Subspace of certain series in a Hilbert space is compact Let $E$ be a Hilbert space and let $\{x_{n}\}$ be an orthonormal basis.  Let $\{c_{n}\}$ be a sequence of positive numbers such that $\sum c_{n}^{2}$ converges.  Let $C$ be the subset of $E$ consisting of all sums $\sum a_{n}x_{n}$ where $\lvert a_{n}\rvert \leq c_{n}$.  Show that $C$ is compact.
My thought was to try showing every sequence had an accumulation point, but it seemed like this quickly became a matter of showing every open cover had a finite subcover.  Also, the underlying set seems to be just $\prod [0, c_{n}]$ which would be compact if it was given the product topology, but I don't believe that is the proper topology.  I'm not really sure where else to go.

You may view $E$ as a subset of $l^{2}$ through the isometry correspondence $e\rightarrow \{(e,x_{l})\}_{l=1}^{\infty}$. Alternatively, it is convenience to view $e\in E$ as a function from the positive integers into $\mathbb{C}$ such that $\int |e(t)|^{2}dK(t) < \infty$, where $K$ is the counting measure on the positive integers. This is set up for an application of the Lebesgue dominated convergence theorem because you have $c\in L^{2}$ such that $|e| \le c$ on the positive integers $\mathbb{N}^{+}$.
If $\{ e_{n} \}\subseteq L^{2}(\mathbb{N}^{+})$ is a sequence in $E$, then you may find a subsequence which converges in the first coordinate (i.e., $\lim_{k} e_{n_{k}}(1)$ exists.) This is because all the first coordinates are bounded by $c_{1}$. Then by choosing a subsequence of this sequence, you find a subsequence of $\{ e_{n}\}$ which converges in the first two coordinates. Continuing, and applying Cantor diagonalization, there is a subsequence $\{ e_{n_{k}}\}_{k=1}^{\infty}$ which converges pointwise everywhere on $\mathbb{N}^{+}$. Let $e$ be the function with $e(j)=\lim_{k}e_{n_{k}}(j)$. Clearly $e\in E$ and, by the Lebesgue dominated convergence Theorem,
$$
               \lim_{k}\|e-e_{n_{k}}\|_{L^{2}}=\lim_{k}\int_{\mathbb{N}^{+}}|e(j)-e_{n_{k}}(j)|^{2}\,dK(j) = 0.
$$
