Let $M$ be a von Neumann algebra over a Hilbert space $\mathcal H$ and $\xi\in \mathcal H$ be a cyclic-separating vector for $M$, that is, $[M\xi]=[M'\xi]=\mathcal H$, where $M'$ is the commutant of $M$. Now consider the subset $\mathcal K$ of $\mathcal H$ defined by
$$\mathcal K :=\left\{ \sum_{i=1}^n \eta_i \gamma_i: \eta_i,\gamma_i\in M\xi\right \},$$
that is, $\mathcal K$ is spanned linearly by all possible products $\eta\gamma$, $\eta,\gamma \in M \xi$. Let us define the product and involution on $M\xi$ by,
$$(x\xi)(y\xi)=xy\xi, ~~\text{and}~~ (x\xi)^*=x^*\xi, ~\forall x,y \in M$$
Consider the inner product in $M\xi$ inherited from $\mathcal H$. Now Wikipedia says that $M\xi$ forms a unital left Hilbert algebra. I am okay with other conditions but I am not able to solve the two things that:
$(1)$ $\mathcal K$ is dense in $M\xi$ with respect to the given inner product.
$(2)$ The map $\theta: M\xi \to M\xi$ by $\theta(x\xi)=x^*\xi$ is pre-closed.
One thing is that, $\mathcal K$ forms an ideal for $M\xi$ also $\xi$ is the identity of $M\xi$, but I am unable to figure out how to show the denseness. Please help me to solve this. Thank you for your valuable time and effort. Thank you.
Since $M$, being a von Neumann algebra, is unital, you have that $$ \mathcal K=\{x\xi:\ x\in M\}=M\xi. $$
A linear operator $T$ operator is pre-closed (closable is the usual modern term) if, among other equivalent conditions, given a sequence $\{x_n\}$ in its domain such that $x_n\to 0$ and $Tx_n\to y$, then $y=0$. So we need to assume that $x_n\xi\to0$ and that $x_n^*\xi\to y\xi$. Now we need to use that $\xi$ is both cyclic and separating. Let $z\in M'$. Then $$ \langle y\xi,z\xi\rangle=\lim_n\langle x_n^*\xi,z\xi\rangle =\lim_n\langle \xi,x_nz\xi\rangle =\lim_n\langle \xi,zx_n\xi\rangle =\lim_n\langle z^*\xi,x_n\xi\rangle\to0. $$ As $M'\xi$ is dense in $H$ (because $\xi$ is separating for $M$), this shows that $y\xi=0$ and hence $\theta$ is closable.