Proof By Proposition 2.7, there exist projections $q_1,q_2,\ldots ,q_n$ such that $\langle q_i\rangle =x_i$ for any i. By Lemma 2.2,

$$ \begin{align*}[q_1]+[q_2]+\cdots+[q_n]\leq [p]. \end{align*} $$

Since A has cancellation of projections, with $v_1v_1^*=q_1$ and $v_1^*v_1\leq p$ , and setting ${p_1=v_1^*q_1v_1}$ , we have

$$ \begin{align*}[q_2]+\cdots+[q_n]\leq [p-p_1]. \end{align*} $$

There exists a partial isometry $v_2$ such that $ v_2v_2^*=q_2$ and $v_2^*v_2\leq p-p_1$ . Set $p_2=v_2^*v_2$ and continue this procedure; we obtain a collection of mutually orthogonal projections $\{p_i\}$ such that

$$ \begin{align*}\langle p_i \rangle=\langle q_i\rangle=x_i,\quad i=1,2,\ldots,n, \end{align*} $$

and

$$ \begin{align*}p_1+p_2+\cdots+p_n\leq p.\\[-30pt] \end{align*} $$

Proof Let us identify $\mathrm {Cu}(C(\Omega ))$ with the semigroup of lower semicontinuous functions $\mathrm {Lsc}(\Omega , \overline {\mathbb {N}})$ . Suppose that $d_{\mathrm {Cu}}(\alpha ,\beta )=0$ . We need only to show that $\alpha $ and $\beta $ agree on the functions for any open set $O\subset \Omega $ (their overall equality is apparent through the additivity and preservation of suprema of increasing sequences).

For any open set $O\subset \Omega $ , there exists a sequence of open subsets $O_n$ such that and $\overline {O_n}\subset O_{n+1}$ for any n. Since $O_n$ is bounded, there exists $r_n>0$ such that $(O_n)_{r_n}\subset O_{n+1}$ , and by the definition of $d_{\mathrm {Cu}}$ , we have and .

Then we have

Note that

which implies , as desired.

Proof Let $d=\mathop {\min }\limits _{\sigma \in S_n}\mathop {\max }\limits _{1\leq i\leq n}d_{\mathrm {Cu}}(\alpha _i,\beta _{\sigma (i)}).$ Then for any $\varepsilon>0$ , there exists $\sigma \in S_n$ such that

$$ \begin{align*}d_{\mathrm{Cu}}(\alpha_i,\beta_{\sigma(i)})<d+\varepsilon,\quad i=1,2,\ldots,n. \end{align*} $$

For any open set $O\subset \Omega $ , we get

Then we have

Hence,

$$ \begin{align*}d_{\mathrm{Cu}}\left(\sum_{i=1}^{n}\alpha_i,\sum_{i=1}^{n}\beta_i\right)\leq d+\varepsilon. \end{align*} $$

Since $\varepsilon $ is arbitrary, the conclusion follows.

Proof We may suppose that $\|x_n\|\leq M$ for all n, so that also $\|x\|\leq M$ . For any ${f\in C(\Omega )}$ and $\varepsilon>0$ , by the Stone–Weierstrass theorem, there exists a polynomial $P(z,\bar {z})$ such that

$$ \begin{align*}\|f-P(z,\bar{z})\|<\frac{\varepsilon}{3}. \end{align*} $$

Note that

$$ \begin{align*} \|(x_n^*)^s x_n^t-(x^*)^s x^t\| \leq & \|(x_n^*)^s x_n^t-x_n(x^*)^{s-1} x^t\|+ \|x_n(x^*)^{s-1} x^t-(x^*)^s x^t\|\\ \leq & M\|(x_n^*)^{s-1} x_n^t-(x^*)^{s-1} x^t\|+ M^{s+t}\|x_n-x\|. \end{align*} $$

By induction, we have

$$ \begin{align*}\|(x_n^*)^s x_n^t-(x^*)^s x^t\|\leq (s+t)M^{s+t}\|x_n-x\|. \end{align*} $$

Therefore, there exists $N_f$ such that if $\|x_n-x\|$ is sufficiently small for all $n\geq N_f$ , we will have

$$ \begin{align*}\|P(x_n,x_n^*)-P(x,x^*)\|<\frac{\varepsilon}{3}. \end{align*} $$

Now we have

$$ \begin{align*} \|f(x_n)-f(x)\| &\leq \|f(x_n)-P(x_n,x_n^*)\|+ \|P(x_n,x_n^*)-P(x,x^*)\|\\ & +\|P(x,x^*)-f(x)\|\\ &< \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon. \end{align*} $$

Since F is finite, $N:=\max \{N_f\mid f\in F\}$ is as desired.

