If a subset A of a Hilbert space H contains an Open ball, then the orthogonal complement of A is the set of zero element.

Question

Question: If A contains an open ball then $A^{\perp}=\left \{ 0 \right \}$

Let A $\subseteq$ a Hilbert space H contains an open ball of size $\epsilon >0.$

Suppose $A^{\perp}=\left \{ v \in A : \left \langle v,a \right \rangle=0 ,\forall a \in A, \forall v \neq 0 \right \}$

Hints will be appreciated.

Thanks in advance.

Answer 1

Hint (1): $A^{\perp}= \textbf{Span}(A)^{\perp}$ and $\textbf{Span}(A)^{\perp} = \{0\} \iff \overline{(\textbf{Span}(A))}= H$.

Hint (2):. Suppose $A^{\perp} \neq \{0\}$. There exists $v \neq 0 \in A^{\perp}$. There exists $B(A_0, \epsilon) \subset A$ by hypothesis. Then $A_0 + \frac{\epsilon}{2}\frac{v}{||v||} \in B(A_0, \epsilon)$. And $\langle A_0 + \frac{\epsilon}{2}\frac{v}{||v||}, v \rangle = \frac{\epsilon}{2}||v|| = \frac{\epsilon}{2}||v|| \neq 0$, therefore yielding a contradiction. The intuitive fact is the following: an open set possesses all the dimensions.