Show that if $(e_n)$ is an orthonormal set in a Hilbert space $H$, the set of all vectors of the form $x=\sum c_ne_n$ is a subspace of $H$.
Hint: Take a Cauchy sequences $(x_r)$, where $x_r=\sum c_{rn}e_n$. Set $C_r=(c_{r1},c_{r2},...)$ and show that $(C_r)$ is a Cauchy sequence in $l_2$. I'm not sure how to proceed. Any hints or solutions are greatly appreciated.
Well, I guess we can assume that our orthonormal set is infinite. If it's finite then it's pretty easy. Let's start with $$ |C_n-C_m| = \left(\sum_{i=1}^\infty |c_{n,i}-c_{m,i}|^2\right)^{1/2}=\left(\sum_{i=1}^\infty |c_{n,i}e_i-c_{m,i}e_i|^2\right)^{1/2}=|x_n-x_m| $$ where the last equality follows from rearranging an absolutely convergent sequence.
We assumed $x_i$ is a Cauchy sequence, so $C_i$ is a cauchy sequence as well. But completeness of $\ell_2$ tells us $C_i \to Y$ for $Y=(y_1,y_2,...)$. Then we want to see that
$$ x_i \to \sum_{i=1}^\infty y_ie_i :=y $$
This is clear as $$ |C_n-Y| = \left(\sum_{i=1}^\infty |c_{n,i}-y_i|^2\right)^{1/2}=\left(\sum_{i=1}^\infty |c_{n,i}e_i-y_ie_i|^2\right)^{1/2}=|x_i-y| $$ again with a rearrangement. So we're done
You can see how the isomorphism @Omnomnomnom mentioned works in this proof. We use the isomorphism move into $\ell^2$ to use its completeness. Then we pullback to see that it worked in the original space.