Commutativity in a Unital Banach Algebra

Question

Let $ A $ be a unital Banach algebra and $ S $ a non-empty subset of $ A $. The centralizer of $ S $ is defined as $$ Z(S) \stackrel{\text{def}}{=} \{ a \in A ~|~ \text{$ as = sa $ for all $ s \in S $} \}. $$ I need to show that if the elements of $ S $ commute with each other, then the elements of $ Z(Z(S)) $ commute with each other.

I know that $ Z(Z(S)) $ is a closed sub-algebra of $ A $ and that $ {\sigma_{A}}(a) = {\sigma_{Z(S)}}(a) $ for any $ a \in Z(S) $. Also, $ S \subseteq Z(Z(S)) $. But I do not see how the commutativity part follows.

Can anybody please give me a hint?

Answer 1

This has nothing to do with banach algebras. It holds in every magma, i.e. set equipped with a binary operation. Also, $S \neq \emptyset$ is a useless and unnatural condition, it should be left out.

Observe that $S \subseteq T$ implies $Z(T) \subseteq Z(S)$ $(\star)$. Since $S$ is commutative, we have $S \subseteq Z(S)$. Applying $(\star)$ this we get $Z(Z(S)) \subseteq Z(S)$. Applying $(\star)$ once again, we get $Z(Z(S)) \subseteq Z(Z(Z(S)))$. But this exactly means that $Z(Z(S))$ is commutative.

Edit: Oh, Haskell and I posted the same proof, simultaneously!

Answer 2

Fact: If $ S $ and $ T $ are subsets of $ A $, then $ S \subseteq T $ implies $ Z(T) \subseteq Z(S) $.

If the elements of $ S $ commute with each other, then $ S \subseteq Z(S) $. Applying the fact above, we obtain $ Z(Z(S)) \subseteq Z(S) $, which then implies that $ Z(Z(S)) \subseteq Z(Z(Z(S))) $. Therefore, the elements of $ Z(Z(S)) $ commute with each other.