Proof (1) Without loss of generality, we may assume that $x_n\rightarrow x$ . Suppose that ${X_n=\mathrm {sp}(x_n)}$ and $X=\mathrm {sp}(x)$ . We need to show that for any $\varepsilon>0$ , there exists $N\in \mathbb {N}$ such that

$$ \begin{align*}d_W(\varphi_{X_n},\varphi_X)<\varepsilon, \quad n\geq N. \end{align*} $$

Let $\delta =\varepsilon /2$ . Since $\Omega $ is compact, there is a finite open cover $\{\Omega _1,\Omega _2,\ldots ,\Omega _m\}$ of $\Omega $ with $\mathrm {diameter}(\Omega _i)\leq \delta $ , $i=1,2,\ldots ,m$ . Let $ \mathcal {F} $ denote the set of unions of some of the sets $\Omega _1,\Omega _2,\ldots ,\Omega _m$ . For any $Y\in \mathcal {F}$ , define

$$ \begin{align*}g_Y(z)= \begin{cases} 1-\mathrm{dist}(z,Y)/ \delta, & \mbox{if } z\in (Y)_\delta,\\ 0 & \mbox{otherwise}. \end{cases} \end{align*} $$

Set

$$ \begin{align*}F=\{g_{Y}(z)\mid Y\in\mathcal{F}\}. \end{align*} $$

Since F is finite, by Lemma 3.5, there exists $N\in \mathbb {N}$ such that

$$ \begin{align*}\|g(x_n)-g(x)\|<\delta,\quad \,g\in F,\,n\geq N. \end{align*} $$

Now for any open set $O\subset \Omega $ , let $Y_O=\mathop {\bigcup }\limits _{O\cap \Omega _i\neq \varnothing }\Omega _i$ . Then $Y_O\in \mathcal {F}$ and

$$ \begin{align*}O\subset Y_O\subset O_\delta\subset (Y_O)_\delta\subset O_{2\delta}. \end{align*} $$

Then we have

$$ \begin{align*}\varphi_{X_n}(f_O)\lesssim_{\mathrm{Cu}} \varphi_{X_n}(f_{Y_O}) \quad\mathrm{and}\quad \varphi_{X}(f_{Y_O})\lesssim_{\mathrm{Cu}} \varphi_{X}(f_{O_{2\delta}}). \end{align*} $$

Note that $g_{Y_O}\in F$ , so for all $n\geq N$ , we have

$$ \begin{align*}\|g_{Y_O}(x_n)-g_{Y_O}(x)\|<\delta. \end{align*} $$

It follows from Lemma 2.2(ii) that

$$ \begin{align*}(g_{Y_O}(x_n)-\delta)_+\lesssim_{\mathrm{Cu}} g_{Y_O}(x). \end{align*} $$

Note that $\mathrm {support}(g_{Y_O})={(Y_O)}_\delta ,$ so that $f_{Y_O}\lesssim _{\mathrm {Cu}} (g_{Y_O}-\delta )_+$ , and we get

$$ \begin{align*}f_{Y_O}(x_n) \lesssim_{\mathrm{Cu}}(g_{Y_O}(x_n)-\delta)_+\lesssim_{\mathrm{Cu}} g_{Y_O}(x)\lesssim_{\mathrm{Cu}} f_{{(Y_O)}_\delta}(x). \end{align*} $$

Therefore,

$$ \begin{align*}\varphi_{X_n}(f_{Y_O})=f_{Y_O}(x_n) \lesssim_{\mathrm{Cu}} f_{({Y_O})_\delta}(x)=\varphi_{X}(f_{{(Y_O)}_\delta}). \end{align*} $$

Now we have

$$ \begin{align*}\varphi_{X_n}(f_O)\lesssim_{\mathrm{Cu}} \varphi_{X_n}(f_{Y_O}) \lesssim_{\mathrm{Cu}} \varphi_{X}(f_{({Y_O})_\delta})\lesssim_{\mathrm{Cu}} \varphi_{X}(f_{O_{2\delta}}). \end{align*} $$

Similarly, for any open $O\subset \Omega $ , we also have

$$ \begin{align*}\varphi_{X}(f_O)\lesssim_{\mathrm{Cu}} \varphi_{X_n}(f_{O_{2\delta}}). \end{align*} $$

Finally, we obtain

$$ \begin{align*}d_W(\varphi_{X_n},\varphi_{X})< 2\delta=\varepsilon. \end{align*} $$

