Let $(R,\cal T)$ be a unital Hausdorff compact [topological ring](https://en.wikipedia.org/wiki/Topological_ring) and let $A$ be an open subset of $R$ containing $1$. Is there a finite set $B$ with $AB=R$?



Kind of multiplicative total boundedness in Hausdorff compact rings
Not in general.


Following your title let us say that a monoid $(M,\cdot,1)$ is totally bounded if for every identity nbd $A$ there exists a finite set $B$ st $AB=M$.
Observe that a (continuous) homomorphic image of a totally bounded monoid is totally bounded.


An example of a compact monoid which is not totally bounded is the monoid $(\{0,1,2,\ldots,\infty\},+,0)$, where $\infty$ is an absorbing point for addition and the topology is seen as the one point compactification of the discrete set $\mathbb{N}$. This is clear by taking $A=\{0\}$.


An example of a compact ring whose multiplicative monoid is not totally bounded
is the ring of $p$-adic integers (for a fixed prime $p$), as the valuation map
$$ (\mathbb{Z}\_p,\cdot,1) \to (\{0,1,2,\ldots,\infty\},+,0) $$
is a surjective continuous homomorphism of monoids and the image is not totally bounded.


