Primitive ideals have standard Borel quotient-norm codings

Choose a countable norm-dense rational-complex star subalgebra $D=\{d_n\}$, by closing a countable dense family under finite rational-complex sums, products and adjoints. For a pure state $\varphi$, cyclicity gives $\mathrm{\parallel }{\pi }_{\varphi }(a){\mathrm{\parallel }}^2={\sup }_d\varphi (d^{\ast }{a}^{\ast }ad)\mathrm{/}\varphi (d^{\ast }d)$, where $d\in D$ and positive denominators are retained: the vectors ${\pi }_{\varphi }(d)\xi$ are dense, and the ratios are their squared norm quotients. Hence strict quotient-norm superlevel sets pull back to unions of the open tests $\varphi (d^{\ast }d)>0$, $\varphi (d^{\ast }{a}^{\ast }ad)>r^2\varphi (d^{\ast }d)$. In the hull-kernel topology $\{J:\mathrm{\parallel }a+J\mathrm{\parallel }>r\}$ is open, since it says the positive cutoff $(\mathrm{\mid }a\mathrm{\mid }-r{)}_{+}$ is not in $J$. These opens generate that topology: every ideal-open is a union of such tests. Thus the kernel map is continuous.

Let $O$ be open in $P(A)$ and $\varphi \in O$. By [F1] there is $a\ge 0$, $\mathrm{\parallel }a\mathrm{\parallel }=\varphi (a)=1$ and $0<ϵ<1$ with $U_{a,ϵ}\cap P(A)\subseteq O$. Its kernel image is exactly $\{J:\mathrm{\parallel }a+J\mathrm{\parallel }>1-ϵ\}$. One inclusion follows from $\psi (a)\le \mathrm{\parallel }{\pi }_{\psi }(a)\mathrm{\parallel }$. For the other, in an irreducible representation with kernel $J$ the norm of a positive operator is the supremum of its expectations on unit vectors, so such a vector yields a pure vector state in $U_{a,ϵ}$ with the same kernel. Therefore the image of $O$ is open, and the map is onto by taking a unit cyclic vector in any irreducible representation. It is consequently continuous, open and surjective.

There is a countable cofinal family of nonzero ideals: from a countable dense family of positive contractions $a_n$ take every nonzero $(a_n-r{)}_{+}$ for positive rational $r$. If $J\mathrm{\ne }0$, approximate a positive norm-one $b\in J$ within $\delta <1\mathrm{/}4$ and choose $\delta <r<1\mathrm{/}2$; its cutoff belongs to $J$ by quotient calculus and is nonzero. Moreover these ideal-opens form a countable base: if $J_0$ avoids an ideal $K$, choose $b\in K^{+}$ with $\mathrm{\parallel }b\mathrm{\parallel }=1$, $\mathrm{\parallel }b+J_0\mathrm{\parallel }>0$, and approximate closely enough that a cutoff lies in $K$ but remains nonzero modulo $J_0$. Its ideal-open contains $J_0$ and is contained in the ideal-open of $K$.

Code a seminorm $q$ by $(q(d_n){)}_n\in {\prod }_n[0,\mathrm{\parallel }{d}_n\mathrm{\parallel }]$, imposing the rational-complex seminorm laws, $q(xy)\le q(x)q(y)$, $q(x^{\ast })=q(x)$ and $q(x^{\ast }x)=q(x{)}^2$. These are countably many closed equations or inequalities. The product has a complete weighted metric by [F4]; finitely approximating its first coordinates and ignoring the small metric tail proves total boundedness, hence compactness by [F4]. Every such $q$ is norm-Lipschitz, since $\mathrm{\mid }q(x)-q(y)\mathrm{\mid }\le q(x-y)\le \mathrm{\parallel }x-y\mathrm{\parallel }$, so it extends uniquely to $A$. Its kernel is a closed ideal. The metric completion of its quotient seminorm exists by [F4]; multiplication extends along Cauchy sequences by submultiplicativity and boundedness of Cauchy sequences, the isometric adjoint extends as well, and the C*-identity passes to limits. It is therefore a C*-algebra; the induced injective star map from the usual C*-quotient $A\mathrm{/}\ker q$ to that completion is isometric: if a positive element lost norm, a continuous spectral cutoff vanishing at0 and supported above the image norm would be nonzero but mapped to0, contradicting injectivity. Thus $q(a)=\mathrm{\parallel }a+\ker q\mathrm{\parallel }$. Conversely every closed ideal gives these laws. This proves the claimed compact metrizable code space of all closed ideals.

If $V_n$ are dense open subsets of $\mathrm{Prim}(A)$, their inverse images are dense open subsets of $P(A)$: every nonempty pure-state open set has nonempty open image by step 2.1, which meets $V_n$. The Polish pure-state space is Baire by [F1,F4], so their intersection meets the preimage of every nonempty primitive open set. Thus $\mathrm{Prim}(A)$ is Baire. The zero algebra gives empty pure and primitive spaces and the same assertion vacuously.

Primitive kernels are prime. Indeed, in an irreducible representation the support projection of any represented ideal is a commuting projection, hence is0 or1. If two ideal images are nonzero, their approximate units converge strongly to1; their products cannot all vanish. Now suppose $A$ is nonzero and prime. For each nonzero ideal $I$, its ideal-open is dense in $\mathrm{Prim}(A)$: any nonempty basic ideal-open comes from nonzero $K$, and primeness makes $IK\mathrm{\ne }0$. A pure norming state detecting a nonzero positive element of $IK$ gives a primitive kernel avoiding both $I$ and $K$. Baire applied to the cofinal countable ideals of step 2.2 gives a primitive kernel avoiding all of them. That kernel must be0, since any nonzero ideal contains one of the cofinal ideals. Applying this to $A\mathrm{/}J$ proves every proper closed prime $J$ is primitive.

Fix a countable dense family $(c_k)$ in the unit ball of $A$, including it in $D$. A proper quotient code is primitive exactly when, for every $a,b\in D$, ${\sup }_{k}q(a{c}_{k}b)=q(a)q(b)$. For a primitive quotient, choose a faithful irreducible representation, vectors nearly attaining the norms of $b$ and $a$, and a contraction linking the normalized output of $b$ to a near-norming input of $a$. Bounded density [F3] approximates this linker on that vector; quotient-norm lifting and density of $(c_k)$ then prove the equality. Conversely, if the quotient is not prime, two nonzero ideals with zero product give nonzero $a,b$ for which all $acb=0$. Continuity in $a,b$ makes this violate a test with $a,b\in D$. Step 4.1 identifies proper prime and primitive quotients. The countable supremum tests are Borel coordinate conditions; excluding the zero quotient is the Borel condition $\mathrm{\exists }dq(d)>0$. Thus primitive codes are Borel.

Coordinate strict superlevels are hull-kernel open by step 1.1, and their countable Boolean combinations give the inverse images of every real Borel interval. Conversely step 2.2 gives a countable ideal-open base, each expressible as a countable union of coordinate norm tests. Hence the code Borel structure equals the hull-kernel-topology Borel structure. The primitive-code subset is standard Borel by [F4]. Together with steps 2.1, 2.2, 3.1 and 4.1 this proves all assertions, without appealing to Choquet's theorem or a standardness claim for arbitrary second-countable $T_0$ spaces.