Countable Complete Orthonormal Set implies countable dense subet

Question

Let $\mathcal H$ be a Hilbert Space, let $B = \{u_j\}_{j=1}^{\infty}$ be a countable orthonormal basis. So we know that if a set is a complete orthonormal basis, the set of all finite linear combinations is dense in $\mathcal H$. Define the set

$$S := \{\sum_{j=1}^{n} (q_j+ir_j)u_j \, : \, n \in \mathbb{N}, q_j,r_j \in \mathbb{Q} \},$$ which is countable since it can be injected into $\mathbb{Q} \times \mathbb{Q} \times \mathbb{N}$. Now, since the set of all finite linear combinations is dense let $x \in \mathcal H$, we have

$$\|x-\sum_{j=1}^{m(x)} \langle x,u_j \rangle u_j \| < \frac{\epsilon}{2}$$ for some $m(x)$ depending only on $x$. Note then that (this is where I'm unsure if this is logically sound) $$\|\sum_{j=1}^{m(x)} \langle x,u_j \rangle u_j\| \leq m(x)|\langle x,u'_j \rangle|$$ where $|\langle x,u'_j \rangle| := \max \limits_{j=1,\ldots, m(x)} \{|\langle x,u_j \rangle| \}$. Ergo, given $x \in \mathcal H$, if we choose $q_j,r_j$ such that $$|\langle x,u_j \rangle - q_j+ir_j| \leq \epsilon (2m(x)|\langle x,u'_j \rangle|)^{-1},$$ we find that $$\|x-\sum_{j=1}^{m(x)} (q_j+ir_j)u_j\| < \epsilon$$ by the triangle inequality.

Does this suffice? Is there a logical issue with choosing epsilon in this manner?

Answer 1

You are confused:

Let $H$ be a Hilbert Space, let $B=\{u_j\}_{j=1}^\infty$ be a countable orthonormal basis. So we know that if a set is a complete orthonormal basis, the set of all finite linear combinations is dense in $H$.

It is true.

Now, since the set of all finite linear combinations is dense let $x\in H$, we have

$\|x−∑_{j=1}^{m(x)}⟨x,uj⟩uj\|<\epsilon/2...(*)$

No. It doesn't mean that the set of all finite linear combination is dense. It is a consequence of countable orthonormality of $B$. The fact that all the finite linear combanations is dense in $H$ is that for each $\epsilon>0$, there are $n(x)\in\mathbb{N}$, scalars $a_1,...,a_{n(x)}$ and elements $u_{j_1},...,u_{j_{n(x)}}\in B$ such that $\|x-\sum_{k=1}^{n(x)}a_ku_{j_k}\|<\epsilon$.

Please, note the differences between tha last expression and $(*)$ (principal difference is that could it be $a_k\neq\langle x,u_{j_k}$).

Now, since $\epsilon$ is fixed, then the $a_k$ and the $u_{j_k}$ are fixed. Let us approximate each scalar $a_i$ with a rational scalar:

Let $\epsilon>0$. Then, there are $n(x)\in\mathbb{N}$, scalars $a_1,...,a_{n(x)}$ and elements $u_{j_1},...,u_{j_{n(x)}}\in B$ such that $\|x-\sum_{k=1}^{n(x)}a_ku_{j_k}\|<\frac{\epsilon}{2}$.

Now, for $\frac{\epsilon}{2n(x)}$ and fixed $k$, there are a complex $q_k$ with rational real and complex parts such that $|a_k-q_k|<\frac{\epsilon}{2n(x)}$

Then

$\begin{eqnarray} \|x-\sum_{k=1}^{n(x)}q_ku_{j_k}\|&=&\|x-\sum_{k=1}^{n(x)}(q_k-a_k+a_k)u_{j_k}\|\\ &=&\|x-\sum_{k=1}^{n(x)}a_ku_{j_k}+\sum_{k=1}^{n(x)}(a_k-q_k)u_{j_k}\|\\ &\le&\|x-\sum_{k=1}^{n(x)}a_ku_{j_k}\|+\|\sum_{k=1}^{n(x)}(a_k-q_k)u_{j_k}\|\\ &<&\dfrac{\epsilon}{2}+\sum_{k=1}^{n(x)}|a_k-q_k|\\ &<&\dfrac{\epsilon}{2}+\dfrac{n(x)\epsilon}{2n(x)}\\ &=&\epsilon \end{eqnarray}$

The desired result follows.