$H \odot K$ complete implies $\dim H$ or $\dim K <\infty$ Let $H,K$ be non-zero Hilbert spaces. Then the algebraic tensor product $H \odot K$ is an inner product space for the unique inner product determined by
$$\langle h \otimes k, h' \otimes k'\rangle = \langle h, h'\rangle_H \langle k, k' \rangle_K$$
I am trying to prove that $H \odot K$ is a Hilbert space (i.e. $H \odot K$ is complete for the norm induced by this inner product) implies that $\dim H <\infty$ or $\dim K < \infty$.
Let us prove the contrapositive, i.e. so suppose that $H$ and $K$ are both infinite dimensional. How can I show that there is a Cauchy sequence in $H \odot K$ that does not converge?
 A: It is very easy to exhibit a non-converging Cauchy sequence in $H\odot K$, but actually proving that it does not
converge is a bit more difficult.  The approach suggested by @Max, although somewhat  sophisticated, is perhaps the easiest
way to pin down all of the details.  Here is a pedestrian way to describe this method.
For each pair of vectors $(x, y)\in H\times K$, consider the bounded linear operator
$$
  T_{x, y}:H\to K
  $$
given by
$$
  T_{x, y}(z) =  \langle x,z\rangle y, \quad\forall z\in H.
  $$
(For simplicity I am assuming that we are working with real vector spaces but this can be fixed in the complex case by
seeing   $T_{x, y}$ as an operator defined on $\bar H$, namely the Hilbert space made out of $H$ by adopting the new scalar
multiplication operation $\lambda \cdot x:= \bar \lambda x$).
It is easy to see that the map
$$
  (x, y)\in H\times K\mapsto   T_{x, y}\in  \mathscr {B}(H,K)
  $$
is bilinear so, by the universal property of (algebraic) tensor products, there exists a unique linear map
$$
  T :H\odot K\to  \mathscr {B}(H,K),
  $$
such that $T (x\otimes y) = T_{x,y}$.
We next claim  that
$$
  \langle T (\xi )x,y\rangle  =  \langle \xi ,x\otimes y\rangle ,
  \tag {1}
  $$
for every $\xi $ in $H\odot K$, $x\in  H$, and $y\in  K$. To prove this write  $\xi =\sum_{i=1}^nx_i\otimes y_i$, and notice that
$$
  \langle T (\xi )x,y\rangle  =
  \sum_{i=1}^n   \langle T_{x_i,y_i}(x),y\rangle  = $$ $$ =
  \sum_{i=1}^n   \langle \langle x_i,x\rangle y_i,y\rangle  =
  \sum_{i=1}^n   \langle x_i,x\rangle \langle y_i,y\rangle  = $$ $$ =
  \sum_{i=1}^n   \langle x_i\otimes y_i,x\otimes y\rangle  =
  \langle \xi ,x\otimes y\rangle,
  $$
proving the claim.
Note that the range of each $T_{x,y}$  is the one-dimensional space spanned by $y$, so $T_{x,y}$ has rank 1.
Furthermore, every $\xi $ in $H\odot K$  may be writen as  $\xi =\sum_{i=1}^nx_i\otimes y_i$, so
$$
  T (\xi ) = \sum_{i=1}^n T_{x_i,y_i},
  $$
so we see that   $T (\xi )$ has rank at most $n$, hence finite, for every $\xi $ in $H\odot K$.
The final contradiction will be achieved by proving that, if $H$ and $K$ are infinite dimensional, and $H\odot K$ is  complete, there exists a vector $\xi $ in $H\odot K$  such that $T (\xi )$ has infinite rank.
Let us therefore assume from now on that $H$ and $K$ are both infinite dimensional, so we may find (necessarily infinite)
orthonormal bases
$\{e_i\}_{i\in  I}$, and $\{f_j\}_{j\in  J}$, for $H$ and $K$ respectively, and it is well known that $\{e_i\otimes f_j\}_{(i, j)\in  I\times J}$ is an
orthonormal basis for $H\otimes K$.
Assuming by contradiction that $H\odot K$ is complete,  observe that  $H\odot K$ coincides with its completion,  so   $H\odot K = H\otimes K$.
Choose infinite subsets $\{i_n:n\in {\mathbb N}\}\subseteq I$, and $\{j_n:n\in {\mathbb N}\}\subseteq J$, and let
$$
  \xi  = \sum_{n=1}^\infty  {1\over n^2} (e_{i_n}\otimes f_{j_n}).
  $$
This is well defined because the series is clearly absolutely convergent and we are working in a complete space!
Using (1) we have that
$$
  T (\xi )e_{i_n} =
  \sum_{j\in J} \langle T (\xi )e_{i_n}, f_j\rangle f_j =
  \sum_{j\in J} \langle \xi , e_{i_n}\otimes f_j\rangle f_j =
  \langle \xi , e_{i_n}\otimes f_{j_n}\rangle f_{j_n} =
  {1\over n^2} f_{j_n}.
  $$
This shows that ${1\over n^2} f_{j_n}$ lies in the range of $T (\xi )$, and hence that $T (\xi )$ has infinite rank,  a contradiction.
