I want to show that if $X$ is a pre-Hilbert space and $A$ is a subset of $X$ with an nonempty interior, then $A^{\perp} = \{ 0 \}$.
I tried to assume the contrary, then there would be an $x \ne 0$ from $A^{\perp}$ such that $ \langle x, a \rangle = 0 $ for some interior point of $A$. Because $a$ is an interior point, there would be an open ball $B_r(a)$ around $a$ with $B_r(a) \subseteq A$.
I know the following facts,
a) $A^{\perp}$ is closed
b) $A$, $A^{\perp}$ are linear, i.e. they are infinite
c) $A \cap A^{\perp} = \{ 0 \}$
d) $(A^{\perp})^{\perp} \subseteq \overline{A}$
e) the scalar product and the norm on $X$ are continous
but I am unable to employ them in any useful way?