$A$ is a finite alphabet. $A^*$ is the set of finite words or the free monoid generated by A.
$A^w$ is the set of infinite words generated by A. Denote $A^\infty=A^*\bigcup A^w$.
$X$ is a set of $A^\infty$. Denote $X_{*}=X\bigcap A^*$, $X_w=X\bigcap A^w$.
A binoid over $A$ is a subset $M$ of $A^\infty$ such that $M_* M\subseteq M,\quad (M_*)^w\subseteq M.$ The product of two elements is just using concatenation to put them together.
The notation and definition above quote from Algebraic Combinatorics on Words (2002), which is written by M.Lothaire.
It seems that if M is a binoid, $M_*$ must be a subsemigroup of $A^*$. But the book also mentioned that such $M_*$ is also a submonoid of $A^*$.
My question is how to ensure the empty word or so-called unit belongs to $M_*$, such that $M_*$ is not only a subsemigroup, but also a submonoid of $A^*$.
Thanks in advance.