Proof. In a nontrivial convex decomposition of \(\widetilde h\), both restrictions to \(B\) equal \(h\) by purity. Uniqueness then makes both states equal to \(\widetilde h\). ◻

Proof. Realize the two states by unit vectors \(\xi,\eta\) in a common irreducible representation \(\pi\). On their finite-dimensional span choose a unitary carrying \(\xi\) to \(\eta\) and a self-adjoint logarithm of that unitary. Kadison transitivity, including its self-adjoint form (Kadison 1957, Theorem 1), supplies a self-adjoint \(a\in B\) whose action on this span is the chosen logarithm. The span reduces \(\pi(a)\), so \(u=e^{ia}\) satisfies \(\pi(u)\xi=\eta\). If \(h\) is represented by \(\xi\) and \(k\) by \(\eta\), then \(k=h\circ\mathop{\mathrm{Ad}}(u^*)\). Conjugating their unique extensions by the same unitary proves the last assertion. ◻

Proof. Every state of \(B\) satisfying the projection constraints extends to a state of \(C\) with the same constraints, hence to \(H\). It must therefore equal \(h\). Every extension of \(h\) also satisfies the constraints and must equal \(H\). ◻

Proof. Put \[L=\{a\in B:h(a^*a)=0\},\qquad N=L\cap L^*.\] Here \(L\) is a closed left ideal and \(N\) is a \(C^*\)-subalgebra. If \(a\in L\), then \(a^*a\in N\), since \(h((a^*a)^2)\le\|a\|^2h(a^*a)=0\).

First, \(h\) is the only state annihilating \(N\). If another state \(k\) annihilates \(N\), then \(L\) is contained in the kernel of \(a\mapsto\pi_k(a)\xi_k\). The map \(Q_h:B\to H_h\), \(Q_h(a)=\pi_h(a)\xi_h\), is onto by one-vector Kadison transitivity. This assertion concerns the whole Hilbert space, even when it is nonseparable. Since \(\ker Q_h=L\), the open mapping theorem for \(B/L\to H_h\) shows that \[T\bigl(\pi_h(a)\xi_h\bigr)=\pi_k(a)\xi_k\] defines a bounded operator \(T:H_h\to H_k\). It intertwines the representations. Thus \(T^*T\) is scalar by irreducibility of \(\pi_h\); its value on \(\xi_h\) shows that \(T^*T=1\). Taking vector coefficients gives \(k=h\).

We next construct the decreasing net. Order \(N_+\) by the usual operator order, a directed order, and set \[e_s=1-(1+s)^{-1},\qquad x_s=1-e_s \qquad(s\in N_+).\] Functional calculus gives \(e_s\in N\), and operator monotonicity of inversion makes \((e_s)\) increasing. For \(b\in N\) and \(s\ge m(bb^*+b^*b)\), \[\|(1-e_s)b\|^2 \le \frac1m\|(1+s)^{-1}s(1+s)^{-1}\| \le\frac1{4m};\] the same estimate holds for \(\|b(1-e_s)\|^2\). Thus \((e_s)\) is an approximate identity of \(N\), and \(h(x_s)=1\).

Let \(q\) be the strong limit of \((e_s)\), the support projection of \(N\). Every normal state supported on \(1-q\) annihilates \(N\), and hence is \(h\); conversely, \(h(1-q)=1\). The corner \((1-q)B^{**}(1-q)\) therefore has exactly one normal state. Normal states separate its self-adjoint elements, so this corner is scalar. It follows that \(1-q=r_h\) is minimal and \(x_s\downarrow r_h\). The identity \(r_har_h=h(a)r_h\) follows by applying \(h\) to this scalar corner.

