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 $C=C_{L}(\langle S\rangle )$.