Semigroup without a weak basis

Question

Let $\mathfrak{S}=(S,*)$ be a semigroup and $X\subseteq S$. Consider the functional closure $\overline{X}$, i.e. minimal set containing $X$ and closed under $*$. Let's say that $X$ is a weak basis (in contrast with the basis that is defined for a stronger concept of the algebraic closure) if $\overline{X}=S$ and $\forall(x\in X):x\notin \overline{X-\{x\}}$. One may prove that finite semigroups, free monoids and groups contain weak bases. Does it hold for all semigroups? If not, are there some counterexamples?

Answer 1

As it was shown in the comments, the assertion actually isn't true for groups and the counterexample is $(\mathbb{Q},+)$. Indeed, assume $a,b$ are elements of the basis $X$ and $a<b$. Suppose $a=x/z$ and $b=y/z$ where $x,y,z\in\mathbb{Z}$ and $z>0$. We can write $1/z=x_1+...+x_n$ where $x_1,...,x_n\in X-\{b\}$. Then $a+(y-x)(x_1+...+x_n)=b$ that gives us a contradiction with the independence of $X$. Thus, $X$ contains just one element $q$. We have $q\neq0$ and thus $0\notin\overline{X}$.