Characterization of the subsemigroup generated by a subset

Let S be a semigroup and let A \subseteq S be a nonempty subset. Recall that the subsemigroup of S generated by A is the set (A) = \{\prod a_i\ |\ a:n \rightarrow A\ \mathrm{for\ some}\ n \in \mathbb{N}^+\}. (Since S is associative, the \prod notation is unambiguous provided we insist that \prod_{i=1}^n a_i = a_1a_2 \cdots a_n.) Let B_A(S) denote the set of all subsemigroups of S which contain A. (Note that I_A(S) is nonempty since it contains S.) Prove that (A) = \bigcap B_A(S).

Note that if T \subseteq S is a subsemigroup containing A, then in particular we have \prod a_i \in T for all a: n \rightarrow A and n \in \mathbb{N}^+. (We can see this by using induction on n.) So (A) \subseteq T, and thus (A) \subseteq \bigcap B_A(S).

Note also that (A) is itself a subsemigroup of S, by generalized associativity. That is, if \prod_{i=1}^n a_i, \prod_{j=1}^m b_j \in (A), then (\prod a_i)(\prod b_i) = \prod_{k=1}^{n+m} c_i where c_k = a_k for 1 \leq k \leq n and c_k = b_{k-n} for n+1 \leq k \leq n+m. So \bigcap B_A(S) \subseteq (A).

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