Notation 2.1. Let A be a $C^*$ -algebra and ${\cal F}\subset A$ be a subset. Let $\epsilon>0$ . Set $a,b\in A$ and write $a\approx _{\epsilon }b$ if $\lVert a-b\rVert < \epsilon $ . We write $a\in _\varepsilon {\cal F}$ if there is $x\in {\cal F}$ such that $a\approx _\varepsilon x$ .

Notation 2.2. Let A be a $C^*$ -algebra and let $S\subset A$ be a subset of A. Denote by $\mathrm {Her}_A(S)$ (or just $\mathrm {Her}(S)$ , when A is clear) the hereditary $C^*$ -subalgebra of A generated by S. Denote by $A^{\textbf{1}}$ the closed unit ball of A, by $A_+$ the set of all positive elements in A, by $A_+^{\textbf{1}}:=A_+\cap A^{\textbf{1}}$ and by $A_{\mathrm {sa}}$ the set of all self-adjoint elements in A. Denote by $\widetilde A$ (or $A^{\sim }$ ) the minimal unitisation of A. When A is unital, denote by $GL(A)$ the set of invertible elements of A and by $U(A)$ the unitary group of A.

Notation 2.3. Let $\epsilon>0$ . Define a continuous function $f_{\epsilon }: {{[0,+\infty )}} \rightarrow [0,1]$ by

$$ \begin{align*} f_{\epsilon}(t) := \begin{cases} 0, &t\in {{[0,\epsilon/2]}},\\ 1, &t\in [\epsilon,\infty),\\ \mathrm{linear}, &{t\in[\epsilon/2, \epsilon].} \end{cases} \end{align*} $$

Proof. The ‘if’ part follows easily by taking ${\cal G}=\{\iota \}$ , where $\iota (t)=t$ for all $t\in [0,1]$ .

We now show the ‘only if’ part.

Suppose that A is e. tracially in ${\cal P}$ . Let $\varepsilon>0$ and let ${\cal F}\subset A$ be a finite subset, and without loss of generality we may assume that ${\cal F}\subset A^{\textbf{1}}$ . Moreover, without loss of generality (omitting an error within $\varepsilon /16$ , say), we may further assume that there is $e_A\in A_+^{\textbf{1}}$ such that

(e3.1) $$ \begin{align} e_Ax=x=xe_A \mathrm{\ for\ all\ } x\in {\cal F}. \end{align} $$

Set $a\in A_+\setminus \{0\}$ , let $\varepsilon>0$ and let ${\cal G}=\{g_1,g_2,\dotsc ,g_n\}\subset C_0([0,1])$ be a finite subset.

By the Weierstrass theorem, there are $m\in \mathbb {N}$ and polynomials $p_i(t)=\sum _{k=1}^m \beta _k^{\left (i\right )}t^k$ such that

(e3.2) $$ \begin{align} \lvert p_i(t)-g_i(t)\rvert<\varepsilon/4 \mathrm{\ for\ all\ }t\in[0,1] \mathrm{\ and\ all\ } i\in\{1,2,\dotsc,n\}. \end{align} $$

Let $M=1+\max \left \{\left \lvert \beta _k^{\left (i\right )}\right \rvert :i=1,2,\dotsc ,n, \ k=1,2,\dotsc ,m\right \}$ and $\delta := \frac {\varepsilon }{32m^3M}$ .

Now, since A is e. tracially in ${\cal P}$ , there exist an element $e\in A_+^{\textbf{1}}$ and a nonzero $C^*$ -subalgebra $B\subset A$ such that B in $\mathcal {P}$ , and the following hold:

1. (1) $\ \lVert ex-xe\rVert <\delta \text { for all } x\in {\cal F}\cup \{e_A\}$ .

2. (2^′) $(1-e)x\in _{\delta } B$ and $\lVert (1-e)x\rVert \ge \lVert x\rVert -\delta $ for all $x\in {\cal F}\cup \{e_A\}$ .

3. (3) $\ e\lesssim a$ .

It remains to show that $g_i(1-e)x\in _{\varepsilon /2} B$ for all $x\in {\cal F}$ , $i=1,2,\dotsc ,n$ .

Claim: For all $x\in {\cal F}$ and all $k\in \{1,2,\dotsc ,m\}$ , we have $(1-e)^kx\in _{\frac {\varepsilon }{16mM}} B$ . In fact,

(e3.3) $$ \begin{align} {{(1-e)^kx\overset{\text{(e3.1)}}{=}(1-e)^ke_A^{k-1}x \overset{(1)}{\approx_{k^2\delta}}\overbrace{(1-e)e_A(1-e)e_A\dotsm (1-e)e_A}^{k-1}(1-e)x \overset{(2')}{\in_{k\delta}} B.}} \end{align} $$

Note that $2k^2\delta \le 2m^2\delta <\varepsilon /16mM$ . The claim follows.

By formula (e3.2) and the claim, for $x\in {\cal F}$ and $i\in \{1,2,\dotsc ,n\}$ we have

(e3.4) $$ \begin{align} g_i(1-e)x\approx_{\varepsilon/4}p_i(1-e)x =\sum_{k=1}^m{\beta}_k^{\left(i\right)} (1-e)^k x\in_{\varepsilon/4} B.\nonumber \\[-26pt] \end{align} $$

Proof. Assume ${\cal P}$ has property (H) and A is e. tracially in ${\cal P}$ . Let $B\subset A$ be a nonzero hereditary $C^*$ -subalgebra of A. Set ${\cal F}\subset B$ and $s\in B_+\setminus \{0\}$ , and $\varepsilon \in (0,1/4)$ .

Without loss of generality, we may assume that ${\cal F}\subset B_+^{\textbf{1}}$ . Let $d\in B_+^{\textbf{1}}$ be such that $dx\approx _{{\varepsilon /32}} x\approx _{{\varepsilon /32}} xd$ and $x\approx _{{{\varepsilon /32}}}dxd$ for all $x\in {\cal F}$ .

Put $\varepsilon _1=\varepsilon /32$ . By [Reference Elliott, Gong, Lin and Niu15, Lemma 3.3], there is $\delta _1\in (0,{{\varepsilon _1}})$ such that for any $C^*$ -algebra E and any $x,y\in E_+^{\textbf{1}}$ , if $x\approx _{\delta _1} y$ , then there is an injective homomorphism $\psi : \mathrm {Her}_{E}\left (f_{{{\varepsilon _1}}/2}(x)\right )\to \mathrm {Her}_{E}(y)$ satisfying $z\approx _{{\varepsilon _1}} \psi (z)$ for all $z\in \mathrm {Her}_{E}\left (f_{{{\varepsilon _1}}/2}(x)\right )^{\textbf{1}}$ .

Note that there is $\delta _2\in (0,\delta _1)$ such that for any $C^*$ -algebra E and any $x,y\in E_+^{\textbf{1}}$ , if $xy\approx _{\delta _2} yx$ , then $x^{1/4}y\approx _{\delta _1/2} yx^{1/4}$ , $x^{1/8}y^{1/2}\approx _{\delta _1/2} y^{1/2}x^{1/8}$ and $x^{1/8}y\approx _{\delta _1/2 }yx^{1/8}$ .

Let $\delta =\delta _2/2$ . Let ${\cal G}=\left \{t,t^{1/4},t^{1/8}\right \}\subset C_0([0,1])$ . Since A is e. tracially in ${\cal P}$ , by Proposition 3.2 there exist a positive element $a\in A_+^{\textbf{1}}$ and a nonzero $C^*$ -subalgebra $C\subset A$ which is in ${\cal P}$ such that

1. (1) $\lVert ax-xa\rVert <{\delta }$ for all $x\in {\cal F}\cup \left \{d,d^{1/2},d^2\right \}$ ,

2. (2) $g(1-a) x\in _{\delta } C$ for all $g\in {\cal G}$ and $\lVert (1-a)x\rVert \ge \lVert x\rVert -\delta $ for all $x\in {\cal F}\cup \left \{d,d^{1/2},d^2\right \}$ and

3. (3) $a\lesssim s$ .

