# Product of open maps open? (not the cartesian)

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.

• Isn't this the same as $\bigcup\limits_{a\in U_1}aU_2$ ? – MPW Mar 15 at 21:24
• @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. – Henno Brandsma Mar 15 at 22:25
• Yes thank you. The multiplications could all be zero for example. – Mahdimatika Mar 19 at 19:58