# Algebraic Tensor product of Hilbert spaces [duplicate]

Let $E$, $F$ be complex Hilbert spaces. The algebraic tensor product of $E$ and $F$ is given by $$E \otimes F:=\left\{\xi=\sum_{i=1}^dv_i\otimes w_i:\;d\in \mathbb{N},\;\;v_i\in E,\;\;w_i\in F \right\}.$$

In $E \otimes F$, we define $$\langle \xi,\eta\rangle=\sum_{i=1}^n\sum_{j=1}^m \langle x_i,z_j\rangle_1\langle y_i ,t_j\rangle_2,$$ for $\xi=\displaystyle\sum_{i=1}^nx_i\otimes y_i\in E \otimes F$ and $\eta=\displaystyle\sum_{j=1}^mz_j\otimes w_j\in E \otimes F$.

The above sesquilinear form is an inner product in $E \otimes F$.

Why $(E \otimes F,\langle\cdot,\cdot\rangle)$ is not a complete space?

Thank you!!

## marked as duplicate by Dietrich Burde, Misha Lavrov, man and laptop, Shailesh, user99914 Jan 20 '18 at 1:22

Essentially, the problem is that you can't have a tensor with infinite $x \otimes y$ terms, but a succession can increase and increase its number of terms preserving convergence.
Formally, choose an orthonormal basis $\{v_i\}, \{w_j\}$, respectively of $E, F$. Consider the succession given by $z_n= \sum_{1 \le k \le n } \frac{1}{2^i} v_i \otimes w_i$. You can check that $|| z_m - z_n || < 1/2^n$ so that it is a cauchy sequence, but there is no limit point.
Infact, every element $t$ of the tensor space has the property to be contained in $E_0 \otimes F_0$ with $E_0,F_0$ of finite dimension. So there exist $x \in E$ such that for every $y \in F$ $\langle t, x \otimes y\rangle = 0$. A virtual limit point $z$ of $z_n$ has not this property. Whenever you choose $x=\sum a_i v_i$ with $a_r \neq 0$, then $\langle v \otimes w_r, z \rangle = \lim \langle v \otimes w_r, z_n \rangle = a_r/2^r \neq 0$.