(2) Set $a=\phi (\mathrm {id})$ , $b=\psi (\mathrm {id})$ . By hypothesis, we have $d_W(a,b)=0$ and $\mathrm {ind}(a)=\mathrm {ind}(b)$ , and so by Theorem 3.6, we get $d_U(a,b)=0$ . This means that there exists a sequence of unitaries $u_n\in A$ such that $u_n^*au_n\rightarrow b$ . Then for any finite subset $F\subset C(\Omega )$ and $\varepsilon>0$ , by Lemma 3.5, there exists $N\in \mathbb {N}$ such that

$$ \begin{align*}\|f(u_n^*au_n)-f(b)\|<\varepsilon,\quad \, f\in F,\,n\geq N. \end{align*} $$

From the Stone–Weierstrass theorem, it can be checked that

$$ \begin{align*}f(u_n^*au_n)=u_n^*f(a)u_n, \quad \, f\in F. \end{align*} $$

Now we get

$$ \begin{align*}\|u_n^*\phi(f)u_n-\psi(f)\|<\varepsilon,\quad \,f\in F. \end{align*} $$

Since $\varepsilon $ is arbitrary, we have $\phi \sim _{aue}\psi $ .

Proof From the definition of $\mathrm {Cu}(A)$ and Lemma 2.7, there exists $a\in A_+$ such that $a\leq 1_A$ and $\langle a\rangle =x$ . By the simplicity of A, there exist $a_1,a_2,\ldots ,a_k$ in A such that $ 1_A=\sum _{i=1}^{k}a_i^*aa_i. $ Then for any $\tau \in T(A)$ ,

$$ \begin{align*}1=\tau(1_A)=\sum_{i=1}^{k}\tau(a_i^*aa_i)=\sum_{i=1}^{k}\tau(a^{1/2}a_i^*a_ia^{1/2})\leq \sum_{i=1}^{k}\|a_i^*a_i\|\cdot \tau(a). \end{align*} $$

Now we get $\tau (a)>0$ , whence from the compactness of $T(A)$ and

$$ \begin{align*}d_\tau(x)\geq \tau(a),\end{align*} $$

we get $\inf \limits _{\tau \in T(A)}d_\tau (x)>0.$

Proof For any open set $O\subset \Omega $ and $\tau \in T(A)$ , let $\mu _{\alpha ^*\tau }$ be the countably additive measure on $\Omega $ such that

If , the proof is trivial. In general, we have $\mu _{\alpha ^*\tau }({B(x,r)})\leq 1$ . Since $R(x,s)\cap R(x,s')=\varnothing $ , if $s\neq s'$ , there are at most finitely many s in $(r/2,r)$ such that

$$ \begin{align*}\mu_{\alpha^*\tau}(R(x,s))>\sigma/2.\end{align*} $$

Since we have ${QT}_2(A)=T(A)$ , by Remark 2.4, ${ QT}_2(A)$ is compact metrizable and has a countable basis, and so we may choose a countable dense subset Y of ${ QT}_2(A)$ .

For any $\tau \in Y$ , we define

$$ \begin{align*}\mathcal{S}_\tau=\{s\,|\, \mu_{\alpha^*\tau}(R(x,s))>\sigma/2\}. \end{align*} $$

Then $\bigcup _{\tau \in Y} \mathcal {S}_{\tau }$ has at most countably many points and

$$ \begin{align*}(r/2,r) \backslash\bigcup_{\tau\in Y} \mathcal{S}_{\tau}\neq \varnothing. \end{align*} $$

Now there exist an $s\in (r/2,r)$ such that $ \mu _{\alpha ^*\tau }(R(x,{s}))\leq \sigma /2$ , i.e.,

$$ \begin{align*}\mu_{\alpha^*\tau}(\Omega\backslash R(x,s))\geq 1-\sigma/2,\quad\, \tau\in Y. \end{align*} $$

That is,

By the density of Y and Theorem 2.5, we have

Let $\{\varepsilon _n\}$ be a strictly decreasing sequence such that

$$ \begin{align*}\varepsilon_n\leq \min\{s-r/2,r-s\},\,\, n=1,2\ldots,\quad \mathrm{and }\quad \lim_{n\rightarrow\infty}\varepsilon_n=0. \end{align*} $$

The sequence is increasing in $\mathrm {Cu}(C(\Omega ))$ with supremum .