By (2), there is $c\in C$ such that $ c\approx _{\delta _1/2} (1-a)^{1/4}d$ . By (1) and the choice of $\delta _2$ , we have $c\approx _{\delta _1}d^{1/2}(1-a)^{1/4}d^{1/2}$ . Then by [Reference Elliott, Gong, Lin and Niu15, Lemma 3.3] and the choice of $\delta _1$ , there is a monomorphism

$$ \begin{align*} \varphi: \mathrm{Her}_A\left(f_{{{\varepsilon_1}}/2}(c)\right)\to \mathrm{Her}_A\left(d^{1/2}(1-a)^{1/4}d^{1/2}\right)\subset B \end{align*} $$

satisfying $\lVert \varphi (x)-x\rVert <{{\varepsilon _1}}$ for all $x\in \mathrm {Her}_C\!\left(f_{{{\varepsilon _1}}/2}(c)\right )^{\textbf{1}}$ . Define $D:=\varphi \left (\mathrm {Her}_C\!\left(f_{{{\varepsilon _1}}/2}(c)\right )\right )\subset B$ . Since C is in ${\cal P}$ and ${\cal P}$ has property (H), $D\cong \mathrm {Her}_C\!\left(f_{\varepsilon _1/2}(c)\right )$ is in ${\cal P}$ . Set $b:=dad\in B_+^{\textbf{1}}$ . Then by (1) and the choice of d, we have

(e3.5) $$ \begin{align} \lVert bx-xb\rVert=\lVert dadx-xdad\rVert \approx_{4{{\varepsilon_1}}} \lVert adxd-dxda\rVert \approx_{2{{\varepsilon_1}}} \lVert ax-xa\rVert<\delta \mathrm{\ for\ all\ }x\in{\cal F}. \end{align} $$

By (2), for any $x\in {\cal F}$ there is $\bar x\in C$ such that $(1-a)^{1/4}x(1-a)^{1/4}\approx _{2{{\varepsilon _1}}} \bar x$ . Then

(e3.6) $$ \begin{align} \begin{split} (1-b)x &= (1-dad)x \approx_{3{{\varepsilon_1}}} (1-a)dxd\\ &\approx_{4{{\varepsilon_1}}} (1-a)^{1/8}d(1-a)^{1/8} \cdot(1-a)^{1/4}x(1-a)^{1/4}\cdot (1-a)^{1/8}d(1-a)^{1/8}\\ &\approx_{4{{\varepsilon_1}}} c\bar xc \approx_{2{{\varepsilon_1}}} (c-{{\varepsilon_1}})_+\bar x (c-{{\varepsilon_1}})_+\\ &\approx_{{{\varepsilon_1}}} \varphi((c-{{\varepsilon_1}})_+\bar x (c-{{\varepsilon_1}})_+)\in D. \end{split} \end{align} $$

In other words,

(e3.7) $$ \begin{align} (1-b)x\in_{\varepsilon} D. \end{align} $$

Therefore, for all $x\in {\cal F}$ ,

(e3.8) $$ \begin{align} \begin{split} \lVert(1-b)x\rVert&=\lVert(1-dad)x\rVert \ge \left\lVert\left(1-a d^2\right)x\right\rVert-\delta \\ &\ge \lVert(1-a)x\rVert-3\varepsilon_1 \ge \lVert x\rVert-\delta-3\varepsilon_1\ge \lVert x\rVert-\varepsilon. \end{split} \end{align} $$

By (3), we have $b=dad\lesssim _A s$ . Note that $b, s\in B$ . Since B is a hereditary $C^*$ -subalgebra, we have $b\lesssim _B s$ . By formulas (e3.5) and (e3.7), we see that B is also e. tracially in ${\cal P}$ .

Notation 4.1. Let ${\cal W}$ be the class of $C^*$ -algebras A such that $W(A)$ is almost unperforated.

Let ${\mathscr Z}$ be the Jiang–Su algebra [Reference Jiang and Su27]. A $C^*$ -algebra A is called ${\mathscr Z}$ -stable if $A\otimes {\mathscr Z}\cong A$ . Let ${\cal C}_{\mathscr Z}$ be the class of separable ${\mathscr Z}$ -stable $C^*$ -algebras.

Proof. Without loss of generality, one may assume that $a,b,c\in A_+^{\textbf{1}}\backslash \{0\}$ and $\epsilon <1/2$ . Then $(n+1)\langle a\rangle \leq n\langle b\rangle $ implies that there exists $r=\sum _{i,j=1}^{n+1}r_{i,j}\otimes e_{i,j}\in A\otimes M_{n+1}$ such that

(e4.1) $$ \begin{align} a\otimes \sum_{i=1}^{n+1}e_{i,i} \approx_{\epsilon/128}r^*\left(b \otimes \sum_{i=1}^{{{n}}}e_{i,i}\right)r. \end{align} $$

Set ${\cal F}:=\{a,b\}\cup \left \{r_{i,j},r_{i,j}^*:i,j=1, 2 ,\dotsc , n+1\right \}$ and $M:=1+\lVert r\rVert $ . Let $\sigma =\frac {\varepsilon }{32M^2(n+1)^4}$ . Since A is e. tracially in ${\cal W}$ , by Proposition 3.2, for any $\delta \in \left (0, \frac {\varepsilon }{256M(n+1)^2}\right )$ , there exist $f\in A_+^{\textbf{1}}\setminus \{0\}$ and a $C^*$ -subalgebra $B\subset A$ which has almost unperforated $W(B)$ such that

1. (1^′) $\lVert fx-xf \rVert <\delta $ for $x\in {\cal F}$ ,

2. (2^′) $(1-f)^{ 1/4} x, (1-f)^{ 1/2} a(1-f)^{ 1/2 }, (1-f)^{1/4}x (1-f)^{ 1/4} \in _\delta B$ for all $x\in {\cal F}$ and

3. (3^′) $f\lesssim c$ .

Put $g=1-f$ . Let

$$ \begin{align*} {\cal G}:=\left\{g^{ {1/4}}x, g^{1/2}xg^{1/2}, g^{1/4}xg^{1/4} : x\in {\cal F}\right\}. \end{align*} $$

Set $x\in {\cal G}\cap A_+$ . By (2 $'$ ), there is $\bar x\in B$ such that $\lVert x-\bar x\rVert <\delta $ . Let $x':=\left (\bar x+\bar x^*\right )/2\in B_{\mathrm {sa}}$ . Then $x\approx _\delta x'$ . Then $x'+\delta \geq x\geq 0$ , which implies $\left \lVert x^{\prime }_-\right \rVert \leq \delta $ . Then $x\approx _\delta x'=x^{\prime }_+-x^{\prime }_-\approx _\delta x^{\prime }_+\in B_+$ . Therefore, there is a map ${\alpha }:{\cal G}\to B$ such that ${\alpha }({\cal G}\cap A_+)\subset B_+$ , and

(e4.2) $$ \begin{align} x \approx_{2\delta} {\alpha}( x ) \mathrm{\ for\ all\ }x\in{\cal G}. \end{align} $$

From (1 $'$ ) and (2 $'$ ), one can choose $\delta $ sufficiently small such that

(e4.3) $$ \begin{align} a\approx_{\varepsilon/16} g^{{1/2}}ag^{{1/2}}+ (1-g)^{1/2}a(1-g)^{1/2} \mathrm{\ and\ } \end{align} $$

(e4.4) $$ \begin{align} \left(g^{{1/2}}{{a}}g^{{1/2}}-\varepsilon/8\right)_+ \approx_{\varepsilon/16} \left(\alpha\left(g^{{1/2}}{{a}} g^{{1/2}}\right)-\epsilon/8\right)_+. \end{align} $$

By (1 $'$ ) and formula (e4.1) (with $\delta $ sufficiently small), one can also assume that

(e4.5) $$ \begin{align} g^{{1/2}}ag^{{1/2}}\otimes \sum_{i=1}^{n+1}e_{i,i} \approx_{\varepsilon/64} R^* \left(g^{{1/4}} b g^{{1/4}}\otimes \sum_{i=1}^{n}e_{i,i}\right)R, \end{align} $$

where $R:=\sum _{i,j=1}^{n+1}\left ({{g^{1/4}}}{{r_{i,j}}} \right )\otimes e_{i,j}$ . By formulas (e4.5) and (e4.2) and $\delta <\frac {\varepsilon }{256M(n+1)^2}$ , one has

(e4.6) $$ \begin{align} \alpha\left(g^{{1/2}}{{a}}g^{{1/2}}\right) \otimes \sum_{i=1}^{n+1}e_{i,i} \approx_{\varepsilon/32} {{\bar R^*}} \left( {\alpha}\left(g^{1/4}{{b}}g^{1/4}\right) \otimes \sum_{i=1}^{n}e_{i,i}\right){{\bar R}}, \end{align} $$

where $\bar R:= \sum _{i,j=1}^{n+1}{\alpha }\left (g^{1/4}{{r_{i,j}}} \right )\otimes e_{i,j}$ . Then by the choice of $\sigma $ ,

(e4.7) $$ \begin{align} \alpha\left(g^{1/2}{a}g^{1/2}\right) \otimes \sum_{i=1}^{n+1}e_{i,i} \approx_{\varepsilon/16} {{\bar R^*}} \left( \left({\alpha}\left(g^{1/4}{{b}}g^{1/4}\right)-\sigma\right)_+ \otimes \sum_{i=1}^{n}e_{i,i}\right){{\bar R}}. \end{align} $$

By formula (e4.7) and [Reference Rørdam40, Proposition 2.2], one has

(e4.8) $$ \begin{align} \left(\alpha\left(g^{{1/2}}{{a}}g^{{1/2}}\right) -\epsilon/8\right)_+\otimes \sum_{i=1}^{n+1}e_{i,i} \lesssim \left({\alpha}\left(g^{1/4} b g^{1/4}\right) -\sigma\right)_+\otimes \sum_{i=1}^{n}e_{i,i}. \end{align} $$

Since $W(B)$ is almost unperforated, one obtains

(e4.9) $$ \begin{align} \left(\alpha\left(g^{{1/2}}{{a}}g^{{1/2}}\right) -\epsilon/8\right)_+ \lesssim \left({\alpha}\left(g^{1/4}{{b}}g^{1/4}\right) -\sigma\right)_+. \end{align} $$

By [Reference Rørdam40, Proposition 2.2] and formulas (e4.4), (e4.9) and (e4.2), it follows that

