Given any orthonormal system in Hilbert space, can we have an orthornormal basis containing it Given a Hilbert space $H$ (not necessarily separable), if $S$ is any orthonormal system, which means for any $x,y\in S$, inner product $\langle x,y \rangle$ is zero if $x$ is not equal to $y$, then how can we find an orthonormal basis of $H$ containing $S$. It seems easy for finite dimensional case in linear algebra.
 A: If you are talking about the orthonormal basis as a maximal orthonormal system, then the answer is yes.
We proceed using Zorn's lemma:
Let $\mathcal{F}$ be the family of all orthonormal subsets of $H$ which contain $S$.
$(\mathcal{F}, \subseteq)$ is a partially ordered set. $\mathcal{F}$ is nonempty, namely $S \in \mathcal{F}$.
Take a chain $\mathcal{L} \subseteq \mathcal{F}$. We have to show that $\mathcal{L}$ has an upper bound in $\mathcal{F}$.
Consider $$M = \bigcup_{E \in \mathcal{L}} E$$
$M$ is obviously an upper bound for $\mathcal{L}$. Certainly $S \subseteq M$. Furthermore, $M$ is orthonormal:
Take $x, y \in M$. There exist $E_x, E_y \in \mathcal{L}$ such that $x \in E_x$ and $y \in E_y$. Since $\mathcal{L}$ is a chain, the sets $E_x$ and  $E_y$ are comparable. Assume $E_x \subseteq E_y$. Then $x, y \in E_y$ so $\langle x, y \rangle = \delta_{x,y}$ because $E_y$ is orthonormal. Therefore, $M \in \mathcal{F}$.
Zorn's lemma implies there exist a maximal element $E$ in $\mathcal{F}$. Hence $E$ is the orthonormal basis you seek.
It can be shown that every element $x \in H$ can be written in the form$$x = \sum_{e \in E} \langle x, e\rangle e$$

On the other hand, if you want an orthonormal Hamel basis containing $S$ (i.e. a maximal linearly independent system, which also happens to be orthonormal), then the answer is no if $H$ is infinite-dimensional.
In fact, an infinite-dimensional Hilbert space cannot possess an orthonormal Hamel basis at all:
If $B$ is an orthonormal Hamel basis for $H$, then let $(e_n)_{n=1}^\infty$ be a sequence in $B$.
The vector 
$$x = \sum_{n=1}^\infty \frac{1}n e_n$$
is in $H$, and cannot be represented as a finite linear combination of elements of $B$. Thus, $B$ cannot be a Hamel basis.