Finally, let \(\rho_s\) be arbitrary states of \(B\). Every weak-star cluster point of the positive functionals \(a\mapsto\rho_s(x_sax_s)\) annihilates \(N\), by the approximate-identity estimates. A nonzero cluster point, after normalization, is therefore \(h\). For self-adjoint \(a\), a failure of (1) would give a subnet and norming states \(\rho_s\) with a cluster point nonzero on \(a-h(a)1\), a contradiction. Real and imaginary parts give the conclusion for arbitrary \(a\). ◻

Proof. The infimum of the \(P_n\) in \(B^{**}\) supports exactly one normal state, namely \(h\). As in the preceding proof, it is the minimal projection \(r_h\). For arbitrary states \(\rho_n\), every cluster point \(\psi\) of \(\rho_n(P_n\,\cdot\,P_n)\) vanishes on \(1-P_m\) for each fixed \(m\). If \(\psi\ne0\), its normalization satisfies every peaking constraint and therefore equals \(h\). The same norming-state argument proves excision. For the last assertion use nesting: \[\|P_n a P_t\|\le \|P_{\min(n,t)}aP_{\min(n,t)}\|. \qedhere\] ◻

Proof. The normal extension of \(\pi_h\) annihilates \(r_k\). Otherwise a unit vector in the range of \(\pi_h^{**}(r_k)\) would represent \(k\), by \(r_kar_k=k(a)r_k\). Every nonzero vector in an irreducible representation is cyclic, so this would imply \(h\sim k\).

Consequently \(h(ay_{\nu_0}a^*)\to0\) as \(\nu_0\) increases. For \(\nu\ge\nu_0\), the inequalities \(y_\nu^2\le y_\nu\le y_{\nu_0}\) give \[\begin{align*} \|x_\lambda a y_\nu\|^2 &\le \|x_\lambda a y_{\nu_0}a^*x_\lambda\|\\ &\le h(ay_{\nu_0}a^*) +\|x_\lambda(ay_{\nu_0}a^*-h(ay_{\nu_0}a^*)1)x_\lambda\|. \end{align*}\] Choose \(\nu_0\) first, then use excision to choose \(\lambda_0\). This proves smallness on a rectangular tail and hence the product-net limit. Lemma 6 gives the stated projection case. ◻

Proof. The space \(H_h\) is infinite-dimensional. Choose orthonormal unit vectors \(\eta_j\) and unitaries \(u_j\in B\) with \(\pi_h(u_j)\xi_h=\eta_j\), as in Lemma 3. The projections \(u_jr_hu_j^*\) are pairwise orthogonal, since \[r_hu_i^*u_jr_h=h(u_i^*u_j)r_h=0\qquad(i\ne j).\] They all have trace \(\tau(r_h)\), so \(m\tau(r_h)\le1\) for every positive integer \(m\). ◻

Proof. For such an expectation, bimodularity gives \(\tau(b^*E(a))=\tau(b^*a)\) for every \(b\in B\). Two expectations therefore have difference \(d\in B\) satisfying \(\tau(d^*d)=0\), so they agree. Faithfulness follows from \(\tau(E(a^*a))=\tau(a^*a)\). ◻

Proof. The \(d_n\) are nonzero orthogonal projections. Their closed linear span is isomorphic to \(c_0(\mathbb N_0)\), and adjoining the unit gives the claimed description of \(D\). Indeed \(p_n=1-\sum_{j<n}d_j\) and \(\|p_n\|=1\), since \(\tau(p_n)>0\). Multiplying by \(p_n\) shows that \(1\) has distance at least one from every linear combination of \(d_0,\ldots,d_{n-1}\).

The values in (3) define a continuous function: by \(d_n\le p_n\) and Lemma 6, \[\left|\frac{\tau(d_nad_n)}{\tau(d_n)}-h(a)\right| \le \|p_n(a-h(a)1)p_n\|\longrightarrow0.\] Every coordinate of \(E\) is a state, so \(E\) is unital and completely positive: positivity of each matrix amplification is tested pointwise in the commutative range. The map fixes \(D\). To check bimodularity, an element of \(D\) acts scalarly on each \(d_n\); at infinity, \(h|_D\) is the evaluation character. Cauchy–Schwarz for \(h\) then gives \(h(ba)=h(b)h(a)\) and \(h(ab)=h(a)h(b)\) for \(b\in D\). These coordinate identities prove bimodularity.

