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My teacher asked me to build an associative topological Hausdorff compact ring $R$ with $1$, which is multiplicatively unbounded. That means there is a neighborhood $U\ni 1$ such that $FU\not=R$ for each finite subset $F$ of $R$.

I am somewhat stuck, because I have a small stock of topological rings, and I see only two main ways to build such an example: to endow a compact topological group with a multiplication or to endow a ring with a compact ring topology. Both of these ways require a concordance of many conditions and therefore it seems to me that my success of the construction “depends on luck, but not on method”.

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Hi: Could I ask a question about the definition of multiplicatively unbounded? As written, it looks like it is of the form $\exists U\forall F(FU\neq R)$. Is this correct? – rschwieb Apr 17 '13 at 13:54
Yes, it is correct. – Alex Ravsky Apr 17 '13 at 14:29
I would have expected an "unbounded" condition to be of the form $\forall\dots\exists\dots$... but in any case thanks for verifying :) – rschwieb Apr 17 '13 at 15:20
Is it always possible to compactify a topological ring into a compact topological ring? And then maybe unboundedness would be inherited by the compactification? This is what I'd hope for if I already had a multiplicatively unbounded ring. – rschwieb Apr 17 '13 at 15:23
In fact, my teacher already showed that each associative topological Hausdorff compact ring $R$ with $1$ is multiplicatively bounded, but I didn't write his answer, because here was no more attention to this question and I have no permission of my teacher to publish his result. :-) I expect to publish this result here later. – Alex Ravsky Apr 29 '13 at 17:11

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