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Source of question : Trying to prove that if the Hilbert $H$ has an orthonormal basis, then it is separable.

Elaboration :

Let $H$ be a Hilbert space and ${e_n : n in mathbb N}$ an orthonormal basis of $H$.

Let $Y_n$ be the space such $forall n in mathbb N$ :

$$Y_n = bigg{sum_{i=1}^nq_ie_i : q_i in mathbb Qbigg}$$

I know that every such $Y_n$ is measurable, because $Y_n$ is (I don't know the word in English - apologies!) with $mathbb Q_n$, because every element of it is written as a linear combination of $q_1,dots,q_n$. Now, I know that if $Y = cup_{n=1}^infty Y_n$, then $Y$ is measurable as a union of measurable sets. Now, $Y$ is dense over $H$, but this is the part that I cannot show.

I know I need to show that $forall x in H$ and $varepsilon >0$, $y in Y$ it is $| x-y| < varepsilon$.

Since ${e_n}$ is an orthonormal basis of $H$, then every $x in H$ can be written as :
$$x = sum_{n=1}^infty langle x, e_nrangle e_n$$

Thus the desired norm becomes :

$$bigg|sum_{n=1}^infty langle x, e_nrangle e_n - ybigg|$$

How should I continue now to prove $|x-y| < varepsilon$ aka that $Y$ is dense over $H$ and complete my proof ?

I'm assuming that this is a real Hilbert space. Following the definition, $H$ is separable if it contains a countable dense subset.

You already found $Ysubset H$; there only remains to prove that it is countable and dense in $H$.

In order to prove the latter, you already noticed that any element $x$ of your Hilbert space may be written as a combination $x=∑_{i=1}^∞⟨x,e_i⟩e_i$. The other important observation is that Parseval's Identity holds, namely $$| x |^2=∑_{i=1}^∞|⟨x,e_i⟩|^2.$$

Since $| x |$ is a finite quantity, it follows that the tail $∑_{i=N}^∞|⟨x,e_i⟩|^2 to 0$ as $Nto infty$. You can then choose $N$ so that the tail is small, say $∑_{i=N}^∞|⟨x,e_i⟩|^2 < epsilon_1$ and approximate the first $N$ coordinates by $q_1, dots, q_N in mathbb Q$ the same way that you would in the proof that $mathbb Q^N$ is dense in $mathbb R^N$.

Write that approximation as $x' = ∑_{i=1}^N q_i e_i$. We know that $x'$ is an element of $ Y_N$, so that it is contained in the proposed subset $Y$. Suppose we choose $q_1, dots q_N$ in such a way that
$$left|∑_{i=1}^N q_i e_i - ∑_{i=1}^N ⟨x,e_i⟩e_iright| < epsilon_2.$$
Then the distance $|x - x'|$ between $x$ and $x'$ will be

$$ left|∑_{i=1}^infty ⟨x,e_i⟩e_i - ∑_{i=1}^N q_i e_iright| leq \ left|∑_{i=1}^N ⟨x,e_i⟩e_i - ∑_{i=1}^N q_i e_i right| + left|∑_{i=N}^infty ⟨x,e_i⟩right| < epsilon_2 + sqrt epsilon_1$$

(using the triangle inequality of the norm). As $epsilon_1$ and $epsilon_2$ are free to choose, given any $epsilon > 0$ you can do so in such a way that $sqrtepsilon_1 + epsilon_2 < epsilon$.

It seems that you know how to prove that $Y$ is countable, so I'll leave that to you.

Actually the correct word here is not 'measurable' but 'countable'! A separable space is one which has a countable dense subset. I assume here you have a real Hilbert space.

As for the proof, given $ epsilon > 0 $, and knowing that the series of squares of coefficients $ sum_{i=1}^{infty} |langle x, e_i rangle|^2 $ converges, there is an $ N $ such that $$ sum_{i=N+1}^{infty} |langle x, e_i rangle|^2 < frac{epsilon}{2} $$ Now we can choose $ y in Y_N $ such that $ || sum_{i=1}^{N} langle x, e_i rangle e_i - y || < frac{epsilon}{2} $. (Simply choose appropriate rational numbers close to the finitely many coefficients $ langle x, e_i rangle , i = 1, 2, cdots, N $). We claim that this $ y $ works because we have $$ || x - y || = || (sum_{i=1}^{N} langle x, e_i rangle e_i - y) + sum_{i=N+1}^{infty} langle x, e_i rangle e_i || le || (sum_{i=1}^{N} langle x, e_i rangle e_i - y) || + || sum_{i=N+1}^{infty} langle x, e_i rangle e_i || $$ $$ < frac{epsilon}{2} + sum_{i=N+1}^{infty} |langle x, e_i rangle|^2 < epsilon $$
This completes the proof.