Finally, \(\sum_n\tau(d_n)=1\) and the trace on \(D\) has no mass at infinity. Summing the finite coordinates gives \(\tau(E(a))=\tau(a)\), since the remaining trace satisfies \(|\tau(p_Na)|\le\|a\|\tau(p_N)\to0\). ◻

Proof. Let \(J\) be a proper closed two-sided ideal of \(C\). The quotient map \(q:C\to C/J\) is injective on \(B\), since \(B\) is simple and \(q(1)\ne0\). Every pure state \(h\) of \(B\) therefore extends through this quotient to a state of \(C\) annihilating \(J\). By [eq:U], this extension is \(h\circ E\). Thus \(h(E(a))=0\) for every positive \(a\in J\) and every pure \(h\) on \(B\). Pure states separate positive elements, so \(E(a)=0\); faithfulness gives \(a=0\). Hence \(J=0\). ◻

Proof. For \(i\in\{F,G\}\), complete \(i\) as a right Hilbert \(D\)-module \(L_i\) with inner product \[\langle x,y\rangle_D=E_D^i(x^*y).\] Left multiplication by \(a\in i\) is bounded, since \[E_D^i(x^*a^*ax)\leq\|a\|^2E_D^i(x^*x),\] and its adjoint is left multiplication by \(a^*\). The expectation gives an orthogonal decomposition \[L_i=D\oplus L_i^\circ, \qquad L_i^\circ=\overline{\ker E_D^i}.\] Both summands are invariant under left and right multiplication by \(D\). Form the Hilbert \(D\)-module \[ K=D\oplus \bigoplus_{\substack{m\geq1,\ i_t\in\{F,G\}\\i_t\ne i_{t+1}}} L_{i_1}^\circ\otimes_D\cdots\otimes_D L_{i_m}^\circ. \tag{7}\] Here \(\otimes_D\) is the interior tensor product of Hilbert modules; the left \(D\)-action on a word acts on its first factor. The first summand \(D\) is called the vacuum summand, with distinguished vector \(1\).

Let \(K^{(i)}\) be the vacuum summand together with all words not starting with \(i\). The decomposition of \(L_i\) identifies \[K=L_i\otimes_D K^{(i)}.\] Represent \(i\) by left multiplication on the first tensor factor. The two representations agree on \(D\). They are faithful: an element \(a\) annihilating the vacuum satisfies \(E_D^i(a^*a)=0\), so its faithful factor trace gives \(a=0\). Let \(Z\) be the \(C^*\)-algebra they generate in the adjointable operators on \(K\).

A centered element of \(i\) acting on \(K^{(i)}\) adds its own tensor factor at the front. Thus a nonempty alternating centered product has zero vacuum compression. Centering arbitrary factor elements shows that \(D\) and these products span a dense subspace of \(Z\). Vacuum compression consequently takes values in the left copy of \(D\) and defines the conditional expectation \(E_D\).

Write \(\tau_D\) for the common trace on \(D\). We next prove that \(\tau_Z=\tau_D\circ E_D\) is tracial and faithful. For traciality, compare \(\tau_Z(aw)\) and \(\tau_Z(wa)\) when \(a\) belongs to one factor and \(w\) is an alternating centered word. If \(w\) has length at least two, both values vanish. Indeed, multiplication at a same-factor end, followed by centering that end product, leaves either an alternating centered word or a \(D\)-coefficient times a nonempty such word. At a different-factor end, first center \(a\) and use \(D\)-bimodularity. For a one-letter word, use the factor trace when the factors agree; when they differ, both values are zero. For \(w\in D\), use the factor trace. Density and repeated cyclic movement of a factor now prove traciality on all of \(Z\).

Suppose \(\tau_Z(z^*z)=0\). For every \(t\in Z\), traciality gives \[0\leq\tau_Z(t^*z^*zt) =\tau_Z(ztt^*z^*) \leq\|t\|^2\tau_Z(zz^*)=0.\] The trace on \(D\) is faithful, so \(E_D((zt)^*(zt))=0\). This is the \(D\)-valued inner product of \(zt1\) with itself, and hence \(zt1=0\). Products of centered letters create all elementary tensors in (7); thus \(Z1\) is dense in \(K\). It follows that \(z=0\), proving faithfulness of \(\tau_Z\).

