in \N_{n+1}: s_k \mathrel{\mathcal R} s_{k+1}$

Then dropping the last term:


 * $\forall k \in \N_n: s_k \mathrel{\mathcal R} s_{k+1}$

so $x \mathrel{\mathcal R^+} s_n$.

It should be clear, then, that $d \left({x, s_n}\right) = n$.

Thus by the inductive hypothesis, $x \mathrel{\mathcal Q} s_n$.

Since $\mathcal R \subseteq \mathcal Q$, $s_n \mathrel{\mathcal Q} s_{n+1} = y$.

Since $x \mathrel{\mathcal Q} s_n$, $s_n \mathrel{\mathcal Q} y$, and $\mathcal Q$ is transitive:


 * $x \mathrel{\mathcal Q} y$

As this holds for all such $x$ and $y$, $\mathcal R^+ \subseteq \mathcal Q$.