(e4.10) $$ \begin{align} \left(g^{{1/2}}{{a}}g^{{1/2}}-\varepsilon/4\right)_+ &\lesssim \left(\alpha\left(g^{{1/2}}{{a}}g^{{1/2}}\right) -\epsilon/8\right)_+ \hspace{6pc} \end{align} $$

(e4.11) $$ \begin{align} & \hspace{6pc} \lesssim \left({\alpha}\left(g^{1/4}{{b}}g^{1/4}\right) -\sigma\right)_+ \lesssim g^{1/4}{{b}}g^{1/4} \lesssim b. \end{align} $$

By (1 $'$ ) and the choice of $\delta $ ,

(e4.12) $$ \begin{align} a\approx_{\varepsilon/16} (1-f)^{1/2}a(1-f)^{1/2}+f^{1/2} a f^{1/2}. \end{align} $$

Choose

(e4.13) $$ \begin{align} a_1&:=\left(g^{{1/2}}a g^{{1/2}}-\varepsilon/2\right)_+ =\left((1-f)^{1/2}a(1-f)^{1/2}-\varepsilon/2\right)_+ \mathrm{\ and\ } \end{align} $$

(e4.14) $$ \begin{align} a_2&:=f^{1/2}af^{1/2}. \hspace{16pc} \end{align} $$

Then by formula (e4.11), one has $a_1\lesssim _A b$ . Note that (3 $'$ ) implies $a_2 \lesssim _A c$ . Thus $a_1$ and $a_2$ satisfy (2) and (3) of the lemma. By formula (e4.12),

$$ \begin{align*} a\approx_{\varepsilon/16} (1-f)^{{1/2}}a(1-f)^{{1/2}}+f^{{1/2}}af^{{1/2}} \approx_{\epsilon/2} a_1+a_2. \end{align*} $$

So (1) of the lemma also holds, and the lemma follows.

Proof. We may assume that A is nonelementary. Set $a,b\in M_m(A)_+\setminus \{0\}$ with $\lVert a\rVert =1=\lVert b\rVert $ for some integer $m\ge 1$ . Set $n\in \mathbb {N}$ and assume $(n+1)\langle a\rangle \leq n\langle b\rangle $ . To prove the theorem, it suffices to prove that $a\lesssim b$ .

Note that if $B\in {\cal W}$ , then for each integer m, we have $M_m(B)\in {\cal W}$ . It follows that $M_m(A)$ is e. tracially in ${\cal W}$ . To simplify notation, without loss of generality one may assume $a, b\in A_+$ .

By [Reference Fu and Lin21, Lemma 4.3], $\mathrm {Her}\left (f_{1/4}(b)\right )_+$ contains $2n+1$ nonzero mutually orthogonal elements $b_0, b_1, \dotsc ,b_{2n}$ such that $\langle b_i\rangle =\langle b_0\rangle $ , $i=1, 2, \dotsc ,2n$ . Without loss of generality, we may assume that $\lVert b_0\rVert =1$ . If $b_0$ is a projection, choose $e_0=b_0$ . Otherwise, by replacing $b_0$ by $g_1(b_0)$ for some continuous function $g_1\in C_0([0,1])$ , we may assume that there is a nonzero $e_0\in A_+$ such that $b_0e_0=e_0b_0=e_0$ . Replacing b by $g(b)$ for some $g\in C_0((0, 1])$ , one may assume that $bb_0=b_0b=b_0$ . Put $c=b-b_0$ . Note that

(e4.15) $$ \begin{align} ce:0=(b-b_0)e_0=be_0-e_0=b_0e_0-e_0=0=e_0c. \end{align} $$

Keep in mind that $b\ge c + e_0$ , $c \perp e_0$ and $2n\langle b_0\rangle \le {{\langle c\rangle =\langle b-b_0 \rangle }}$ . One has

(e4.16) $$ \begin{align} (2n+2)\langle a\rangle \le 2n \langle b\rangle \le 2n(\langle b-b_0\rangle +\langle b_0\rangle)\le 2n\langle c\rangle +\langle c\rangle=(2n+1)\langle c \rangle. \end{align} $$

By Lemma 4.2, for any $\varepsilon \in (0,1/2)$ there exist $a_1,a_2\in A_+$ such that

1. (i) $a\approx _{\epsilon /2}a_1+a_2$ ,

2. (ii) $a_1\lesssim _A c$ and

3. (iii) $a_2\lesssim _A e_0$ .

By (i)–(iii) and applying [Reference Rørdam40, Proposition 2.2] (recall $be_0=e_0b=e_0$ ), one obtains

(e4.17) $$ \begin{align} (a-\varepsilon)_+\lesssim a_1+a_2 \lesssim c+e_0\le b. \end{align} $$

Since this holds for every $\varepsilon \in (0,1/2)$ , one concludes that $a\lesssim b$ .

Proof. It follows from [Reference Rørdam42, Theorem 4.5] and Theorem 4.3.

Proof. Fix $a\in \mathrm {Ped} (A)_+^{\textbf{1}}$ and let $C=\mathrm {Her}(a)$ . We will show that every $2$ -quasitrace of C is a trace. We may assume that C is nonelementary. Set $\tau \in QT(C)$ . Fix $x, y\in C_{\mathrm {sa}}$ with $\lVert x\rVert ,\lVert y\rVert \leq 1/2$ . Set $\varepsilon \in (0,1/2)$ . Let ${\cal F}:=\{x,y,x+y\}$ . Let $n\in \mathbb {N}$ be such that $\varepsilon>1/n$ . By [Reference Fu and Lin21, Lemma 4.3], there exist mutually orthogonal norm $1$ positive elements $c_1,c_2,\dotsc , c_n\in A_+\backslash \{0\}$ such that $c_1\sim c_2\sim \dotsb \sim c_n$ . Then $d_{\tau } ( c_1 )\leq 1/n<\varepsilon $ .

Let $\delta \in (0,\varepsilon )$ be such that for any $d\in C_+^{\textbf{1}}$ and $z\in C_{\mathrm {sa}}^{\textbf{1}}$ , if $\lVert [d,z]\rVert <\delta $ , then

(e4.18) $$ \begin{align} z\approx_\varepsilon (1-d)^{1/2}z(1-d)^{1/2}+d^{1/2}zd^{1/2} \end{align} $$

and

(e4.19) $$ \begin{align} \tau(z)\approx_\varepsilon \tau\left((1-d)^{1/2}z(1-d)^{1/2}\right)+\tau\left(d^{1/2}zd^{1/2}\right). \end{align} $$

(Note [Reference Blackadar and Handelman4, II.2.6] that $\left \lVert \left [(1-d)^{1/2}z(1-d)^{1/2},d^{1/2}zd^{1/2}\right ]\right \rVert $ can be sufficiently small depending on $\delta $ .) Note that ${\cal T}$ has property (H). Since A is simple and e. tracially in ${\cal T}$ , by Proposition 3.5 C is also e. tracially in ${\cal T}$ . There exist an element $e\in C_+^{\textbf{1}}$ and a nonzero $C^*$ -subalgebra $B\subset C$ such that B is in $\mathcal {T}$ , and the following are true:

1. (1) $\lVert ez-ze\rVert <\delta $ for all $z\in \cal F$ .

2. (2) $(1-e)^{1/2}z(1-e)^{1/2}\in _{\delta /2} B$ for all $z\in {\cal F}$ .

3. (3) $e\lesssim c_1$ .

We may choose $e_B\in \mathrm {Ped}(B)_+^{\textbf{1}}$ such that

1. (2^′) $(1-e)^{1/2}z(1-e)^{1/2}\in _{\delta } B_1:=\overline {e_BBe_B}$ for all $z\in {\cal F}$ .

Note that for $z\in {\cal F}$ , $e^{1/2}ze^{1/2}$ is self-adjoint. One has $\left (e^{1/2}ze^{1/2}\right )_+,\left (e^{1/2}ze^{1/2}\right )_-\in \mathrm {Her}_A(e)$ . Then

(e4.20) $$ \begin{align} \left\lvert\tau\left(e^{1/2}ze^{1/2}\right)\right\rvert &= \left\lvert\tau\left(\left(e^{1/2}ze^{1/2}\right)_+\right)-\tau\left(\left(e^{1/2}ze^{1/2}\right)_-\right)\right\rvert \hspace{6pc} \end{align} $$

(e4.21) $$ \begin{align} & \hspace{7pc} \leq d_\tau\left(\left(e^{1/2}ze^{1/2}\right)_+\right)+d_\tau\left(\left(e^{1/2}ze^{1/2}\right)_-\right) \leq 2d_\tau(e)\leq 2\varepsilon. \end{align} $$

Then by (1), the choice of $\delta $ and formulas (e4.18) and (e4.19), for $z\in {\cal F}$ ,

(e4.22) $$ \begin{align} \hspace{3pc} \tau(z) &\approx_{2\varepsilon} \tau\left((1-e)^{1/2}z(1-e)^{1/2}\right)+\tau\left(e^{1/2}ze^{1/2}\right) \end{align} $$

(e4.23) $$ \begin{align} \text{(by formula (e4.21))}&\approx_{2\varepsilon} \tau\left((1-e)^{1/2}z(1-e)^{1/2}\right). \hspace{7pc} \end{align} $$

