Let $A$ be a locally convex algebra, or even just a topological algebra, and let $U_1,U_2\in A$ be open, is the product $$ U_1\cdot U_2=\left\{ a\cdot b\mid a\in U_{ 1} ,b\in U_{ 2} \right\} $$

open? I assume of course continuity of the multiplication, but is this enough? Or is there a counter example. I am particularly interested in the case, where $A$ is locally convex, however I think even then it is not true in general, because even if we chose an element $x\in A$ that deviates little from $ab\in U_1U_2$, for example $p(x-ab)<\varepsilon$ for chosen seminorm $p$ and $\varepsilon>0$, there is no guarantee, that $x$ can be written as a product. Would it work, if we assumed,that every element can at least be written as a sum of products and ask again, wether the linear span of $U_1\cdot U_2$ is continuous?

Thanks for your time.

  • $\begingroup$ Isn't this the same as $\bigcup\limits_{a\in U_1}aU_2$ ? $\endgroup$ – MPW Mar 15 at 21:24
  • $\begingroup$ @MPW, question is, is map $x \to a \cdot x$ open if $a$ is a non-invertible element? For invertible $a$ it's a homeomorphism (like in a group), but here we need not have a group, I think. $\endgroup$ – Henno Brandsma Mar 15 at 22:25
  • $\begingroup$ Yes thank you. The multiplications could all be zero for example. $\endgroup$ – Mahdimatika Mar 19 at 19:58

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