Binary Operation on Subset is Binary Operation

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Theorem

Let $S$ be a set.

Let $\circ$ be a binary operation on $S$.


Let $T \subseteq S$.

Let $\circ {\restriction}_T$ be the restriction of $\circ$ to $T$.


Then $\circ {\restriction}_T$ is a binary operation on $T$.


Proof

Let $\Bbb U$ be a universal set.

Let $\circ: S \times S \to \Bbb U$ be a binary operation on $S$.

Let $T \subseteq S$.

Let $\left({a, b}\right) \in T \times T$.

By definition of ordered pair and cartesian product:

$a \in T$ and $b \in T$

As $T \subseteq S$, it follows that:

$a \in S$ and $b \in S$

Thus:

$\left({a, b}\right) \in S \times S$

As $\circ$ is a binary operation on $S$, it follows that:

$\circ \left({a, b}\right) \in \Bbb U$

But by definition of restriction of $\circ$:

$\circ \left({a, b}\right) = \circ {\restriction}_T \left({a, b}\right)$

Thus:

$\circ {\restriction}_T \left({a, b}\right) \in \Bbb U$

As $a$ and $b$ are arbitrary elements of $T$, it follows that this holds for all $a, b \in T$.

Hence the result by definition of binary operation.

$\blacksquare$