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Let $(R,\cal T)$ be a unital Hausdorff compact topological ring and let $A$ be an open subset of $R$ containing $1$. Is there a finite set $B$ with $AB=R$?

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    $\begingroup$ If $A$ is an open proper ideal, there is no such $B$, finite or infinite. $\endgroup$ Jan 20, 2016 at 15:51
  • $\begingroup$ Yes. I edited the question. $\endgroup$ Jan 21, 2016 at 4:10
  • $\begingroup$ I might be missing something but doesn´t it follow immediately from compactness? $\endgroup$ Jan 25, 2016 at 13:12
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    $\begingroup$ @RamirodelaVega: How? Notice that even if $A$ is open, $Ab$ is not necessarily open (e.g., when $b=0$). $\endgroup$ Jan 25, 2016 at 13:33
  • $\begingroup$ @EmilJeřábek: Of course! I was under the (obviously wrong) impresion that $x \mapsto xb$ was a homeomorphism... as if it was a group operation. $\endgroup$ Jan 25, 2016 at 18:47

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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.

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  • $\begingroup$ This seems correct, but maybe a little on the terse side? To match with the $A$ of the OP, let's denote the valuation as $v: \mathbb{Z}_p \to \mathbb{N} \cup \{\infty\}$ and put $A = v^{-1}(0)$. Then if $B \subset \mathbb{Z}_p$ is finite, we have $v(B A) = v(\bigcup_{b \in B} bA) = \bigcup_{b \in B} v(b) + v(A) = \{v(b): b \in B\}$, which is finite, so $BA$ can't possibly be all of $\mathbb{Z}_p$. $\endgroup$
    – Todd Trimble
    Sep 11, 2016 at 0:21
  • $\begingroup$ @Todd it's a counterexample, there is no much to be said after giving it... $\endgroup$
    – Uri Bader
    Sep 11, 2016 at 7:20
  • $\begingroup$ Yet, I considerably expanded it. $\endgroup$
    – Uri Bader
    Sep 11, 2016 at 7:36

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