Question: How do I show Unital simple $C^*$-algebra with tracial topological rank $0$ has stable rank $1$?
(Showing finiteness would also be enough)
Def. $C^*$-algebra $A$ is said to have property (SP) if for every nonzero positive element $a\in A$, $\mathbf{closure}[aAa]$ contains at least one nonzero projection.
Def. Unital simple $C^*$-algebra $A$ is said to have tracial topological rank $0$ if it has property (SP) and for every finite subset $\mathcal F\subset A$ and $\epsilon>0$ and projection $p\in A$ there is a finite dimensional subalgebra $B$ with $1_B=q$ such that:
$\begin{align}&\text{(1) for every $x\in\mathcal F$, $\|qx-xq\|<\epsilon$};\\&\text{(2) for every $x\in \mathcal F$, there is $y\in B$ such that $\|qxq-y\|<\epsilon$;}\\&\text{(3) there is $v\in A$ such that $v^*v=1-q$ and $vv^*\leq p$.}\end{align}$
Def. $C^*$-algebra $A$ is said to have stable rank $1$ if invertible elements are dense in $A$. $A$ is said to have real rank $0$ if invertible self-adjoint elements are dense in self-adjoint elements.
An Introduction to the Classification of Amenable $C^*$-algebras.
Theorem 3.6.11 Every unital simple $C^*$-algebra $A$ with tracial topological rank $0$ has stable rank $1$ and real rank $0$.
It is proved in Theorem 3.6.10 that every unital separable simple $C^*$-algebra with tracial topological rank less than $1$ has stable rank one. However, it only gives the proof for real rank in Theorem 3.6.11, and the stable rank part in non-separable case is discussed nowhere.
Separability is used to show finiteness of $A$. With finiteness we can show for every $x$ and $\epsilon>0$ there is some $y\in A$ ,non-zero positive $a\in A^+$, and unitary $u\in U(A)$ such that $\|x-y\|<\epsilon$ and $a(uy)=(uy)a=0$. This shows that we can assume $xa=ax=0$ for some non-zero positive $a\in A^+$.
To show $RealRank(A)=0$, separability isn't needed since we can simply find $f,g\in C(\mathbb R:\mathbb R)$ such that
$\begin{align}&(1)\|f(\lambda)-\lambda\|_\infty<\epsilon\\&(2)f(\lambda)=0\text{ when }\lambda\in (-\epsilon,\epsilon)\\ &(3)\text{supp}(g)\subseteq (-\epsilon,\epsilon)\end{align}$
and set $y=f(x)$, $a= g(x)$.
With $ax=xa=0$, we can show $x$ is approximated by invertible elements or by invertible self-adjoint elements and the proofs are similar. By property (SP), $a$ can be chosen to be projection. By simpleness, there are orthogonal equivalent projections $b,c\leq a$.
$(1-b)A(1-b)$ is also of tracial topological rank $0$, by proposition 3.6.5. Therefore if we let $\mathcal F=\{x\}$ and $p=c$, then there is a finite dimensional subalgebra $B\leq (1-b)A(1-b)$ with $1_B=q$ that satisfies the conditions.
$x\approx (1-b-q)x(1-b-q)\oplus qxq\in (1-q)A(1-q)\oplus B $ so it is sufficient to show that $(1-b-q)x(1-b-q)$ is approximated by invertibles in $(1-q)A(1-q)$ and $qxq$ is approximated by invertibles in $B$.
Since $B$ is finite dimensional, that $qxq$ is approximated by invertibles is obvious.
By condition(3), $[1-b-q]\leq [c]=[b]$, therefore there is $b'\leq b$ such that $[1-b-q]=[b']$. We know $(1-q)A(1-q)=(1-b-q)A(1-b-q)\oplus b'Ab'\oplus (b-b')A(b-b')\simeq M_2((1-b-q)A(1-b-q))\oplus (b-b')A(b-b')$
, therefore $(1-b-q)x(1-b-q)=\left(\begin{array}{}(1-b-q)x(1-b-q)&0\\0&0\end{array}\right)\oplus 0\approx \left(\begin{array}{}(1-b-q)x(1-b-q)&\epsilon\\\epsilon&0\end{array}\right)\oplus \epsilon$.
Of course separability doesn't follow from $TracialTopologicalRank(A)=0$. I have no idea how to deal with the non-separable case.