Recently I've become interested in generating sets of the unit interval $[0,1]$ when considered as a semigroup equipped with the usual multiplication. More precisely we are looking for subsets $S\subseteq[0,1]$ such that the semigroup generated by $S$, that is, $$ \langle S\rangle_{\rm SG}:=\Big\{\prod_{j=1}^nx_j:n\in\mathbb N, x_1,\ldots,x_n\in S\Big\} $$ equals $[0,1]$, possibly after taking the closure. First, a few observations:
Observation 1. Any (exact or approximative) generating set has to be infinite.
Indeed if $S\subseteq[0,1]$ is finite, then there exists $\varepsilon>0$ such that $S\cap(1-\varepsilon,1)=\emptyset$ (else $1$ would be an accumulation point of $S\setminus\{1\}$ which means $S$ would be infinite). But $[0,1-\varepsilon)$ is a subsemigroup of $([0,1],\cdot)$ so the open set $(1-\varepsilon,1)$ will be missing from $\langle S\rangle_{\rm SG}$; not even taking the closure can rectify this situation.
Observation 2. Any (exact) generating set has to be uncountably infinite.
Because $\langle S\rangle_{\rm SG}=\bigcup_{n\in\mathbb N}\{\prod_{j=1}^nx_j:x_1,\ldots,x_n\in S\}$, if $S$ is countable then the generated semigroup is the union of countably many countable sets, hence countable itself --- so it can never be equal to $[0,1]$.
Observation 3. For all $\varepsilon\in(0,1)$ one has $\langle \{0\}\cup[1-\varepsilon,1]\rangle_{\rm SG}=[0,1]$. In other words there exists an exact generating set of $([0,1],\cdot)$ of arbitrarily small measure.
Obviously $(1-\varepsilon)^n\to 0$ as $n\to\infty$, and all the points between $(1-\varepsilon)^n$ and $1$ can be written as the product of at most $n$ elements from $[1-\varepsilon,1]$: The latter is path-connected and multiplication is continuous so the ($n$-fold) product of this interval with itself has to be path-connected, as well. Therefore $\langle [1-\varepsilon,1]\rangle_{\rm SG}=(0,1]$ and adding $\{0\}$ yields the whole interval $[0,1]$.
So far, so straightforward. Now let us shift our focus towards approximate generating sets. Complementing the second observation one finds:
Observation 4. There exist countable approximate generating sets of $([0,1],\cdot)\,$.
This is a no-brainer: just intersect $[0,1]$ with $\mathbb Q$ (which is itself a semigroup) and take the closure to get back $[0,1]$.
In my opinion the more interesting question here is: how "small" can an approximate generating set become? "Small" in quotation marks because neither the Lebesgue measure nor the concept of countability can detect differences in sizes here; either one has to take a different measure to make this precise, or one understands "small" with respect to the partial order $(\mathcal P([0,1]),\subseteq)$ or something similar(?)
Anyway---recalling the first observation---$1$ has to be an accumulation point of any approximate generating set (after taking out said $1$). Thus my first idea was to consider $$ S_2:=\Big\{1-\frac1{2^n}:n\in\mathbb N\Big\}=\Big\{\frac12,\frac34,\frac78,\frac{15}{16},\ldots\Big\} $$ and to ask whether $\overline{\langle S_2\rangle_{\rm SG}}=[0,1]$.
In order to get a feeling for this problem I wanted to see how, e.g., $x=\frac13$ is approximated by $\{\prod_{j=1}^nx_j:x_1,\ldots,x_n\in S_2\}$ as $n$ grows. What I tried was to find a sequence of nested intervals with endpoints in $\langle S_2\rangle_{\rm SG}$ which converges to $\frac13$ (and possibly discover some structure in this construction which can be generalized to any $x\in[0,1]$).
Starting with the smallest element of $S_2$, obviously, $\frac13\in[0,\frac12]$. For the next step, as there is no other fraction with denominator $2$ in the interior of this interval we go to the next "available" denominator (in this case: $4$). And, indeed, $\frac14=\frac12\cdot\frac12\in\langle S_2\rangle_{\rm SG}$ which yields $\frac13\in[\frac14,\frac12]$. Repeating this process, there are no fractions with denumerator $4$ "left" in this interval we go from $4$ to $8$; we find $\frac38=\frac12\cdot\frac34\in(\frac14,\frac12)$. This yields $\frac13\in[\frac14,\frac38]$. The next intervals are: $$ \Big[\frac{9}{32},\frac38\Big],\Big[\frac{9}{32},\frac{45}{128}\Big],\Big[\frac{21}{64},\frac{45}{128}\Big],\Big[\frac{21}{64},\frac{45}{128}\Big],\Big[\frac{21}{64},\frac{343}{1024}\Big],\Big[\frac{675}{2048},\frac{343}{1024}\Big],\ldots $$
Note how the denominator $16$ was skipped because the only possible candidate, $\frac{7}{16}\in[\frac14,\frac38]$ cannot be written as a finite product of numbers of the form $1-\frac1{2^n}$ --- same for the denominators $64$, $256$, $512$ (and probably many more). Another thing I noticed was that the procedure does not really "care" about the number $\frac13$ or any of its properties, other than its location inside $[0,1]$. Instead the main question in this procedure appears to be the following:
Given $x,y\in\langle S_2\rangle_{\rm SG}$, $x<y$ does there exist $z\in\langle S_2\rangle_{\rm SG}$ such that $x<z<y$?
Looking at the above example I believe this to be true but so far I did not see any pattern to exploit here. Moreover I felt like there was nothing special about the numbers I used to define $S_2$, hence why I'd even go so far as to conjecture the following:
Conjecture. Let $(a_n)_{n\in\mathbb N}\subseteq\mathbb R_+$ be strictly monotonically increasing such that $a_n\to 1$ as $n\to\infty$. Then $\overline{\langle \{a_n:n\in\mathbb N\}\rangle_{\rm SG}}=[0,1]$.
Any comments and ideas are welcome!