Finally, let \(P_F\) be the orthogonal projection onto \(D\oplus L_F^\circ=L_F\). This submodule reduces the \(F\)-action. A core \(w=b_1a_1\cdots a_{m-1}b_m\) as in (6) sends \(L_F\) into words starting with \(G\), so \(P_FwP_F=0\). Since \(P_F\) commutes with \(F\), arbitrary endpoint coefficients satisfy \[P_Fa_0wa_mP_F=a_0P_FwP_Fa_m=0.\] Successive centering shows that \(F\) and the products (6) span a dense subspace. Their compressions belong to the faithful left copy of \(F\) on \(L_F\), which is norm closed. Compression therefore defines a conditional expectation \(E_F:Z\to F\). The vacuum lies in \(L_F\), giving \(E_D=E_D^F\circ E_F\) and trace preservation. Faithfulness follows from faithfulness of \(\tau_Z\). ◻

Proof of Proposition 12. Take \[F=M_2(A),\qquad G=M_2(D_1),\qquad D=D_1\oplus D_2.\] Embed \(D\) in \(F\) by \((d_1,d_2)\mapsto\operatorname{diag}(d_1,d_2)\) and in \(G\) by \((d_1,d_2)\mapsto\operatorname{diag}(d_1,\theta^{-1}(d_2))\). The normalized matrix traces agree on \(D\), and the maps \[E_D^F(a)=(E_1(a_{11}),E_2(a_{22})),\qquad E_D^G(b)=(b_{11},\theta(b_{22}))\] are trace-preserving conditional expectations. Apply Lemma 13 to obtain \(Z\), and retain the module notation used in its proof. The diagonal matrix projections \(e_i=e_{ii}\) are common to the two factors. Set \[ s=e_{21}^G,\qquad H(A)=e_1Ze_1,\qquad v=e_{12}^Fs, \tag{8}\] and identify \(A\) with \(e_1Fe_1\). Then \(v^*v=vv^*=e_1\). For \(d\in D_1\), the element \(sds^*\) is the second-corner copy of \(\theta(d)\) in \(F\), so \(vdv^*=\theta(d)\) in the first corner.

The matrix units of \(F\) identify \(Z\) with \(M_2(e_1Ze_1)\). Under this identification, the matrix algebra over \(C^*(A,v)\) contains \(F\) and \(s=e_{21}^Fv\). It contains \(G\) as well, because \(G\) is generated by \(D\) and \(s,s^*\). Hence \(H(A)=C^*(A,v)\). The trace on the corner is \(2\tau_Z\), and its restriction to \(A\) is \(\tau\). The restriction of \(E_F\) gives the required faithful trace-preserving expectation onto \(A\).

To check reduced words, write \(q=e_{12}^F\), so \(v=qs\) and \(v^*=s^*q^*\). Between consecutive \(G\)-letters the internal \(F\)-coefficients are \[\begin{array}{c|cccc} (\epsilon_t,\epsilon_{t+1}) &(+,+)&(+,-)&(-,+)&(-,-)\\ \hline \text{coefficient} &a_tq&a_t&q^*a_tq&q^*a_t. \end{array}\] The first and last are off diagonal. The middle two have zero \(D\)-expectation precisely under the respective conditions in (5). Since \(s,s^*\) are centered, the expanded word is a product of the form (6), and its expectation onto \(F\), hence onto \(A\), is zero.

For uniqueness, split a coefficient at an opposite-sign turn as \((a-E_i(a))+E_i(a)\). The first term is centered at that turn, while the second cancels two stable letters using the unitary relation. Repeating this operation expresses every polynomial as an \(A\)-term plus nonempty reduced words. Thus the stipulated data determine the trace of every polynomial, and in particular all moments of \(x^*x\) for each polynomial \(x\).