By (2 $'$ ), there are $\bar x, \bar y\in (B_1)_{\mathrm {sa}}$ such that

(e4.24) $$ \begin{align} (1-e)^{1/2}x(1-e)^{1/2}\approx_{2\delta} \bar x, \qquad (1-e)^{1/2}y(1-e)^{1/2}\approx_{2\delta} \bar y. \end{align} $$

Then

$$ \begin{align*} \tau(x+y) &\overset{\text{formula (e 4.23)}}{\approx_{{{4\varepsilon}}}} \tau\left((1-e)^{1/2}(x+y)(1-e)^{1/2}\right) \\ &\overset{\text{formula (e 4.24)}}{\approx_{{4\delta}}} \tau\left(\bar x+\bar y\right) \\ \text{(}\tau\ \text{is a trace on}\ B_1\text{)}&\overset{\phantom{\text{formula (e 4.24)}}}{=} \tau(\bar x)+\tau(\bar y) \\ &\overset{\text{formula (e 4.24)}}{\approx_{{4\delta}}} \tau\left((1-e)^{1/2}x(1-e)^{1/2}\right)+ \tau\left((1-e)^{1/2}y(1-e)^{1/2}\right) \\ &\overset{\text{formula (e 4.23)}}{\approx_{{4\varepsilon}}} \tau(x)+ \tau(y). \end{align*} $$

Since $\varepsilon $ and $\delta $ are arbitrary small, we have $\tau (x+y)=\tau (x)+\tau (y)$ , and therefore $\tau $ is a trace on C.

Proof. Note that A has infinite dimension. Set $a\in \mathrm {Ped}(A)_+\setminus \{0\}$ with $\lVert a\rVert =1$ .

Since $\overline {f_{1/4}(a)Af_{1/4}(a)}$ is an infinite-dimensional simple $C^*$ -algebra, one may choose $c, d\in \overline {f_{1/4}(a)Af_{1/4}(a)}_+\setminus \{0\}$ such that $cd=dc=0$ .

Since A is e. tracially in ${\cal PI}$ , there exist a sequence of positive elements $e_n\in A_+$ with $\lVert e_n\rVert \le 1$ and a sequence of $C^*$ -subalgebra $B_n\subset A$ such that $B_n$ in ${\cal PI}$ , and the following are true:

1. (1) $a\approx _{1/2^n} e_n^{1/2}ae_n^{1/2}+(1-e_n)^{1/2}a(1-e_n)^{1/2}$ .

2. (2) $(1-e_n)^{1/2}a(1-e_n)^{1/2}\in _{1/2^n} B_n$ and $\left \lVert (1-e_n)^{1/2}a(1-e_n)^{1/2}\right \rVert \ge \lVert a\rVert -{1/2^n}$ .

3. (3) $e_n\lesssim c$ .

By (2), there is $b_n\in {B_n}_+$ such that $b_n\approx _{1/2^n} (1-e_n)^{1/2}a(1-e_n)^{1/2}$ . Then by (1),

(e4.25) $$ \begin{align} a\approx_{2/2^n} b_n+e_n^{1/2}ae_n^{1/2}. \end{align} $$

Note that $\inf _n\{\lVert b_n\rVert \}\ge \lVert a\rVert /2>0$ . Choose $0<\varepsilon <\lVert a\rVert /16$ .

By [Reference Phillips37, Lemma 1.7], for all sufficiently large n we have

(e4.26) $$ \begin{align} {{0\neq(b_{n}-2\varepsilon)_+\lesssim \left(b_{n}+e_{n}^{1/2}ae_{n}^{1/2}-2\varepsilon\right)_+\lesssim a.}} \end{align} $$

Note that $(b_{n}-2\varepsilon )_+\in \mathrm {Ped}(B_n)_+\setminus \{0\}$ . Then there are $d_1, d_2\in \mathrm {Ped}(B_n)_+\setminus \{0\}$ such that $d_1\perp d_2$ , $d_1+d_2\lesssim d_2\lesssim (b_{n}-2\varepsilon )_+$ and

(e4.27) $$ \begin{align} d_1+d_2\lesssim (b_{n}-2\varepsilon)_+\lesssim a. \end{align} $$

It follows that a is infinite, and therefore A is purely infinite.

Proof. This is a theorem of Rørdam [Reference Rørdam42, Corollary 5.1]. Since we do not assume that A is exact and will use only $2$ -quasitraces, some more explanation is in order. The explanation, of course, follows exactly the same lines as the proof of [Reference Rørdam42, Corollary 5.1].

Set $a\in \mathrm {Ped}(A)_+^{\textbf{1}}$ and $B:=\overline {aAa}$ . Then B is algebraically simple (see, for example, [Reference Blackadar3, II.5.4.2]). Assume that B has no nonzero $2$ -quasitraces.

Consider $W(B)$ . Note that $W(B)\subset W(A)$ , and $W(B)$ has the property that if $x\in W(B)$ and $y\in W(A)$ such that $y\le x$ , then $y\in W(B)$ . It follows that $W(B)$ is almost unperforated. Since B is algebraically simple, every element in $W(B)$ is a strong order unit.

Set $t, t'\in W(B)$ (with t a strong order unit). The statement (and the proof) of [Reference Rørdam40, Proposition 3.1] imply that if there is no state on $W(B)$ (with the strong order unit t), then there must be some integer $n\in \mathbb {N}$ and $u\in W(B)$ such that

(e4.28) $$ \begin{align} n t' +u\le nt +u. \end{align} $$

Then by [Reference Rørdam40, Proposition 3.2] (see the proof also), as $W(B)$ is almost unperforated,

(e4.29) $$ \begin{align} t'{{~\leq~}}t. \end{align} $$

On the other hand, by [Reference Blackadar and Handelman4, II.2.2], every lower semicontinuous dimension function on $W(B)$ is induced by a $2$ -quasitrace on B. Since B is assumed to have no nonzero $2$ -quasitraces, combining with [Reference Rørdam40, Proposition 4.1] (as well as the paragraph before it) shows that there is no state on $W(B)$ . Therefore formula (e4.29) implies that for any $b, c\in B_+\setminus \{0\}$ , we have $b\lesssim c$ . It follows that B is purely infinite and so is A.

To see the last part of the statement, suppose that there are $b, c\in \mathrm {Ped}(A)_+^{\textbf{1}}\setminus \{0\}$ such that $bc=cb=0$ and $b+c\lesssim c$ . Let $a=b+c$ and $B=\overline {aAa}$ . Note that $a\in \mathrm {Ped}(A)_+$ . Then B has nonzero $2$ -quasitraces.

Therefore

(e4.30) $$ \begin{align} d_\tau(c)\ge d_\tau(b+c)\mathrm{\ for\ all\ } \tau\in QT(B). \end{align} $$

On the other hand, for any $\tau \in QT(B)$ and any $1>\varepsilon >0$ ,

(e4.31) $$ \begin{align} \tau(f_{\varepsilon}(b+c))=\tau(f_{\varepsilon}(b)+f_{\varepsilon}(c))=\tau(f_{\varepsilon}(b))+\tau(f_{\varepsilon}(c)). \end{align} $$

Fix $1>\varepsilon _0>0$ such that $f_{\varepsilon _0}(b)\not =0$ . Since B is algebraically simple, $\tau \left (f_{\varepsilon _0}(b)\right )>0$ for all $2$ -quasitraces $\tau $ . Fix $\tau \in QT(B)$ . Then, by equation (e4.31),

(e4.32) $$ \begin{align} d_\tau(b+c)\ge \tau\left(f_{\varepsilon_0}(b)\right)+d_\tau(c)>d_\tau(c). \end{align} $$

This contradicts formula (e4.30). It follows that no such pairs b and c exist. Thus A is finite.

Since $M_n(A)$ has the same relevant property as A, we conclude that A is stably finite.

Proof. By Theorem 4.3, $W(A)$ is almost unperforated. It follows from Remark 2.5 that $\mathrm {Cu}(A)$ is almost unperforated. Fix $e\in \mathrm {Ped}(A)_+\setminus \{0\}$ and let $B:=\mathrm {Her}(e)$ . As in the proof of Proposition 4.9, every lower semicontinuous dimension function on $W(B)$ is induced by a $2$ -quasitrace of B. Set $a, b\in (A\otimes {\cal K})_+$ such that $d_\tau (a)<d_\tau (b)$ for all $\tau \in QT(B)$ . By [Reference Elliott, Robert and Santiago17, Propositions 4.2, 4.6], $a\lesssim b$ .

Proof. It follows from [Reference Rørdam42, Theorem 4.5] that every $C^*$ -algebra B in ${\cal C}_{\mathscr Z}$ has almost unperforated $W(B)$ . It follows from Theorem 4.3 and Remark 2.5 that $\mathrm {Cu}(A)$ is almost unperforated. By Proposition 4.9, if A is not purely infinite, then it is stably finite, and by the proof of Corollary 4.10, A has strict comparison for positive elements.

Proof. Let ${\cal F}\subset A$ a finite subset, $\varepsilon>0$ , $s\in A_+\setminus \{0\}$ and an integer n be given. Suppose that there are a finite subset ${\cal F}'$ and an element $e_F\in A_+^{\textbf{1}}$ such that $e_Fy=y=ye_F$ for all $y\in {\cal F}'$ , and if $x\in {\cal F}$ , there is $y\in {\cal F}'$ such that $\lVert y-x\rVert <\varepsilon /4$ . Without loss of generality, we may assume that ${\cal F}\subset A^{\textbf{1}}$ . We may further assume that ${\cal F}'\subset A^{\textbf{1}}$ .

