Left and Right Operation are Closed for All Subsets

Theorem
Let $S$ be a set.

Let:
 * $\leftarrow$ be the left operation on $S$
 * $\rightarrow$ be the right operation on $S$.

That is:
 * $\forall x, y \in S: x \leftarrow y = x$
 * $\forall x, y \in S: x \rightarrow y = y$

Let $\mathcal P \left({S}\right)$ be the power set of $S$.

Then for all $T \in \mathcal P \left({S}\right)$, both $\leftarrow$ and $\rightarrow$ are closed on $T$.

Thus, for all $T \in \mathcal P \left({S}\right)$:
 * $\left({T, \leftarrow}\right)$ is a subsemigroup of $\left({S, \leftarrow}\right)$
 * $\left({T, \rightarrow}\right)$ is a subsemigroup of $\left({S, \rightarrow}\right)$.

Proof
From Element under Right Operation is Left Identity we have that $\left({S, \rightarrow}\right)$ is a semigroup, whatever the nature of $S$.

From Element under Left Operation is Right Identity we have that $\left({S, \leftarrow}\right)$ is a semigroup, whatever the nature of $S$.

Let $T \in \mathcal P \left({S}\right)$.

Then:
 * From Element under Right Operation is Left Identity, $\left({T, \rightarrow}\right)$ is a semigroup, and therefore a subsemigroup of $\left({S, \rightarrow}\right)$.
 * From Element under Left Operation is Right Identity, $\left({T, \leftarrow}\right)$ is a semigroup, and therefore a subsemigroup of $\left({S, \leftarrow}\right)$.

This applies whatever $S$ is and whatever the subset $T$ is.