For a faithful tracial state \(t\) on any \(C^*\)-algebra, \[ \|x\|=\lim_{r\to\infty}t((x^*x)^r)^{1/(2r)}. \tag{9}\] To see this, restrict \(t\) to the commutative algebra generated by \(x^*x\) and the unit. Its spectral measure has full support, by faithfulness, and its \(r\)th-root moments tend to the spectral maximum. Consequently identical polynomial trace moments give identical polynomial norms. The generator correspondence is therefore isometric and extends to the asserted isomorphism. This argument requires no expectation in the comparison extension beyond the stated trace-vanishing rule. ◻

Proof. Because \(D_i\subset A'\), the restrictions \(E_i|_{A'}\) remain conditional expectations onto \(D_i\). The subalgebra \(C^*(A',v)\) inherits a faithful trace, the unitary relation and the reduced-word trace vanishing. Proposition 12 identifies it with \(H(A')\). The reduction of polynomials in the preceding proof shows that \(E_A\) maps this subalgebra into \(A'\) and has the stated restriction. ◻

Proof. Consider \(C=H_k(H_j(O))\). Corollary 14 places both \(H_j(O)\) and \(H_k(O)\) inside \(C\). Compose the expectations \(C\to H_j(O)\to O\), and call the result \(E_O\). Its restriction to each smaller adjunction is that adjunction’s expectation onto \(O\), as can be checked on its reduced words.

We prove that the two smaller adjunctions are free over \(O\): an alternating product of elements from their respective kernels of \(E_O\) has zero \(E_O\)-expectation. Approximate each centered \(H_k(O)\)-element by finite sums of nonempty reduced \(k\)-words over \(O\). Such approximations follow by reducing a polynomial approximation and subtracting its \(O\)-expectation. Keep the centered \(H_j(O)\)-elements as coefficients.

When two \(k\)-words are joined across a centered element \(b\in H_j(O)\), the new internal coefficient has the form \(aba'\), where \(a,a'\in O\). Its expectation onto \(O\) is \(aE_O(b)a'=0\), so its expectation onto either \(D_i^k\) is also zero. The resulting word is therefore a nonempty reduced \(k\)-word over \(H_j(O)\), including at every new opposite-sign turn. Initial and final \(H_j(O)\)-coefficients are allowed. Its expectation onto \(H_j(O)\) vanishes. Passing to norm limits proves the freeness assertion. An alternating product with no \(k\)-block is a single centered \(j\)-element and is already killed by \(E_O\).

The individual expectations and this freeness determine all joint \(O\)-valued moments. Indeed, center the factors of a joint word; the fully alternating centered term vanishes, and every other term has fewer blocks after its \(O\)-coefficients are absorbed into neighbors. The resulting recursion is symmetric in \(j\) and \(k\). Applying the same argument in the reverse construction gives equal scalar trace moments for every polynomial in \(O,v_j,v_k\). Faithfulness and (9) give the desired isomorphism. ◻

Proof. The compatible norm-one expectations extend continuously to \(O\) and retain complete positivity, bimodularity and trace preservation. Denote these extensions by \(E_i\). The compatible traces extend faithfully to both completions. For example, the quotient of \(R\) by its trace-null ideal is injective, hence isometric, on every earlier faithfully tracial algebra. It is therefore isometric on their dense union and on \(R\), so that ideal is zero. The same argument applies to \(O\).

We have \(R=C^*(O,v)\), and the unitary relation persists. Consider a nonempty reduced word with coefficients in \(O\). Approximate its finitely many coefficients in one common earlier \(O_\beta\). At a turn requiring \(E_i(a)=0\), replace an approximation \(b\) by \(b-E_i^\beta(b)\). This remains in \(O_\beta\), is exactly centered, and differs from \(a\) by at most twice the original error. Thus the given word is a norm limit of reduced words at earlier stages, all with trace zero. Proposition 12 identifies \(R\) with \(H(O)\). This finite-coefficient argument requires no assumption on the cofinality of \(\lambda\). ◻

Proof. At a successor, unique extensions compose, and the intermediate lift is pure by 2. At a limit, two extensions agree on each earlier algebra, hence on their dense union. Existence follows from state extension. Transfinite induction gives the first assertion. For the second, a unitary implementing equivalence in the original algebra also conjugates the unique lifts; use 3. ◻

