Centralizer of subset of Lie ring is Lie subring

Statement
Suppose $$L$$ is a Lie ring and $$S$$ is a subset of $$L$$. Then, the following two statements hold:


 * 1) The centralizer $$C_L(S)$$ of $$S$$ in $$L$$ is a subring of $$L$$.
 * 2) $$C_L(S) = C_L(\langle S \rangle)$$ where $$\langle S \rangle$$ denotes the Lie ring generated by $$S$$.

Proof
Given: A Lie ring $$L$$, a subset $$S$$. $$C = C_L(S)$$.

To prove: $$C$$ is a subring of $$L$$, and $$C = C_L(\langle S \rangle)$$.

Proof:


 * 1) $$C$$ is an additive subgroup: For $$x,y \in C$$ and $$s \in S$$, $$[x+y,s] = [x,s] + [y,s] = 0$$. Thus, $$x + y \in S$$. Similarly, $$[-x,s] = -[x,s] = 0$$ and $$[0,s] = 0$$. Thus, $$C$$ is an additive subgroup.
 * 2) $$C$$ is closed under the Lie bracket: For $$x,y \in C$$ and $$s \in S$$, the Jacobi identity gives $$[[x,y],s] + [[y,s],x] + [[s,x],y] = 0$$. The second term is zero because $$[y,s] = 0$$, and the third term is zero because $$[s,x] = -[x,s] = 0$$. Thus, $$[[x,y],s] = 0$$.
 * 3) $$C = C_L(\langle S \rangle)$$: Consider $$C_L(C)$$. This contains $$S$$, and is a Lie subring by applying the above result to $$C$$ in place of $$S$$. Thus, $$\langle S \rangle \subseteq C_L(C)$$. In particular, $$C \subseteq C_L(\langle S \rangle)$$. On the other hand, $$C_L(\langle S \rangle) \subseteq C_L(S) = C$$, so in fact $$C = C_L(\langle S \rangle)$$.