Let $\delta \in (0,\varepsilon /8)$ be a positive number such that for any elements $z\in A^{\textbf{1}}$ and ${{w}}\in A_+^{\textbf{1}}$ , $\lVert zw-wz\rVert <\delta $ implies that

(e5.1) $$ \begin{align} {{\left\lVert(1-w)^{1/2}z-z(1-w)^{1/2}\right\rVert<\varepsilon/8.}} \end{align} $$

Put ${\cal F}_1={\cal F}\cup \{e_F\}\cup {\cal F}'$ . Suppose that there are $\theta \in A_+^{\textbf{1}}$ and D as in the statement of the proposition, such that (i), (ii) and (iii) hold for $\delta $ (in place of $\varepsilon $ ) and ${\cal F}_1$ (in place of ${\cal F}$ ).

Then in Definition 5.2(3) holds.

Define ${\beta }: A\to A$ by ${\beta }(a):= (1-\theta )^{1/2}a(1-\theta )^{1/2}$ for all $a\in A$ . It is a c.p.c. map. For each $x\in {\cal F}_1$ , define $x_1:=\theta ^{1/2} x\theta ^{1/2} \,\in \mathrm {Her}(\theta )$ . Then $\lVert x_1\rVert \le \lVert x\rVert $ . Note that by the choice of $\delta $ , for all $x\in {\cal F}\cup {\cal F}'$ ,

(e5.2) $$ \begin{align} e_F{\beta}(x)=e_F(1-\theta)^{1/2}x(1-\theta)^{1/2}\approx_{\varepsilon/8}(1-\theta)^{1/2}e_Fx(1-\theta)^{1/2}\approx_{\varepsilon/8} {\beta}(x) \approx_{\varepsilon/4} {\beta}(x)e_F. \end{align} $$

Moreover, for all $x\in {\cal F}_1$ ,

(e5.3) $$ \begin{align} {\beta}(x)=(1-\theta)^{1/2}x(1-\theta)^{1/2} {{ \approx_{\varepsilon/8} (1-\theta)}}x\in_{\delta} D\otimes 1_n. \end{align} $$

So Definition 5.2(2) holds. Also by the choice of $\delta $ , for all $x\in {\cal F}_1$ ,

(e5.4) $$ \begin{align} x=\theta x+(1-\theta)x\approx_{\varepsilon/4} \theta^{1/2}x\theta^{1/2}+(1-\theta)^{1/2}x(1-\theta)^{1/2}=x_1+{\beta}(x). \end{align} $$

Hence Definition 5.2(1) holds. Thus A is tracially approximately divisible.

Proof. Let $\delta>0$ be given. Choose ${{N(\delta )}}\ge 1$ such that

(e5.6) $$ \begin{align} \max\left\{\left\lvert{{\left(1-t^{1/n}\right)^2t}}\right\rvert: t\in [0,1]\right\}<\delta^2 \mathrm{\ for\ all\ } n\ge {{N(\delta)}}. \end{align} $$

Then for any $C^*$ -algebra A, any $e\in A_+^{\textbf{1}}$ and any $x\in A$ satisfying $x^*x\le e$ and $xx^*\le e$ ,

(e5.7) $$ \begin{align} \left\lVert\left(1-e^{1/n}\right)x\right\rVert = \left\lVert\left(1-e^{1/n}\right)xx^*\left(1-e^{1/n}\right)\right\rVert^{1/2} \leq \left\lVert\left(1-e^{1/n}\right)e\left(1-e^{1/n}\right)\right\rVert^{1/2} <\delta \end{align} $$

for all $n\geq N(\delta )$ . Similarly, we also have $\left \lVert x\left (1-e^{1/n}\right )\right \rVert <\delta $ for all $n\geq N(\delta )$ . The lemma follows.

Proof. Let B be a hereditary $C^*$ -subalgebra of A, ${\cal F}\subset B^{\textbf{1}}$ be a finite subset, $\varepsilon>0$ , $s\in B_+\setminus \{0\}$ be a positive element and $n\ge 1$ be an integer. Suppose also that there exists a finite subset ${\cal F}'\subset B^{\textbf{1}}$ such that $y\in _{\varepsilon /4}{\cal F}'$ for all $y\in {\cal F}$ , and there exists an element $e_F\in B_+^{\textbf{1}}$ such that $e_Fx=x=xe_F$ for all $x\in {\cal F}'$ . Let $g_0, g_1\in C_0([0,1])$ be such that $0\le g_0, g_1\le 1$ , $g_0(0)=0$ , $g_0(t)=1$ for $t\in [1-\varepsilon /64,1]$ and $g_0$ is linear on $[0, 1-\varepsilon /64];$ and $g_1(t)=0$ if $t\in [0,1-\varepsilon /64]$ , $g_1(1)=1$ and $g_1$ is linear on $[1-\varepsilon /64, 1]$ . Put $b_0:=g_0(e_F)$ and $b_1:=g_1(e_F)$ . Then

(e5.8) $$ \begin{align} b_0b_1=b_1=b_1b_0,\qquad b_0\ge e_F, \qquad \lVert b_0-e_F\rVert<\varepsilon/64. \end{align} $$

Since for all $x\in {\cal F'}$ we have $e_Fxx^*=xx^*=xx^*e_F$ and $e_Fx^*x=x^*x=x^*xe_F$ , by the spectral theory, we have $b_ixx^*=xx^*=xx^*b_i$ and $b_ix^*x=x^*x=x^*xb_i$ , $i=0,1$ . It follows that

(e5.9) $$ \begin{align} b_ix=x=xb_i\mathrm{\ and\ } b_ix^*=x^*=x^*b_i\mathrm{\ for\ all\ } x\in {\cal F}', \quad i=0,1. \end{align} $$

Let ${\cal F}_1= \{b_1\}\cup {\cal F}'$ . Choose $\delta>0$ in [Reference Elliott, Gong, Lin and Niu15, Lemma 3.3] associated with $\varepsilon /64$ (in place of $\varepsilon $ ) and $\sigma =\varepsilon /64$ . Set $\eta =\min \{\delta /4, \varepsilon /256\}$ .

We choose $N:= N(\eta )\ge 1$ as in Lemma 5.4.

Let $0<\delta _1<\eta /2$ . Moreover, we choose $\delta _1$ sufficiently small that if $C_1\subset C_2$ is any pair of $C^*$ -algebras and $c\in C_2$ with $0\le c\le 1$ and $c\in _{\delta _1} C_1$ , and if $0\le c_1, c_2\le 1$ and $c_1c_2\approx _{\delta _1} c_2\approx _{\delta _1}c_2c_1$ , then

(e5.10) $$ \begin{align} c^{1/N}\in_{\eta} (C_1)_+^{\textbf{1}}\qquad \text{and} \qquad c_1c_2^{1/N}c_1\approx_\eta c_2^{1/N}. \end{align} $$

Since A is tracially approximately divisible, there are $\theta _a\in A_+^{\textbf{1}}$ , a $C^*$ -subalgebra $D_a\otimes M_n\subset A$ and a c.p.c. map ${\beta }: A\to A$ such that

1. (1) $x\approx _{\delta _1/2} x_1+ {\beta }(x) $ such that $\lVert x_1\rVert \le 1$ and $x_1\in \mathrm {Her}(\theta _a)$ for all $x\in {\cal F}_1$ ,

2. (2) ${\beta }(x)\in _{\delta _1/2} D_a\otimes 1_n$ and $ b_0{\beta }(x)\approx _{\delta _1/2} {\beta }(x)\approx _{\delta _1/2} {\beta }(x)b_0 \text { for all } x\in {\cal F}_1$ and

3. (3) $\theta _a\lesssim s$ .