Since $\alpha $ and $d_\tau $ preserve the suprema of increasing sequences,

For any $\tau \in { QT}_2(A)$ , by [Reference Dadarlat and Toms14, Lemma 3.1], there exist an integer $N_{\tau }\in \mathbb {N}$ and an open neighborhood $V_{\tau }$ of $\tau $ such that

Then $\{V_\tau \,|\,\tau \in { QT}_2(A)\}$ forms an open cover of ${ QT}_2(A)$ , and so from the compactness of ${ QT}_2(A)$ , there are finitely many sets $\{V_{\tau _1},V_{\tau _2},\ldots ,V_{\tau _k}\}$ covering ${ QT}_2(A)$ . Now we set

$$ \begin{align*}N_0=\max\{N_{\tau_1},N_{\tau_2},\ldots,N_{\tau_k}\}. \end{align*} $$

For any $n\geq N_0$ , we have

Then for any $0<\varepsilon \leq \varepsilon _{N_0}$ , we have $s\pm \varepsilon \in (r/2,r)$ and

Proof By Proposition 4.1, we set

For each $i\in \{1,2,\ldots ,m\}$ , by Lemma 4.2 for $x_i$ , $\delta /2$ , and $\sigma /(2m+1)$ , there exist ${s_i\in (\delta /4,\delta /2)}$ and $\varepsilon _i$ such that $s_i\pm \varepsilon _i\in (\delta /4,\delta /2)$ and

$$ \begin{align*}\mu_{\alpha^*\tau}(R(x_i,s_i)_{\varepsilon_i})\leq \frac{\sigma}{2m+1}<\frac{\sigma}{2m},\quad \tau\in { QT}_2(A). \end{align*} $$

Set $R= \bigcup _{i=1}^m R(x_i,s_i)$ , then

$$ \begin{align*}\mu_{\alpha^*\tau}(R)\leq \sum_{i=1}^{m}\mu_{\alpha^*\tau}(R(x_i,s_i)_{\varepsilon_i})<\frac{\sigma}{2m}\cdot m=\frac{\sigma}{2}. \end{align*} $$

Since $\Omega \backslash R$ is open, there exists a positive function $f_{\Omega \backslash R}\in C(\Omega )$ corresponding to $\Omega \backslash R$ (see 2.8) such that

$$ \begin{align*}d_\tau(\alpha (\langle{f}_{\Omega\backslash R}\rangle))=\mu_{\alpha^*\tau}(\Omega\backslash R)>1-\frac{\sigma}{2},\quad \tau \in { QT}_2(A). \end{align*} $$

Let $\{\sigma _n\}$ be a strictly decreasing sequence such that

$$ \begin{align*}\sigma_n\leq \min\{\varepsilon_1,\varepsilon_2,\ldots,\varepsilon_m\},\,\, n=1,2,\ldots,\quad \mathrm{and }\quad \lim_{n\rightarrow\infty}\sigma_n=0. \end{align*} $$

Set

$$ \begin{align*}W_n=\mathrm{supp}\{(f_{\Omega\backslash R}-\sigma_n)_+\}. \end{align*} $$

Then is an increasing sequence in Cu $(C(\Omega ))$ with supremum . Since $\alpha $ preserves suprema, we have

Hence,

Since ${ QT}_2(A)$ is compact, with a similar method of the proof of Lemma 4.2, there exists $N_0$ such that

Now for fixed integers $n_0>n_1> N_0$ , we have $\sigma _{n_0}<\sigma _{n_1}$ and

$$ \begin{align*}W_{n_0}\cup R_{\sigma_{n_0}}\cup\{x \mid f_{\Omega\backslash R}(x)=\sigma_{n_0}\}=W_{n_1}\cup R_{\sigma_{n_1}}\cup\{x \mid f_{\Omega\backslash R}(x)=\sigma_{n_1}\}=\Omega. \end{align*} $$

As $W_{n_0}\supset W_{n_1}$ , we then have

$$ \begin{align*}\{x \mid f_{\Omega\backslash R}(x)=\sigma_{n_0}\}\subset R_{\sigma_{n_1}}\subset\bigcup_{i=1}^m R(x_i,s_i)_{\varepsilon_i}. \end{align*} $$

Now we set

$$ \begin{align*}\eta:\,= \sigma_{n_0},\quad U:\,=W_{n_0}. \end{align*} $$

Note that

We also have

$$ \begin{align*}U\cup R_{\eta}\cup \{x\mid f_{\Omega\backslash R}(x)=\eta\}=\Omega\subset U\cup \bigcup_{i=1}^m R(x_i,s_i)_{\varepsilon_i}. \end{align*} $$

