Closure for Finite Collection of Relations and Operations

Theorem
Let $\RR_1, \RR_2, \ldots \RR_n$ be relations.

Let $\SS_1, \SS_2, \ldots \SS_m$ be operations.

Let $T$ be a small class.

Let the image of $\RR_i$ over any small class $x$ be small classes for $1 \le i \le n$.

Let the image of $\SS_i$ over any Cartesian product $x \times x$ be small classes for $1 \le i \le m$.

Then there exists a small class $X$ such that:


 * $(1): \quad$ $T \subseteq X$
 * $(2): \quad$ Each $\RR_i$ is closed $X$.  Each $\SS_i$ is closed  $X \times X$.
 * $(3): \quad$ $X$ is the smallest small class satisfying conditions $(1)$ and $(2)$ above.

If $Y$ satisfies conditions $(1)$ and $(2)$, then $X \subseteq Y$.

Proof
Let $R \sqbrk x$ denote the image of $x$ under $R$.

Set:
 * $\ds \map G x = x \cup \bigcup_{i \mathop = 1}^n \RR_i \sqbrk x \cup \bigcup_{i \mathop = 1}^m \SS_i \sqbrk x$

Using the Principle of Recursive Definition, construct the function $F$ as follows:


 * $\map F 0 = T$
 * $\map F {n + 1} = \map G {\map F n}$

Define $X$ as follows:


 * $\ds X = \bigcup_{n \mathop \in \omega} \map F n$

Proof of Closure
Take any $\RR_i$.

Suppose $x \mathrel {\RR_i} y$ and $x \in \map F n$ for some $n$.

Then $y \in \map F {n + 1}$ by definition of image.

Thus $y \in X$ and $R_i$ is closed $X$.

Similarly, take any $\SS_i$.

Suppose $\paren {x \mathrel {S_i} y} = z$ and that $x \in X$ and $y \in X$.

It follows that $x, y \in \map F n$ for some $n$.

So:
 * $z \in \map F {n + 1}$

Therefore:
 * $z \in X$

Proof that $X$ is the Smallest Relational Closure
Suppose $T \subseteq Y$ and that $R_i$ and $S_i$ are closed $Y$.

$\map F n \subseteq Y$ shall be proven by induction:

For all $n \in \omega$, let $\map P n$ be the proposition:
 * $\map F n \subseteq Y$

Basis for the Induction
$\map P 0$ is the case:
 * $\map F 0 \subseteq Y$

which has been proved above.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\map F k \subseteq Y$

Then we need to show:
 * $\map F {k + 1} \subseteq Y$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \omega: \map F n \subseteq R$

Hence by Indexed Union Subset:
 * $T \subseteq Y$