Give an example of Euclidean space. In the question it is asking what will be if we take out one condition of theorem. 
$\textbf{Theorem}$: Let {$\phi_{n}$} be orthonormal system in a complete Euclidean Space R. Then {$\phi_{n}$} is complete if and only if  R contains no nonzero element orthogonal to all elements of { $\phi_{n}$}.   
$\textbf{Question}$: Give an example of Euclidean Space $R$ and orthonormal system {$\phi_{n}$} in $R$ such that R contains no nonzero element orthogonal to every $\phi_{n}$, even though {$\phi_{n}$} fails to be complete. 
So, if we can give such example, by above theorem, $R$ can not be complete. 
$\textbf{Definition 1}$:{$\phi_{n}$} is complete if a linear combinations of elements of {$\phi_{n}$} are everywhere dense in $R$
$\textbf{Definition 2}$ : Euclidean space $R$ if $R$ is linear space with scalar product.
To be honest, I couldn't find any example myself.  I always find this quite challenging when you drop one condition of theorem, it is obviously doesn't hold. Otherwise it wouldn't be theorem. Thanks in advance.

Take separable Hilbert space $H$ with basis $e_1, ..., e_n, ...$  Now take the subspace generated (algebraically) by $e_2,e_3, ..., e_n, ...$ and the vector $e_1 + 1/2 e_2 + ... + 1/n e_n + ...$. This is a Euclidean space. The system $e_2, ... , e_n, ...$ is maximal orthonormal, but it is not complete, because the subspace itself is dense in $H$ and $e_2, ..., e_n,...$ is not a complete system in $H$. 