Choose $d(x)\in (D_a\otimes 1_n)^{\textbf{1}}$ such that

(e5.11) $$ \begin{align} \lVert{\beta}(x)-d(x)\rVert<\delta_1\mathrm{\ for\ all\ } x\in {\cal F}_1. \end{align} $$

Let $b_2={\beta }(b_1)^{1/N}$ . By equation (e5.9), ${\beta }(b_1)\ge {{{\beta }(x)^*{\beta }(x)}}$ and ${\beta }(b_1)\ge {\beta }(x){\beta }(x)^*$ for all $x\in {\cal F}'$ (see, for example, [Reference Blackadar, Kumjian and Rørdam5, Corollary 4.1.3]). By condition (2) here, the choice of N and application of Lemma 5.4,

(e5.12) $$ \begin{align} b_2{\beta}(x)={\beta}(b_1)^{1/N}{\beta}(x) \approx_\eta {\beta}(x)\qquad \mathrm{\ for\ all\ } x\in {\cal F}'. \end{align} $$

Recall that ${\beta }(b_1)\in _{\delta _1} D_a\otimes 1_n$ . By the choice of $\delta _1$ , we may choose $d\in (D_a\otimes 1_n)_+$ such that

(e5.13) $$ \begin{align} \lVert d-b_2\rVert<\eta. \end{align} $$

Then, with $b:=b_0b_2b_0$ , by the second part of formula (e5.10),

(e5.14) $$ \begin{align} \lVert d-b\rVert<2\eta \qquad\text{and}\qquad f_{\varepsilon/64}(d)d\approx_{\varepsilon/64} d\approx_{2\eta} b. \end{align} $$

By the choice of $\eta $ , applying [Reference Elliott, Gong, Lin and Niu15, Lemma 3.3] yields an isomorphism

$$ \begin{align*} \varphi: \overline{f_{\varepsilon/64}(d)(D_a\otimes M_n)f_{\varepsilon/64}(d)}\to \overline{bAb} \subset B \end{align*} $$

such that

(e5.15) $$ \begin{align} \lVert \varphi(y)-y\rVert<\varepsilon/64\lVert y\rVert\mathrm{\ for\ all\ } y\in \overline{f_{\varepsilon/64}(d)(D\otimes 1_n)f_{\varepsilon/64}(d)}. \end{align} $$

Note that $\overline {f_{\varepsilon /64}(d)(D_a\otimes M_n)f_{\varepsilon /64}(d)}\cong D_1\otimes M_n$ and $\overline {f_{\varepsilon /64}(d)(D_a\otimes 1_n)f_{\varepsilon /64}(d)}\cong D_1\otimes 1_n$ for some $C^*$ -subalgebra $D_1\subset D_a$ . Let $D_b=\varphi (D_1)$ . Define a c.p.c. map ${\alpha }: B\to {{B}}$ by

(e5.16) $$ \begin{align} {\alpha}(y):= b{\beta}(y)b\mathrm{\ for\ all\ } y\in {{B}}. \end{align} $$

Then, for all $x\in {\cal F}_1$ , by formulas (e5.14) and (e5.11),

(e5.17) $$ \begin{align} {\alpha}(x)&=b{\beta}(x)b\approx_{2\left(2\eta+\varepsilon/64\right)} f_{\varepsilon/64}(d) d{\beta}(x)d f_{\varepsilon/64}(d) \hspace{2pc} \end{align} $$

(e5.18) $$ \begin{align} \hspace{3pc} &\approx_{\delta_1} f_{\varepsilon/64}(d)d d(x) df_{\varepsilon/64}(d)\in_{\varepsilon/64} D_b\otimes 1_n\subset \overline{bAb}\subset B. \end{align} $$

If $y\in {\cal F}$ , choose $x\in {\cal F}'$ such that $\lVert y-x\rVert <\varepsilon /4$ . Then ${\alpha }(y)\approx _{\varepsilon /4} {\alpha }(x)\in _{\varepsilon /4} D_b\otimes 1_n$ . Define $y_1=b_0x_1b_0$ . Then, by conditions (1) and (2) and equation (e5.12),

(e5.19) $$ \begin{align} y&\approx_{\delta_1/2+\varepsilon/4} b_0(x_1+{\beta}(x))b_0=y_1+b_0{\beta}(x)b_0 \end{align} $$

(e5.20) $$ \begin{align} &\approx_{2\eta} y_1+b_0b_2{\beta}(x)b_2b_0 \hspace{6pc} \end{align} $$

(e5.21) $$ \begin{align} &\approx_{2\delta_1}y_1+b_0b_2b_0{\beta}(x)b_0b_2b_0=y_1+{\alpha}(x) \end{align} $$

(e5.22) $$ \begin{align} &\approx_{\varepsilon/4} y_1+{\alpha}(y) \quad \text{for all }y\in {\cal F}. \hspace{3.5pc} \end{align} $$

Note that $\delta _1/2+\varepsilon /4+2\eta +2\delta _1+\varepsilon /4<\varepsilon $ . Put $\delta _2:=3\delta _1/2+2\eta $ . Then $0<\delta _2<5\varepsilon /256$ . Also for all $y\in {\cal F}$ ,

(e5.23) $$ \begin{align} e_F{\alpha}(y)\approx_{\varepsilon/4}e_F{\alpha}(x)&=e_F b_0b_2b_0{\beta}(x)b_0b_2b_0\approx_{\delta_2} e_F{\beta}(x)\approx_{\varepsilon/64} b_0{\beta}(x)\approx_{\delta_1} {\beta}(x) \end{align} $$

(e5.24) $$ \begin{align} &\approx_{\delta_1} {\beta}(x)b_0\approx_{\varepsilon/64}{\beta}(x)e_F\approx_{\delta_2+\varepsilon/4}{\alpha}(y)e_F \end{align} $$

(recall $\lVert b_0-e_F\rVert <\varepsilon /64$ ). Put $\theta _b=b_0\theta _a b_0$ . Then $y_1\in \overline {\theta _bB\theta _b}$ . Moreover,

(e5.25) $$ \begin{align} \theta_b\lesssim \theta_a\lesssim s. \end{align} $$

From formulas (e5.22), (e5.18), (e5.24) and (e5.25), the theorem follows.

Proof. Set $a\in {{\overline {d_1Ad_1+\dotsb +d_nAd_n}}}$ and fix $\varepsilon>0$ . Then there exist $a_1,\dotsc ,a_n\in A$ and $\delta>0$ such that $a\approx _{\varepsilon } f_\delta (d_1)a_1f_\delta (d_1)+\dotsb +f_{\delta }(d_n)a_nf_\delta (d_n)$ . Set $x_1,\dotsc ,x_n\in A$ such that $x_i^*x_i=f_\delta (d_i)$ and $x_ix_i^*\in \overline {e_iAe_i}$ , $i=1,\dotsc ,n$ (see [Reference Rørdam40, Proposition 2.4]). For $i,j\in \{1,\dotsc ,n\}$ and $i\leq j$ , $e_id_j=0$ implies $x_j^*x_jx_ix_i^*=0$ , thus

(e5.26) $$ \begin{align} x_jx_i=0 \quad(i\leq j). \end{align} $$

Claim 1: ${{(x_1+x_2+\dotsb +x_n)^{n+1}=0}}$ .

Proof of Claim 1: Note that $(x_1+x_2+\dotsb +x_n)^{n+1}$ is a sum of $n^{n+1}$ terms with the form $x_{k_1}x_{k_2}\dotsm x_{k_{n+1}} (k_1,\dotsc ,k_{n+1}\in \{1,\dotsc ,n\})$ . Assume $x_{k_1}x_{k_2}\dotsm x_{k_n+1}\neq 0$ ; then $x_{k_i}x_{k_{i+1}}\neq 0$ ( $i=1,\dotsc ,n$ ). By equation (e5.26), it follows that $k_{i+1}\leq k_i-1$ ( $i=1,\dotsc ,n$ ). In particular, $k_{n+1}\le k_n-1$ . Then $k_{n+1}\le k_n-1\le k_{n-1}-2$ . An induction implies that $k_{n+1}\leq k_1-n\leq 0$ , which gives a contradiction. Thus all $n^{n+1}$ terms of the form $x_{k_1}x_{k_2}\dotsm x_{n+1}$ are zero. It follows that $(x_1+x_2+\dotsb +x_n)^{n+1}=0$ .

Claim 2: $\left (f_\delta (d_1)a_1x_1^*+\dotsb +f_{\delta }(d_n)a_nx_n^*\right )^{n+1}=0$ .

Proof of Claim 2: Let $y_i=f_\delta (d_i)a_ix_i^*$ ( $i=1,\dotsc ,n$ ). For $i\leq j$ , using equation (e5.26), we have

(e5.27) $$ \begin{align} y_iy_j = f_\delta(d_i)a_ix_i^*f_\delta\left(d_j\right)a_jx_j^* = f_\delta(d_i)a_ix_i^*\left(x_j^*x_j\right)a_jx_j^*=f_\delta(d_i)a_i\left(x_jx_i\right)^*x_ja_jx_j^* =0. \end{align} $$

Then, as in the proof of Claim 1, we have $(y_1+\dotsb +y_n)^{n+1} =0$ . Claim 2 follows.

Let $x=x_1+\dotsb +x_n$ and let $y=y_1+\dotsb +y_n=f_\delta (d_1)a_1x_1^*+\dotsb +f_{\delta }(d_n)a_nx_n^*$ . Then by Claims 1 and 2, both x and y are nilpotent elements. For $i,j\in \{1,\dotsc ,n\}$ and $i\neq j$ , $e_ie_j=0$ implies $x_ix_i^*x_jx_j^*=0$ , thus $ x_i^*x_j=0. $ Then $yx=f_\delta (d_1)a_1f_\delta (d_1)+\dotsb +f_{\delta }(d_n)a_nf_\delta (d_n)\approx _\varepsilon a$ .

Proof. We assume that A is infinite-dimensional. Fix an element $x\in A$ and fix $\varepsilon>0$ . We may assume that x is not invertible. Since A is finite, x is not one-sided invertible. To show that x is a norm limit of invertible elements, it suffices to show that $ux$ is a norm limit of invertible elements for some unitary $u\in {\widetilde A}$ . Note that since A is simple, ${\widetilde A}$ is prime. Thus, by [Reference Rørdam39, Proposition 3.2, Lemma 3.5], we may assume that there is $a'\in \widetilde A_+\backslash \{0\}$ and $a'x=xa'=0$ . There is $e\in A_+$ such that $a'ea'\not =0$ . Put $a=a'ea'$ .

