Magma Subset Product with Self

Theorem
Let $\left({S, \circ}\right)$ be a magma.

Let $T \subseteq S$.

Then $\left({T, \circ}\right)$ is a magma $T \circ T \subseteq T$, where $T \circ T$ is the subset product of $T$ with itself.

Proof
By definition:
 * $T \circ T = \left\{{x = a \circ b: a, b \in T}\right\}$

Necessary Condition
Let $\left({T, \circ}\right)$ be a magma.

Then $T$ is closed.

That is:
 * $\forall x, y \in T: x \circ y \in T$

Thus:
 * $x \circ y \in T \circ T \implies x \circ y \in T$

Sufficient Condition
Let $T \circ T \subseteq T$.

Then:
 * $x \circ y \in T \circ T \implies x \circ y \in T$

That is, $T$ is closed.

Therefore $\left({T, \circ}\right)$ is a magma by definition.