Hilbert $C^*$ Modules, dense submodules

Question

I read a paper in which the author uses the propetry that if $A$ is a submodule of a

Hilbert $C^\ast$ module ($C$ is a $C^*$ algebra) such that

$A^\bot=0$ then $A$ is dense. I don't know how to prove it since in a gneral Hilbert module over a $C^*$ algebra $(A^\bot)^\bot$ is greater than $Closure(A)$ and the standard Hilbert space proof does not work.

The paper in question is: UNBOUNDED OPERATORS ON HILBERT C∗-MODULES OVER C∗-ALGEBRAS OF COMPACT OPERATORS by Guljas you can find on the web.

Answer 1

Unless I am misunderstanding your question, I don't believe the statement is true in general. For example, take $C = C(X)$ for some $X$. Take your $C^*$-module to be $C$ itself with the standard inner product $\langle f, g \rangle = f^*g$. Let $x_0 \in X$ and define $$ A = \{ f \in C \mid f(x_0) =0 \}. $$ Then $A^\perp = 0$ but $A$ is not dense in $C$.