Let $B_0=\{z\in A: az=za=0\}$ . Then $x\in B_0$ , and $B_0$ is a hereditary $C^*$ -subalgebra of A. There is $e_b'\in {B_0}_+$ with $\lVert e_b\rVert =1$ such that $e_b'xe_b'\approx _{\varepsilon /64}x$ . So $f_{\varepsilon /64}\left (e_b'\right )xf_{\varepsilon /64}\left (e_b'\right )\approx _{\varepsilon /16}x$ . Put $e_b=f_{\varepsilon /64}\left (e_b'\right )$ and $B=\mathrm {Her}(e_b)$ . Without loss of generality, we may further assume that $x\in B$ .

Since we assume that A is infinite-dimensional, $\overline {aAa}$ contains nonzero positive elements $a_0,a_1$ such that $a_0a_1=0$ .

Since A is simple, there is $c\in A$ such that $e_bc(a_1)^{1/2}\not =0$ (see the proof of [Reference Cuntz12, 1.8]).

Note that since $e_b\in \mathrm {Ped}(B)$ , we have $\mathrm {Ped}(B)=B$ (see, for example, [Reference Blackadar3, II.5.4.2]). It follows that there are $y_1,y_2,\dotsc ,y_m\in B$ such that

(e5.28) $$ \begin{align} \sum_{i=1}^m y_i^*e_bca_1c^*e_by_i=e_b. \end{align} $$

It then follows that $\langle e_b\rangle \le m\langle a_1\rangle $ . Put $n=2m$ .

For any $z_1,z_2,\dotsc ,z_n\in B_+$ which are n mutually orthogonal and mutually equivalent positive elements,

$$ \begin{align*}n\langle z_1\rangle \le \langle e_b\rangle \le m\langle a_1\rangle. \end{align*} $$

Since $W(A)$ is almost unperforated,

(e5.29) $$ \begin{align} z_1\lesssim a_1. \end{align} $$

Since B is a hereditary $C^*$ -subalgebra of A, by Theorem 5.5, B is also tracially approximately divisible. There are $b\in B_+^{\textbf{1}}$ , a $C^*$ -subalgebra $D\otimes M_n\subset B$ and a c.p.c. map ${\beta }: A\to A$ such that

1. (1) $x\approx _{\varepsilon /8} x_0+{\beta }(x)$ , where $x_0\in \overline {bAb}$ ,

2. (2) ${\beta }(x)\in _{\varepsilon /8} D\otimes 1_n$ and

3. (3) $b\lesssim a_0$ .

Thus, there is $x_1\in D\setminus \{0\}$ such that

(e5.30) $$ \begin{align} \lVert x-(x_0+ x_1\otimes 1_n)\rVert<\varepsilon/4. \end{align} $$

Choose a positive element $d\in D$ such that

(e5.31) $$ \begin{align} \lVert dx_1d-x_1\rVert<\varepsilon/4. \end{align} $$

By the choice of n, we have $d\otimes e_{1,1}\lesssim a_1$ , where $\left \{e_{i,j}\right \}$ forms a system of matrix units for $M_n$ .

Define $g_1:= a_0$ , $g_2 := a_1$ , $g_{2+i}:= d\otimes e_{i,i}$ ( $i=1,\dotsc ,n-1$ ).

Define $h_1:= b$ , $h_{1+i}:= d\otimes e_{i,i}$ ( $i=1,\dotsc ,n$ ).

Note that $h_i\lesssim g_i$ ( $i=1,\dotsc ,n+1$ ) and $g_ih_j=0$ , if $i\le j$ and $i,j=1,\dotsc ,n+1$ . Note that $x_0+dx_1d\otimes 1_n\in \overline {h_1Ah_1}+\overline {h_2Ah_2}+\dotsb +\overline {h_{n+1}Ah_{n+1}}$ . Then by Lemma 5.6, there are nilpotent elements $v,w\in A$ such that $x_0+dx_1d\otimes 1_n\approx _{\varepsilon /4}vw$ . Choose $\delta>0$ such that $vw\approx _{\varepsilon /4}(v+\delta )(w+\delta )$ . Since $v,w$ are nilpotent elements, $v+\delta $ and $w+\delta $ are invertible. Then, combining formulas (e5.30) and (e5.31),

(e5.32) $$ \begin{align} x\approx_{\varepsilon/4} x_0+x_1\otimes 1_n \approx_{\varepsilon/4} x_0+dx_1d\otimes 1_n\approx_{\varepsilon/2} (v+\delta)(w+\delta) \in GL\left(\widetilde A\right). \end{align} $$

Therefore we have shown that $x\in \overline {GL\left (\widetilde A\right )}$ . Thus, in the case that A is unital, A has stable rank one. Since, by Theorem 5.5, this works for every hereditary $C^*$ -subalgebra of A, A almost has stable rank one in the case that A is not unital.

Proof. We assume that A is infinite-dimensional. Let A be a simple $C^*$ -algebra which is e. tracially in ${\cal C}_{\mathscr Z}$ . By [Reference Rørdam42, Theorem 4.5], every ${\mathscr Z}$ -stable $C^*$ -algebra B has almost unperforated $W(B)$ (see Remark 2.5). Then, by Theorem 4.3, $W(A)$ is almost unperforated. Let $\varepsilon>0$ , ${\cal F}\subset A^{\textbf{1}}$ a finite subset, $a\in A_+\setminus \{0\}$ and $n\ge 1$ an integer be given. Since A is infinite dimensional, choose $a_1, a_2\in \mathrm {Her}(a)_+\setminus \{0\}$ such that $a_1a_2=a_2a_1=0$ .

There are $e_A\in A_+^{\textbf{1}}$ and $\delta>0$ such that

(e5.33) $$ \begin{align} f_\delta(e_A)x\approx_{\varepsilon/4} x\approx_{\varepsilon/4} xf_\delta(e_A)\mathrm{\ for\ all\ } x\in {\cal F}. \end{align} $$

Note that by Theorem 5.5, $A_1:=\overline {f_{\delta /2}(e_A)Af_{\delta /2}(e_A)}$ is also a ( $\sigma $ -unital) simple $C^*$ -algebra which is e. tracially in ${\cal C}_{\mathscr Z}$ (as ${\cal C}_{\mathscr Z}$ has property (H); see [Reference Toms and Winter46, Corollary 3.1]).

Note also that $f_{\delta /2}(e_A)af_{\delta /2}(e_A)\lesssim a$ . To simplify notation, by replacing x by $f_\delta (e_A)xf_\delta (e_A)$ for all $x\in {\cal F}$ , a by $f_{\delta /2}(e_A)af_{\delta /2}(e_A)$ and $a_i$ by $f_{\delta /2}(e_A)a_if_{\delta /2}(e_A)$ ( $i=1,2$ ), without loss of generality we may assume that $x, a, a_1, a_2\in A_1$ . We may also assume, without loss of generality,

(e5.34) $$ \begin{align} e_1x=x=xe_1\mathrm{\ for\ all\ } x\in {\cal F} \end{align} $$

for some strictly positive element $e_1\in A_1^{1} $ . Note that $f_{\delta /2}(e_A)\in \mathrm {Ped}(A)$ . Therefore $A_1$ is algebraically simple and $f_{\delta /2}(e_A)$ is a strictly positive element of $A_1$ . There are an integer $l\ge 1$ and $x_i\in A_1$ , $i=1,2,\dotsc ,l$ , such that

(e5.35) $$ \begin{align} \sum_{i=1}^l x_i^*a_1x_i=e_1. \end{align} $$

Set ${\cal F}_1={\cal F}\cup \{e_1\}$ . Choose $0<\eta <\varepsilon /2$ such that if $\theta '\in A_+^{\textbf{1}}$ with $\lVert \theta 'x-x\theta '\rVert <\eta $ , then

(e5.36) $$ \begin{align} (\theta')^{1/2}x\approx_{\varepsilon/2} x(\theta')^{1/2}\mathrm{\ for\ all\ } x\in {\cal F}_1. \end{align} $$

There exist $\theta _1\in A_+^{\textbf{1}}$ and a ${\mathscr Z}$ -stable $C^*$ -subalgebra B of $A_1$ such that

1. (i) $\lVert \theta _1x-x\theta _1\rVert <\eta /64$ and $\left \lVert (1-\theta _1)^{1/2}x-x(1-\theta _1)^{1/2}\right \rVert <\eta /64$ for all $x\in {\cal F}_1$ ,

2. (ii) $(1-\theta _1)^{1/2}x(1-\theta _1)^{1/2}, (1-\theta _1)^{1/2}x, x(1-\theta _1)^{1/2}, (1-\theta _1)x, x(1-\theta _1), (1-\theta _1)x(1-\theta _1)\in _{\eta /64} B$ for all $x\in {\cal F}_1$ and

3. (iii) $\theta _1\lesssim a_2$ .

