positive elements in $\mathbb{M}_n(A)$ are unltraweakly dense in the positive part of $\mathbb{M}_n(A^{**})$

Question

I try to read the book of C*-algebra and finite- dimensional approximations. In the proof of Theorem 2.3.8, i can't understand the question that positive elements in $\mathbb{M}_n(A)$ are unltraweakly dense in the positive part of $\mathbb{M}_n(A^{**})$, where A is a C*-algebra. 1. we known ultraweak top. coincides with weak operator top. on bounded set, but how to understand this quesiton by using the kaplansky density theorem. 2. which topology is applicable for the kaplansky desity theorem besides strong operator top. and w operator top？ please help me.

Answer 1

It is true in general that $M_n (A)^+ $ is ultraweakly dense in $M_n(A'')^+$. The first thing is to notice that $M_n (A'')=M_n (A)''$. So we can simply work with an inclusion $A\subset A''$ with $A $ (wot, sot, ultra wot) dense.

Note first that, by using Kaplanski, the unit ball of $A $ is sot (and so, wot and ultrawot) dense in the unit ball of $A''$. This allows us to work with bounded nets. Next, $A^{\rm sa} $ is wot dense in $(A'')^{\rm sa} $. Indeed, if $x=x^*$ and $x_n\to x $ (wot), then $(x_n+x_n^*)/2\to x $ (wot).

Now choose $x\in (A'')^+$ with $\|x\|\leq1$. By the above, there is a net of selfadjoints $x_n $ with $x_n\to x $ (wot). If we consider the nets of positive and negative parts, $\{x_n^+\} $ and $\{x_n^-\} $ (where $x_n=x_n^+-x_n^-$ and $x_n^+x_n^-=0$), by wot-compactness of the unit ball we may replace them by convergent subnets. Write $y=\lim_{wot} x_n^+$ and $z=\lim_{wot } x_n^-$. Then $y^{1/2}=\lim_{sot}(x_n^+)^{1/2} $ and similarly for $z $. We also have $(x_n^+)^{1/2}(x_n^-)^{1/2}=0$. As the sot preserves products of bounded nets, we get $y^{1/2}z^{1/2}=0$; but then $yz=0$. As $x=y-z $ it follows that $z=x^-=0$. Thus $$x=y=\lim_{wot}x_n^+. $$