Define

$$ \begin{align*}O_1:=U\cap B(x_1,s_1) , \end{align*} $$

$$ \begin{align*}O_2: =(U\backslash O_1) \cap B(x_2,s_2), \end{align*} $$

$$ \begin{align*}\cdots \end{align*} $$

$$ \begin{align*}O_m:= (U\backslash\cup_{i=1}^{m-1}O_i) \cap B(x_m,s_m). \end{align*} $$

Note that all the $O_i$ are open sets in U. Let us delete the empty sets and rewrite those remaining as $\{O_1,O_2, \ldots ,O_N\}$ ; then $ U=\bigcup _{i=1}^N O_i. $

Let us now show that $\{O_1, O_2, \ldots ,O_N\}$ is an almost $\delta $ -cover with respect to $\alpha $ . For any $x\in \Omega $ , if $x\in U$ , it is trivial that $\mathrm {dist}(x,U)=0$ ; if $x\in \bigcup _{i=1}^m R(x_i,s_i)_{\varepsilon _i}$ , there exists $i_0$ such that $x\in B(x_{i_0},\delta /2)$ . Since

$$ \begin{align*}\sum_{i=1}^{m}\mu_{\alpha^*\tau}(R(x_i,s_i)_{\varepsilon_i}) <\frac{\sigma}{2}<\mu_{\alpha^*\tau} \left(B\left(x_{i_0},\frac{\delta}{2}\right)\right)\!, \end{align*} $$

we have $B(x_{i_0},\delta /2)\backslash \bigcup _{i=1}^m R(x_i,s_i)_{\varepsilon _i}\neq \varnothing $ , and hence, there exists $y\in B(x_{i_0},\delta /2)\cap U$ such that $ \mathrm {dist}(x,y)<\delta $ . From the construction of $O_i$ , for any $i\geq 1$ , $O_i$ is contained in $B(x_j,s_j)$ for some j, and so diameter $(O_i)\leq \delta $ . If $i\neq j$ , then $O_i$ and $O_j$ can be separated by $R_\eta $ , and so dist $(O_i,O_j)>0$ . Then (i)–(iii) hold.

Now we check (iv). Given any $O_i$ , there exists j such that

Set

$$ \begin{align*}Y_1:=B\left(x_j,\frac{\delta}{2}\right)\cap U,\quad Y_2:= B\left(x_j,\frac{\delta}{2}\right)\cap\bigcup_{i=1}^m R(x_i,s_i)_{\varepsilon_i}. \end{align*} $$

Then we have

$$ \begin{align*}Y_1 \subset B\left(x_j,\frac{\delta}{2}\right)\subset Y_1\cup Y_2\subset (O_i)_\delta. \end{align*} $$

Recall that

Then

and hence,

Now we have

Since

$$ \begin{align*}\Omega\backslash \overline{U}\subset \bigcup_{i=1}^m R(x_i,s_i)_{\varepsilon_i}, \end{align*} $$

we have

Since A has strict comparison, by Proposition 2.6 and the inclusion $Y_1\subset (O_i)_\delta \cap U$ , we have

Finally, notice that for any i, we have shown that $B(x_{i},\delta /2)\cap U\neq \varnothing $ , and then $B(x_{i},\delta /2)\subset (O_{j_i})_\delta $ for some $j_i$ . Combining this with assumption (3), $\{(O_{j_1})_\delta ,(O_{j_2})_\delta ,\ldots , (O_{j_m})_\delta \}$ is also an open cover of $\Omega $ and has a unique almost connected component. Note that $\Omega =\bigcup _{i=1}^m(O_{j_i})_\delta =\bigcup _{k=1}^NO_{k}$ , then for any $k\in \{1,2,\ldots ,N\}$ , there exists $(O_{j_i})_\delta $ such that $O_k\cap (O_{j_i})_\delta \neq \varnothing $ . Thus any two elements in $\{(O_{1})_\delta ,(O_{2})_\delta ,\ldots ,(O_{N})_\delta \}$ are almost connected through $\{(O_{j_1})_\delta ,(O_{j_2})_\delta ,\ldots , (O_{j_m})_\delta \}$ . In general, $\{(O_{1})_\delta , (O_{2})_\delta ,\ldots ,(O_{N})_\delta \}$ has a unique almost connected component, that is, (v) holds.

Proof Suppose that $\{O_1,O_2,\ldots , O_N\}$ is an almost $\delta $ -cover respect to $\alpha $ .