Let

$$ \begin{align*} {\cal F}_2&=\left\{(1-\theta_1)^{1/2}x(1-\theta_1)^{1/2}, (1-\theta_1)^{1/2}x, x(1-\theta_1)^{1/2}, (1-\theta)x, x(1-\theta_1), (1-\theta_1)x(1-\theta_1): \right. \nonumber \\ &\left. \quad x\in {\cal F}\right\}. \end{align*} $$

For each $f\in {\cal F}_2$ , fix $b(f)\in B$ such that $\lVert b(f)\rVert \le 1$ and

(e5.37) $$ \begin{align} \lVert f-b(f)\rVert<\eta/32. \end{align} $$

Let ${\cal G}=\{b(f): f\in {\cal F}_2\}$ . We write $B=C\otimes {\mathscr Z}$ . Since ${\mathscr Z}$ is strongly self-absorbing, without loss of generality we may assume that there is a finite subset ${\cal G}_1\subset C$ such that ${\cal G}=\{y\otimes 1_{\mathscr Z}: y\in {\cal G}_1\}\subset C\otimes 1_{\mathscr Z}$ . To further simplify notation, without loss of generality we may assume that there exists a strictly positive element $e_C\in C$ such that

(e5.38) $$ \begin{align} e_by=y=ye_b\mathrm{\ for\ all\ } y\in {\cal G}_1, \end{align} $$

where $e_b=e_C\otimes 1_{\mathscr Z}$ .

For any integer n, choose m such that $m> l$ and n divides m. Let $\psi : M_m\to {\mathscr Z}$ be an order $0$ c.p.c. map such that

(e5.39) $$ \begin{align} 1_{\mathscr Z}-\psi(1_m)\lesssim_{\mathscr Z} \psi\left(e_{1,1}\right) \end{align} $$

(see [Reference Rørdam and Winter43, Proposition 5.1(iv) implying (ii)]). Define $\varphi : M_m\to B$ by $\varphi (c):= e_C\otimes \psi (c)$ for all $c\in M_m$ . Set

(e5.40) $$ \begin{align} \theta_2:=e_b-\varphi(1_m)=e_C\otimes 1_{\mathscr Z}-e_C\otimes \psi(1_m)=e_C\otimes (1_{\mathscr Z}-\psi(1_{{m}}))\lesssim_{B} e_C\otimes \psi\left(e_{1,1}\right). \end{align} $$

Note that

(e5.41) $$ \begin{align} \theta_2 g=g\theta_2\mathrm{\ for\ all\ } g\in {\cal G}. \end{align} $$

It follows from formulas (e5.37) and (e5.41) that for any $y\in {\cal F}_2$ ,

(e5.42) $$ \begin{align} \theta_2y\approx_{\varepsilon/32} y\theta_2. \end{align} $$

Define $D:= \overline {e_Cce:C\otimes \psi \left (e_{1,1}\right )}$ and $D'$ the $C^*$ -subalgebra generated by

(e5.43) $$ \begin{align} \{e_Cce:C\otimes \psi(z): c\in C\mathrm{\ and\ } z\in M_m\}. \end{align} $$

Recall that $\psi $ gives a homomorphism $H: C^*(\psi (1_m))\otimes M_m \to {\mathscr Z}$ such that $H(\iota \otimes g)=\psi (g)$ for all $g\in M_m $ , where $\iota (t)=t$ for $t\in \mathrm {sp}(\psi (1_m))$ (see [Reference Winter and Zacharias49, Corollary 4.1]). It follows that $D'\cong D\otimes M_m$ . Define ${\beta }_1: A\to A$ by

(e5.44) $$ \begin{align} {\beta}_1(y):= \left(1_{\widetilde{A}}-\theta_2\right)^{1/2}y\left(1_{\widetilde{A}}-\theta_2\right)^{1/2}\mathrm{\ for\ all\ } y\in A \end{align} $$

(where $1_{\widetilde {A}}$ denotes the identity of $\widetilde {A}$ when A is not unital and is the identity of A if A has one). Note also that $(1-\theta )^{1/2}$ is an element which has the form $1+f_1(\theta )$ for $f_1(t)=(1-t)^{1/2}-1\in C_0([0,1])^{\textbf{1}}$ . If $g=y\otimes 1_{\mathscr Z}\in {\cal G}$ , then (noting that $y\in {\cal G}_1\subset C$ , and seeing equation (e5.38)),

(e5.45) $$ \begin{align} {\beta}_1(g)&=(1-\theta_2)g =g -e_C\otimes (1_{\mathscr Z}-\psi(1_m))g \hspace{9pc} \end{align} $$

(e5.46) $$ \begin{align} &=(e_C\otimes 1_{\mathscr Z})g-e_C\otimes(1_{\mathscr Z}- \psi(1_m))g \hspace{7.6pc} \end{align} $$

(e5.47) $$ \begin{align} &=(e_C\otimes \psi(1_m)) (y\otimes 1_{\mathscr Z})=\left(e_C^{1/2}ye_C^{1/2}\right)\otimes \psi(1_m)\in D\otimes 1_m. \end{align} $$

Define a c.p.c. map ${\beta }: A\to A$ by

(e5.48) $$ \begin{align} {\beta}(x):= {\beta}_1\left((1-\theta_1)^{1/2}x (1-\theta_1)^{1/2}\right)\mathrm{\ for\ all\ } x\in A. \end{align} $$

For $x\in {\cal F}$ , let $f=(1-\theta _1)^{1/2}x(1-\theta _1)^{1/2}$ . Then, by formula (e5.37),

(e5.49) $$ \begin{align} {\beta}(x)={\beta}_1\left((1-\theta_1)^{1/2}x (1-\theta_1)^{1/2}\right)\approx_{\eta/32} {\beta}_1(b(f))\in D\otimes 1_m. \end{align} $$

Put $\theta =\theta _1+(1-\theta _1)^{1/2}\theta _2(1-\theta _1)^{1/2}$ . We have

(e5.50) $$ \begin{align} 0\le \theta\le \theta_1+(1-\theta_1)^{1/2}(1-\theta_1)^{1/2}=1. \end{align} $$

For $x\in {\cal F}$ , let $f'=(1-\theta _1)x$ . Recall that we assume that $b(f')=y'\otimes 1_{\mathscr Z}$ for some $y'\in C^{\textbf{1}}$ . Then for $x\in {\cal F}$ , applying formulas (e5.37) and (e5.41) repeatedly, we have

(e5.51) $$ \begin{align} (1-\theta)x&=(1-\theta_1)x- (1-\theta_1)^{1/2}\theta_2(1-\theta_1)^{1/2} x \hspace{6.5pc} \end{align} $$

(e5.52) $$ \begin{align} &\approx_{\eta/32} (1-\theta_1)x- (1-\theta_1)x\theta_2=(1-\theta_1)x(1-\theta_2) \end{align} $$

(e5.53) $$ \begin{align} &\approx_{\eta/32} b(f')(1-\theta_2)=(1-\theta_2)^{1/2} b(f')(1-\theta_2)^{1/2} \end{align} $$

(e5.54) $$ \begin{align} \hspace{2pc} &={\beta}_1(b(f'))\approx_{\eta/32} {\beta}_1\left((1-\theta_1)^{1/2}x(1-\theta_1)^{1/2}\right)={\beta}(x). \end{align} $$

From equations (e5.49) and (e5.54), we have

(e5.55) $$ \begin{align} (1-\theta)x \in_{\eta/8} D\otimes 1_m\mathrm{\ for\ all\ } x\in {\cal F}. \end{align} $$

Recall that $(1-\theta _1)^{1/2}x, x(1-\theta _1)^{1/2}, (1-\theta _1)^{1/2}x(1-\theta _1)^{1/2} \in {\cal F}_2$ . Hence, for $x\in {\cal F}$ , by (i) above and formula (e5.42),

(e5.56) $$ \begin{align} \hspace{-0.5pc} \theta x &=\left(\theta_1+(1-\theta_1)^{1/2}\theta_2(1-\theta_1)^{1/2}\right) x\approx_{2\eta/64} x\theta_1+(1-\theta_1)^{1/2} \theta_2 x(1- \theta_1)^{1/2} \end{align} $$

(e5.57) $$ \begin{align} & \hspace{1pc} \approx_{\eta/32} x\theta_1+(1-\theta_1)^{1/2}x(1-\theta_1)^{1/2}\theta_2 \approx_{\eta/32} x \theta_1+\theta_2(1-\theta_1)^{1/2}x(1-\theta_1)^{1/2} \end{align} $$

(e5.58) $$ \begin{align} &\approx_{\eta/32} x\theta_1+(1-\theta_1)^{1/2} x\theta_2(1-\theta_1)^{1/2} \hspace{12pc} \end{align} $$

(e5.59) $$ \begin{align} &\approx_{\eta/64} x\theta_1+ x(1-\theta_1)^{1/2}\theta_2(1-\theta_1)^{1/2}=\theta x. \hspace{10pc} \end{align} $$

Note that by formulas (e5.40) and (e5.35), in $W(A)$ we have

(e5.60) $$ \begin{align} m\langle \theta_2\rangle &=m\langle e_C\otimes (1_{\mathscr Z}- \psi(1_m))\rangle \hspace{11.5pc} \end{align} $$

(e5.61) $$ \begin{align} & \hspace{2pc} \le m\left\langle e_C\otimes \psi\left(e_{1,1}\right)\right\rangle \le \langle e_C\otimes \psi(1_m)\rangle\le \langle e_C\otimes 1_{\mathscr Z}\rangle\le l \langle a_1 \rangle. \end{align} $$

Therefore (recall that $W(A)$ is almost unperforated), since $l<m$ ,