TITLE: Tensor Product: Hilbert Spaces
QUESTION [6 upvotes]: This question has been modified...
Problem
Given Hilbert spaces.
In general, their algebraic tensor product isn't complete:
$$\mathcal{H}\hat{\otimes}\mathcal{K}=\mathcal{H}\otimes\mathcal{K}\iff\dim\mathcal{H}<\infty\lor\dim\mathcal{K}<\infty$$
How to prove this from scratch?
Attempt
Choose orthonormal bases:
$$\mathcal{S}\otimes\mathcal{T}:=\{\sigma\otimes\tau:\sigma\in\mathcal{S},\tau\in\mathcal{T}\}$$
One obtains some candidates:
$$\sigma_k\otimes\tau_l\in\mathcal{S}\otimes\mathcal{T}:\quad\sum_{kl=1}^\infty\frac{1}{kl}\sigma_k\otimes\tau_l\quad\sum_{k=1}^\infty\frac{1}{kl}\delta_{kl}\sigma_k\otimes\tau_l$$
However the former one drops out:
$$\sum_{kl=1}^\infty\frac{1}{kl}\sigma_k\otimes\tau_l=\left(\sum_{k=1}^\infty\frac{1}{k}\sigma_k\right)\otimes\left(\sum_{l=1}^\infty\frac{1}{l}\tau_l\right)$$
So it is not obvious at all wether the latter one works out!
Reference
Build-up on: Vector Spaces: Tensor Product

REPLY [3 votes]: Let us choose orthonormal bases $\{e_i: i \in I\}$ of $\mathcal H$ and $\{f_j : j\in J \}$ of $\mathcal K$.
Assume $\mathcal K$ is finite dimensional, thus $J$ is finite. Let $x\in \mathcal H \hat\otimes \mathcal K$ and write it as
$$
x=\sum_{j\in J} x_j \otimes f_j \quad \text{with}\quad x_j =\sum_{i\in I} (e_i\otimes e_j, x) \, e_i\,,
$$
where $(\,\cdot\,,\,\cdot\,)$ is the scalar product to be chosen linear in the second argument.
This shows that $x\in \mathcal H\otimes \mathcal K$.
Conversely, if both are infnite dimensional, we can identify a subset of $I$ and $J$ with $\mathbb N$.
Then 
$$y=\sum_{n\in \mathbb N} \frac1{n!}e_n\otimes f_n \in \mathcal H\hat\otimes \mathcal K.
$$
Show that $y\not \in \mathcal H\otimes \mathcal K$.