Let

$$ \begin{align*}U=\bigcup_{i=1}^N O_i,\,\,\,\rho=\frac{1}{4}\min\{\delta,\mathrm{dist}(O_i,O_j),\, i\neq j, \,1\leq i,j\leq N\}. \end{align*} $$

Then the facts that and $\alpha $ preserves the compact containment relation imply that

Since $\mathrm {Cu}(A)$ is algebraic (see 2.9), for each i, there exists an increasing sequence of compact elements $\{x_i^n\}_n$ with supremum . From the compact containment relation, there exists $n_i\in \mathbb {N}$ such that . For convenience, we use $x_i$ to denote $x_i^{n_i}$ ; then,

Now we have

By Proposition 2.8, there exists a collection of mutually orthogonal projections $\{p_i\}$ such that

$$ \begin{align*}\langle p_i \rangle=x_i,\quad i=1,2,\ldots,N \end{align*} $$

and

$$ \begin{align*}p_1+p_2+\cdots+p_N\leq p. \end{align*} $$

Set $p_0=p-\sum _{i=1}^{N}p_i$ . Note that

and

By weak cancellation in $\mathrm {Cu}(A)$ (Definition 2.5), we have

Now choose $z_0\in \Omega \backslash \overline {U}$ and $z_i\in O_i$ ( $1\leq i\leq N$ ). Define

$$ \begin{align*}\phi(f)= \sum_{i=0}^{N}f(z_i)p_i,\quad f\in C(\Omega). \end{align*} $$

Then we need to show $d_{\mathrm {Cu}}(\mathrm {Cu}(\phi ),\alpha )<9\delta $ .

For any nonempty open set $V\subset \Omega $ , we have $V_\delta \cap U\neq \varnothing $ . Now we consider the following two cases.

Case 1: There exists $k\in \{1,2,\ldots ,N\}$ such that $ O_k\subset V_{8\delta }\backslash V_{3\delta }. $

Define index sets

$$ \begin{align*}I_0=\{i\mid V\cap (O_i)_\rho\neq \varnothing,\,1\leq i\leq N\},\end{align*} $$

$$ \begin{align*}I_1=\{i\mid O_i\cap (O_k)_\delta\neq \varnothing,\,1\leq i\leq N\}. \end{align*} $$

If $i\in I_1$ , then $O_i\cap V_{2\delta }=\varnothing $ , we have $I_0\cap I_1=\varnothing $ . We also note that

$$ \begin{align*}\bigcup_{i\in I_0} (O_i)_{2\rho}\cup (\bigcup_{i\in I_1} O_i)\subset V_{9\delta}. \end{align*} $$

Then we have

Note that

$$ \begin{align*}V\subset (V\cap \, \Omega\backslash \overline{U})\cup (V\cap U_\rho).\end{align*} $$

Now we have

Case 2: There doesn’t exist $k\in \{1,2,\ldots ,N\}$ such that $ O_k\subset V_{8\delta }\backslash V_{3\delta }. $

In this case, $U\cap (V_{7\delta }\backslash V_{4\delta })=\varnothing $ . Now we define index sets

$$ \begin{align*}J_0=\{i\mid O_i\subset V_{4\delta},\,1\leq i\leq N\},\end{align*} $$

$$ \begin{align*}J_1=\{i\mid O_i\subset \Omega\backslash V_{7\delta},\,1\leq i\leq N\}. \end{align*} $$

Thus, we have $J_0\cup J_1=\{1,2,\ldots ,N\}$ and $J_0\cap J_1=\varnothing $ .

By (i), if $\Omega \backslash { V_{6\delta }}\neq \varnothing $ , then $J_1\neq \varnothing $ . Then for arbitrary $i\in J_0, i'\in J_1$ , we have $\mathrm {dist}((O_{i})_\delta , (O_{i'})_\delta )>\delta $ , this means that $(O_{i})_\delta $ and $(O_{i'})_\delta $ can’t be almost connected, and this contradicts (v). Then we must have $V_{6\delta }=\Omega $ . It is clear that

and

Combining these two cases, we have

$$ \begin{align*}d_{\mathrm{Cu}}(\mathrm{Cu}(\phi),\alpha)<9\delta.\\[-35pt] \end{align*} $$

Proof Since $\Omega $ is compact, for $\delta =\varepsilon /9$ , there exist $x_1,x_2,\ldots ,x_m\in \Omega $ such that

$$ \begin{align*}\Omega=\bigcup_{i=1}^mB(x_i,{\delta/4}). \end{align*} $$

Denote

$$ \begin{align*}\Lambda=\{1,2,\ldots,m\}, \end{align*} $$

Then $\mathcal {F}$ has finitely many almost connected (Definition 4.1) components $\mathcal {F}_1,\ldots ,\mathcal {F}_{l}$ .

