Hilbert-Modules and Cohens factorization

Question

Let $B$ be a $C^*$-algebra and $E$ a right Hilbert-$B$-Module. If I take an approximate unit $(e_i)_i$ in $B$ then for all $x \in E$ we have $x\cdot e_i \rightarrow x$ in $E$ by a simple calculation. I think by Cohens factorization Theorem we should have $E \cdot B:=\{x \cdot b | x \in E, b \in B\}=E$. But on the Wikipedia page it only says that $E \cdot B$ is dense. Now I am a little worried that I got that wrong. I would be grateful if somebody could clarify that for me. Thank you

Answer 1

Theorem: Let B be a $C^∗$-algebra, X a right Banach B-module. Then XB is a closed submodule of X. This is Theorem II.5.3.7 in Blackadar, Operator Algebras: Theory of C*-Algebras and Von Neumann Algebras, Volume 13, page 90, (it refers to 4.1 in Pedersen, Gert K. Factorization in $C^∗$-algebras. Exposition. Math. 16 (1998), no. 2, 145–156. which I cannot access)It is also theorem 4.6.4 in $C^*$-algebras and Finite-dimensional Approximations By Nathanial Patrick Brown, Narutaka Ozawa.