Proof. Otherwise realize \(h,k\) by unit vectors \(\xi,\eta\) in one irreducible representation \(\pi\) of \(B_\lambda\). For each earlier \(\alpha\) in the given tail, the cyclic subspaces \[M_\alpha=\overline{\pi(B_\alpha)\xi},\qquad N_\alpha=\overline{\pi(B_\alpha)\eta}\] are reducing and carry inequivalent irreducible representations of \(B_\alpha\). The orthogonal projection onto \(N_\alpha\), restricted to \(M_\alpha\), is an intertwiner, so it is zero. Thus \(\langle\pi(a)\xi,\eta\rangle=0\) for every \(a\in B_\alpha\). Continuity gives the same equality for every \(a\in B_\lambda\), contradicting cyclicity of \(\xi\). ◻

Proof. Simplicity and faithfulness follow from the permanence facts above applied to the matrix stages. Separability and infinite dimension are immediate from those stages. A state taking value one on \(p_n\) must restrict to the vector state of the first basis vector on \(M_{2^n}\). These vector states are compatible, so they determine exactly one state on the dense union and its completion. This proves peaking and purity. The trace formula is the normalized matrix trace formula.

A nonzero finite-dimensional irreducible representation of an algebra containing \(A_0\) restricts to a unital representation of \(A_0\). Its kernel is zero by simplicity, which would embed the infinite-dimensional algebra \(A_0\) in a finite-dimensional algebra. ◻

Proof. Let \(k\in T\) be active. Both sequences for its bridge lie in \(A[T\cap k]\), by (19). The original expectations onto their generated abelian algebras restrict to this smaller base, since their ranges already lie in it. The restriction property 14 identifies the algebra obtained by adjoining \(v_k\) with that reduced HNN extension.

The sequences peak to pure states of the smaller base by 4, applied to its inclusion in \(X_k\). That same lemma gives each of these particular states its unique extension to \(X_k\). If the two smaller states were equivalent, 3 and uniqueness would make their extensions equivalent, contrary to the choice of the original active step. The smaller base contains \(A_0\), so 20 excludes finite-dimensional irreducibles. All hypotheses of 17 therefore hold in the inherited iteration.

Continuity follows because a polynomial uses only finitely many indices. Simplicity follows by induction from 11 and the limit observation above. Faithfulness follows either from the restricted trace on \(X_\delta\) or from the same permanence argument. Finally, a countable set of stable unitaries together with the separable algebra \(A_0\) generates a separable algebra. ◻

Proof. We will insert the omitted indices, in their original order, after \(A[S]\). Recall that the ordinal \(j\) is the set of indices below \(j\), whereas \(j+1\) also contains the index \(j\) itself. Fix an active \(j\notin S\), and for \(j+1\le\beta\le\delta\) set \[ O_\beta=A[j\cup(S\cap\beta)],\qquad R_\beta=A[(j+1)\cup(S\cap\beta)]. \tag{20}\] These index sets are dependency-closed. Initially, \(O_{j+1}=X_j\) and \(R_{j+1}=X_{j+1}\); afterward both chains perform precisely the inherited steps with indices in \(S\).

The omitted pair stays inequivalent on \(O_\beta\). Both \((p_n)\) and \((q_n^j)\) peak on every \(O_\beta\) by 4, since they peak in \(X_\delta\). Their states are initially inequivalent. Suppose \(k>j\) is an active member of \(S\), and they are still inequivalent on \(O_k\). The \(q_n^k\)-sequence also lies in \(O_k\) and peaks there. Each of these three particular states has a unique extension to \(X_k\), by 4; this does not use the theorem being proved.

In \(X_k\), the \(q_n^j\)-state is equivalent to the \(p_n\)-state, because step \(j\) has already been performed. The \(q_n^k\)-state is inequivalent to that state, because step \(k\) is active. Equivalence on \(O_k\) would pass to these particular unique extensions. Consequently, on \(O_k\) the \(q_n^k\)-state is equivalent to neither the \(p_n\)-state nor the \(q_n^j\)-state. The \(k\)-bridge joins only the classes of its designated pair, by 17. It therefore does not join the two states of the omitted \(j\)-bridge. At limits their inequivalence persists by 19. Transfinite induction proves the claim.

