Let $E$, $F$ be two complex Hilbert spaces and $\mathcal{L}(E)$ (resp. $\mathcal{L}(F)$) be the algebra of all bounded linear operators on $E$ (resp. $F$).


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 $E \otimes F$.


It is well known that $(E \otimes F,\langle\cdot,\cdot\rangle)$ is not a complete space. Let $E \widehat{\otimes} F$ be the completion of $E \otimes F$ under the inner product $\langle\cdot,\cdot\rangle$.


If $T\in \mathcal{L}(E)$ and $S\in \mathcal{L}(F)$, then the tensor product of $T$ and $S$ is denoted $T\otimes S$ and defined as
$$\big(T\otimes S\big)\bigg(\sum\_{k=1}^d x\_k\otimes y\_k\bigg)=\sum\_{k=1}^dTx\_k \otimes Sy\_k,\;\;\forall\,\sum\_{k=1}^d x\_k\otimes y\_k\in E \otimes F,$$
which lies in $\mathcal{L}(E \otimes F)$. The extension of $T\otimes S$ over the Hilbert space $E \widehat{\otimes} F$, denoted by $T \widehat{\otimes} S$, is the tensor product of $T$ and $S$ on the tensor product space, which lies in $\mathcal{L}(E\widehat{\otimes}F)$.


Let $\operatorname{Im} (X)$ and $\overline{\operatorname{Im} (X)}$ denote respectively the range of an operator $X$ and the closure of its range.



> 
> Let $T,M\in \mathcal{L}(E)$ and $S,N\in \mathcal{L}(F)$ be such that
> 
> 
> * $\operatorname{Im} (T)\subseteq \overline{\operatorname{Im} (M)}$.
> * $\operatorname{Im} (S)\subseteq\overline{\operatorname{Im} (N)}$.
> 
> 
> I want to prove that
>  $$\operatorname{Im}(T \widehat{\otimes} S)\subseteq \overline{\operatorname{Im}(M \widehat{\otimes} N)}.$$
> 
> 
> 


Note that I show that
$$\overline{\operatorname{Im} (M)}\otimes\overline{\operatorname{Im} (N)}\subseteq\overline{\operatorname{Im}(M \otimes N)}.$$



The tensor product of two bounded operators
I think this is easy. It follows from the simple fact that if $E\_0$ is a dense linear subspace of $E$ then $T(E\_0)$ is dense in $T(E)$. Thus Im$(T)\otimes$ Im$(S) =$ Im$(T \otimes S) = T\hat{\otimes} S(E\otimes F)$ is dense in Im$(T\hat{\otimes} S)$, and it is easy to see that Im$(T)\otimes$ Im$(S)$ is contained in the closure of Im$(M\otimes N)$.