For each $i\in \{1,2,\ldots ,l\}$ , we also define

$$ \begin{align*}\Lambda_i=\{j\mid B(x_j,{\delta/4})\in \mathcal{F}_{i}\},\,\,\Lambda_0=\Lambda\backslash \bigcup_{i=1}^l \Lambda_i, \end{align*} $$

$$ \begin{align*}\Omega_i= \bigcup_{j\in \Lambda_i}B(x_j,{\delta/4}),\,\, \Omega_0=\bigcup_{j\in \Lambda_0} B(x_j,{\delta/4}). \end{align*} $$

(One may say that $\Omega _1,\ldots ,\Omega _l$ are “separated” by $\Omega _0$ .)

Since $\Omega _i\cap \Omega _j\neq \varnothing $ for all $i,j\in \{1,2,\ldots ,l\}$ with $i\neq j$ , we have

Thus,

Now we will prove that is compact for each $i\in \{1,2,\ldots ,l\}$ .

For each i, let $\{a_{n,i}\}_n$ be a $\ll $ -increasing sequence in $\mathrm {Lsc}(\Omega ,\overline {\mathbb {N}})$ with supremum . Set $b_{n}=a_{n,1}+a_{n,2}+\cdots +a_{n,l},$ then $\{b_{n}\}_n$ is also a $\ll $ -increasing sequence with supremum

$$ \begin{align*}\sup_n b_{n}= \sup_n a_{n,1}+ \sup_n a_{n,2}+\cdots+\sup_n a_{n,l}.\end{align*} $$

Since $\alpha $ preserves suprema of increasing sequences, then we have

From the compactness of , there exists $k\in \mathbb {N}$ such that , i.e.,

Since we have (in $\mathrm {Lsc}(\Omega ,\overline {\mathbb {N}})$ ) for any $m=1,2,\ldots ,l$ , then

Since is compact, we also have

From the weak cancellation of $\mathrm {Cu}(A)$ , we have

This means that is compact in $\mathrm {Cu}(A)$ .

Since we have , by Proposition 2.8, there exists a collection of mutually orthogonal projections $\{p_i\}$ such that

and

$$ \begin{align*}p_1+p_2+\cdots+p_l\leq 1_{A}. \end{align*} $$

Let $h(t)\in \mathrm {Lsc}(\Omega ,\overline {\mathbb {N}})$ . For any open set $V\subset \Omega $ , define

$$ \begin{align*}h|_{V}(t)= \begin{cases} h(t), & \mbox{if } t\in V \\ 0, & \mbox{if}\,\, t\notin V. \end{cases} \end{align*} $$

For each $i\in \{1,2,\ldots ,l\}$ , define $\alpha _i$ as follows:

$$ \begin{align*}\alpha_i(h(t))=\alpha(h|_{\Omega_i}(t)). \end{align*} $$

It can be checked that $\alpha _1,\alpha _2,\ldots ,\alpha _l$ are Cu-morphisms from $\mathrm {Lsc}(\Omega ,\overline {\mathbb {N}})$ to $\mathrm {Cu}(A)$ . We also have

$$ \begin{align*}\alpha_1+\alpha_2+\cdots+\alpha_l=\alpha. \end{align*} $$

For each i, we apply Lemmas 4.3 and 4.4 for $\Omega _i$ , $\delta $ , $p_i$ and $\alpha _i$ (the key point is that is compact); this gives $\phi _i:C(\Omega )\rightarrow p_iAp_i$ such that

$$ \begin{align*}d_{\mathrm{Cu}}(\mathrm{Cu}(\phi_i),\alpha_i)<9\delta. \end{align*} $$

Denote $ \phi =\sum _{i=1}^{l}\phi _i. $ Since $\phi _1,\phi _2, \ldots ,\phi _l$ have mutually orthogonal ranges, we have

$$ \begin{align*}\mathrm{Cu}(\phi_1)+\mathrm{Cu}(\phi_2)+\cdots+\mathrm{Cu}(\phi_l)=\mathrm{Cu}(\phi). \end{align*} $$

By Proposition 3.4, we obtain

$$ \begin{align*}d_{\mathrm{Cu}}(\mathrm{Cu}(\phi),\alpha)<9\delta=\varepsilon.\\[-34pt] \end{align*} $$

