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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?

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1 Answer 1

up vote 1 down vote accepted

Hint: the orthogonal of $A$ is the same thing as the orthogonal of the vector sub-space generated by $A$ (use finite sums to see that). What about a vector sub-space in a normed space which has a non-empty interior?

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... then in equals the whole space, okay now its simple :) –  Stefan Nov 29 '12 at 18:11

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