How does having Fourier Expansion apply in the proof of separability of a Hilbert space?

Question

A Hilbert space is separable if and only if it has a countable orthonormal basis.

Proof. Assume that $\{ e_j\}$ is an orthonormal basis for the Hilbert space $H$. By Theorem (i), it is easy to show that the countable subset of elements $x$ with $\langle x,e_j \rangle$ in the set of rationals for all $j$ is dense. Thus, $H$ is separable.

Theorem (i). Every element $x$ of a Hilbert space $H$ with an orthonormal basis $\{ e_j\}$ can be expressed in terms of the Fourier expansion

$$x=\sum\limits_{j=1}^\infty \langle x,e_j \rangle e_j.$$

Question: I don't understand the sentence "By Theorem (i), it is easy to show that the countable subset of elements $x$ with $\langle x,e_j \rangle$ in the set of rationals for all $j$ is dense."

Answer 1

Let $A$ be the countable subset of elements $x$ with $ \langle x,e_j \rangle$ in the set of rationals for all $j$, i.e. $$A= \{ x \in H : \langle x , e_j \rangle \in \mathbb{Q} \quad \forall j \in \mathbb{N} \}$$ Let $y \in H$ and let $\epsilon > 0$ be given. Then by Theorem (i), $y$ can be written as $y = \sum_{j=1}^{\infty} \langle y , e_j \rangle$. Now for each $j \in \mathbb{N}$ choose a rational number $q_j \in \mathbb{Q}$ s.t. $| \langle y , e_j \rangle - q_j | < \frac{\epsilon}{2^{j}}$ which is guaranteed by the density of $\mathbb{Q}$ in $\mathbb{R}$. Define $x \in A$ as $$x : = \sum_{j=1}^{\infty} q_j e_j$$ Now observe that $$\| y-x \| = \| \sum_{j=1}^{\infty} (\langle y , e_j \rangle - \langle x , e_j \rangle ) e_j \| \leq \sum_{j=1}^{\infty} |\langle y , e_j \rangle - q_j | \| e_j \| \leq \sum_{j=1}^{\infty} \frac{\epsilon}{2^j} = \epsilon$$ Thus $A$ is dense in $H$.