Proof From Theorem 4.5, there exists a sequence of homomorphisms $\phi _n$ with finite dimensional range such that $d_{\mathrm {Cu}}(\mathrm {Cu}(\phi _n),\alpha )\rightarrow 0$ . Let $x_n=\phi _n(\mathrm {id})$ and $\varepsilon>0$ . As the range of $\phi _n$ is finite dimensional, we have $[\lambda -x_n]=0$ in $K_1(A)$ for all $\lambda \notin \operatorname {sp}(x_n)$ . By Theorem 3.6 and Remark 2.5, there exists $N _1>0$ such that

$$ \begin{align*}d_U(x_{n},x_{m})\leq 2d_W(x_{n},x_{m})<\frac{\varepsilon}{2},\quad n,m\geq N_1. \end{align*} $$

Then for $\varepsilon / 2^2$ , there exists $N_2> N_1$ such that

$$ \begin{align*}d_U(x_{n},x_{m})<\frac{\varepsilon}{2^2},\quad n,m\geq N_2. \end{align*} $$

Similarly, for any k, there exists $N_k> N_{k-1}$ such that

$$ \begin{align*}d_U(x_{n},x_{m})<\frac{\varepsilon}{2^k},\quad ,n,m\geq N_k. \end{align*} $$

Then for each $k\geq 1$ , there exists a unitary $u_k\in A$ such that

$$ \begin{align*}\|x_{N_k}- u_k^* x_{N_{k+1}}u_k\|<\frac{\varepsilon}{2^k}.\end{align*} $$

Write

$$ \begin{align*} \widetilde{x}_1 &=: x_{N_1}, \\\widetilde{x}_2 &=: u_1^*x_{N_2}u_1, \\&\kern5pt \vdots \\\widetilde{x}_k &=: (u_{k-1}\cdots u_2u_1)^*x_{N_k}u_{k-1}\cdots u_2u_1,\\&\kern5pt \vdots \end{align*} $$

Then $\{\widetilde {x}_k\}$ is a Cauchy sequence. We may assume that $\widetilde {x}_k\rightarrow x$ . Note that all the $\widetilde {x}_k$ and x are normal and $\sigma (\widetilde {x}_k),\sigma (x)\subset \Omega $ .

Define $\phi :C(\Omega )\rightarrow A$ by $\phi (f)=f(x)$ . By Lemma 3.7(i), we have

$$ \begin{align*}d_W(\phi_{N_k},\phi)=d_W(x_{N_k},x)=d_W(\widetilde{x}_k,x)\rightarrow 0. \end{align*} $$

From the properties of $d_{\mathrm {Cu}}$ (see 3.3), we have

$$ \begin{align*} d_{\mathrm{Cu}}(\mathrm{Cu}(\phi),\alpha) &\leq d_{\mathrm{Cu}}(\mathrm{Cu}(\phi_{N_k}),\alpha)+d_{\mathrm{Cu}}(\mathrm{Cu}(\phi_{N_k}),\mathrm{Cu}(\phi)) \\&= d_{\mathrm{Cu}}(\mathrm{Cu}(\phi_{N_k}),\alpha)+d_W(\phi_{N_k},\phi) \rightarrow 0. \end{align*} $$

Then the $*$ -homomorphism $\phi : C(\Omega ) \rightarrow A$ satisfies $d_{\mathrm {Cu}}(\mathrm {Cu}(\phi ),\alpha )=0$ , and so by Proposition 3.3, we have $\alpha =\mathrm {Cu}(\phi )$ .

Suppose that $\psi : C(\Omega ) \rightarrow A$ also satisfies $\mathrm {Cu}(\psi )=\alpha $ . As $K_1(A)$ is trivial, we obtain $\mathrm {ind}(\phi (\mathrm {id}))=\mathrm {ind}(\psi (\mathrm {id}))$ . By Lemma 3.7 (ii), we obtain $\phi \sim _{aue}\psi $ . Thus, $\phi $ is unique up to approximate unitary equivalence.

Proof See the proof of [Reference Ciuperca, Elliott and Santiago11, Corollary 1.2].

Proof Since A has real rank zero and stable rank one, $\mathrm {Cu}(A)$ is algebraic (Theorem 2.9) and has weak cancellation (Definition 2.5). Moreover, $\mathrm {Cu}(A)$ is the limit of Cu-semigroups of the form $\mathrm {Lsc}(X,\overline {\mathbb {N}})$ , and so $\mathrm {Cu}(A)$ is unperforated (because each $\mathrm {Lsc}(X,\overline {\mathbb {N}})$ is). By [Reference Antoine, Perera and Thiel2, Corollary 5.5.13], there exists an AF algebra B such that $\mathrm {Cu}(A)\cong \mathrm {Cu}(B)$ . Then by Corollary 5.8, this lifts to an isomorphism $A \cong B.$