Takesaki volume II chapter VII lemma 1.15 Consider the following fragments from Takesaki's second volume "Theory of operator algebras" in chapter VII on weight theory, lemma 1.15. Here $\mathcal{M}$ is a von Neumann algebra and $\mathscr{S}$ is the unit ball of $\mathscr{M}$.

Why exactly can we approximate a point in the strong closure of $E^*p\cap r\mathscr{S}$ by a sequence (as opposed to a net)? I imagine it has something to do with metrizability of $p\mathscr{M}p$? Another explanation is that $\mathscr{M}_p$ is a typo and it should be $\mathscr{M}p \cap \mathscr{S}$ that should be metrizable? But in that case I don't see why this must be metrizable.
Any help/clarification in the matter is highly appreciated!
 A: Given the comment of @Andromeda to my first answer, let me provide the following alternative proof, which only uses results that are proven earlier in the Takesaki books.
Let $a_n$ be a bounded sequence in $\mathscr{M}p$ such that $d(a_n,a) \to 0$. We have to prove that $a_n \to a$ strongly. Write $b_n = a_n - a$. Since $\omega(b_n^* b_n) \to 0$, it follows from Proposition 5.3 in Chapter III of Volume I of Takesaki that the sequence $|b_n|$ in $p \mathscr{M} p$ converges to $0$ strongly. Let $b_n = v_n |b_n|$ be the polar decomposition of $b_n$ in $\mathscr{M}$. Then also $b_n \to 0$ strongly. Thus $a_n \to a$ strongly.
A: Yes, the metric $d$ in Takesaki's proof metrizes the $\sigma$-strong topology on $\mathscr{M}p \cap \mathscr{S}$. Here $\mathscr{S}$ denotes the unit ball of $\mathscr{M}$ and it thus suffices to prove that if $a_n,a \in \mathscr{M}p \cap \mathscr{S}$ such that $d(a_n,a) \to 0$, then $a_n \to a$ strongly in the standard representation of $\mathscr{M}$.
By definition, $p$ is the support of the normal state $\omega$. Put $q = 1-p$ and choose a faithful normal semifinite weight $\varphi$ on $q\mathscr{M} q$. Then define the faithful normal semifinite weight $\psi$ on $\mathscr{M}$ by $\psi(x) = \omega(pxp) + \varphi(qxq)$. Let $L^2(\mathscr{M},\psi)$ be the associated GNS Hilbert space, which is our model for a standard representation of $\mathscr{M}$. For every $x \in \mathscr{M}$ with $\psi(x^* x) < +\infty$, denote by $\widehat{x}$ the corresponding vector in $L^2(\mathscr{M},\psi)$. Denote by $\mathscr{M}_{an}$ the $*$-algebra of analytic elements w.r.t. the modular automorphism group $\sigma^\psi$.
Since $d(a_n,a) \to 0$, we have $a_n \widehat{p} \to a \widehat{p}$. For every $b \in \mathscr{M}_{an}$, we have that
$$a_n \widehat{p b} = a_n \, J\sigma^\psi_{i/2}(b)^* J \, \widehat{p} =  J\sigma^\psi_{i/2}(b)^* J \, a_n \, \widehat{p}  \to J\sigma^\psi_{i/2}(b)^* J \, a \, \widehat{p} = a \, \widehat{p b} \; .$$
Since $\{\widehat{p b} \mid b \in \mathscr{M}_{an}\}$, is a dense subspace of $p L^2(M,\psi)$ and since $a_n$ is a bounded sequence, it follows that $a_n \xi = a_n 
p \xi \to a p \xi = a \xi$ for all $\xi \in L^2(M,\psi)$. Thus, $a_n \to a$ strongly.