The omitted adjunction commutes past the inherited steps. Write \(H_j\) for the reduced HNN adjunction with the fixed data \((p_n),(q_n^j)\). We claim, respecting the trace and named generators, \[ R_\beta\cong H_j(O_\beta) \qquad(j+1\le\beta\le\delta). \tag{21}\] This is true initially. At an active \(k\in S\) above \(j\), the data for both adjunctions lie in \(O_k\). The expectation from \(R_k=H_j(O_k)\) onto \(O_k\), followed by either \(k\)-data expectation, is a trace-preserving expectation onto that data algebra. By 9, it agrees with the expectation supplied there by the corresponding peaking sequence. The same observation applies with \(j\) and \(k\) interchanged. Thus the reordering isomorphism 15 applies and gives \[H_k\bigl(H_j(O_k)\bigr) \cong H_j\bigl(H_k(O_k)\bigr),\] proving the next instance of (21). Other successor steps are identities. Figure 1 displays the two orders at an active index \(k\in S\).

At a limit, the isomorphisms already constructed agree on overlaps because they fix the named generators. The expectations from the \(O_\beta\) onto the two fixed abelian algebras generated by \((p_n)\) and \((q_n^j)\) also agree on overlaps, by 9. Applying 16 to these fixed data identifies the closure of the preceding \(H_j(O_\beta)\) with \(H_j\) of the limit base, proving (21) there.

At \(\beta=\delta\) we have \(O_\delta=A[S\cup j]\) and \(R_\delta=A[S\cup(j+1)]\). The two sequences still peak to inequivalent states on \(O_\delta\), by the first part of the proof. Thus 17 applies to (21), giving (U) for \[A[S\cup j]\subset A[S\cup(j+1)].\] If \(j\in S\) or \(j\) is inactive, this inclusion is the identity. The continuous chain \(\bigl(A[S\cup j]\bigr)_{j\le\delta}\) starts at \(A[S]\) and ends at \(X_\delta\); continuity follows because a polynomial involves only finitely many indices outside \(S\). Now 18 proves the theorem. ◻

Proof. Construct increasing countable dependency-closed sets \(S_n\), starting with \(S_0=\varnothing\). Choose a countable dense subset \(D_n\) of \(A[S_n]\). For each \(a\in D_n\) and each integer \(m\ge1\), 5 supplies a positive contraction \(x_{a,m}\) in \(X_\delta\) satisfying \[h(x_{a,m})=1,\qquad \bigl\|x_{a,m}(a-h(a)1)x_{a,m}\bigr\|<1/m.\] Each witness has countable support. Adjoin all these supports to \(S_n\) and close under dependencies to obtain \(S_{n+1}\).

Set \(S=\bigcup_n S_n\) and \(B=A[S]\). Suppose \(h|_B=t h_1+(1-t)h_2\) for states \(h_1,h_2\) and \(0<t<1\). Each \(h_i\) takes value one on every witness \(x_{a,m}\). A positive contraction of state value one fixes the state’s GNS vector, so \(h_i(a)=h_i(x_{a,m}a x_{a,m})\) and \(h_i(x_{a,m}^2)=1\). Hence \[|h_i(a)-h(a)| =\bigl|h_i\bigl(x_{a,m}(a-h(a)1)x_{a,m}\bigr)\bigr|<1/m.\] Letting \(m\) tend to infinity shows that both states agree with \(h\) on every \(D_n\). Their union is dense in \(B\), so \(h|_B\) is pure. Its unique extension is \(h\) by 22. ◻

Proof. There are at most \[\max(\aleph_0,|\delta|)^{\aleph_0} \le\mathfrak c^{\aleph_0}=\mathfrak c\] countable subsets of \(\delta\). For each one, the corresponding algebra \(A[S]\) is separable and has at most \(\mathfrak c\) states: values on a countable dense subset determine a state. By 23, every pure state of \(X_\delta\) is the unique extension of one such restriction. There are at most \(\mathfrak c\cdot\mathfrak c=\mathfrak c\) possibilities. ◻