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Let $\{ e_r \}_{r>0}$ an uncountable orthonormal system in a Hilbert space $H$, prove that for every $v \in H$ $\langle v, e_r \rangle \neq 0$ for at most countably many $r>0$.

If we assume $\{ e_r \}_{r>0}$ is complete then because we are in a Hilbert space it means that $\{ e_r \}_{r>0}$ is a basis and hence for every $v \in H$ we have $v = \sum_{n=0}^{\infty}a_ie_i$ hence $\langle e_r,v \rangle \neq 0$ only for $r \in \mathbb{N}$ indeed.

How to go about the other case? tried in many ways but cannot find a proof

user10364768
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    In the other case, we can always add vectors until it is complete, and then apply the argument you presented. – Alex Dec 17 '19 at 18:42
  • Thanks, not sure though, how do we know that only countably many additions to the system can form a basis? – user10364768 Dec 17 '19 at 19:26
  • Even if infinitely many additions are required, in the final result only finitely many will be non-zero, due to your argument. See the answer I posted below. – Alex Dec 17 '19 at 19:34

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The fact that $\{e_r\}$ is orthonormal means that we can complete it to a basis $\{e_r\}\cup\{f_s\}$. Then since we are in a Hilbert space, your reasoning implies that $\langle e_r,v \rangle \neq 0$ or $\langle f_s,v \rangle \neq 0$ only for a countable number of indices $r,s$. But, since $\{e_r\}\subseteq\{e_r\}\cup\{f_s\}$, then in particular the number of indices $r$ for which $\langle e_r,v \rangle \neq 0$ is also countable.

Alex
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