State Preparation, Isometry Synthesis, and Limits

Choose $x\in O$ and $r>0$ with the Euclidean ball $B(x,r)\subseteq O$. Put $s=r\mathrm{/}(2\sqrt{m})>0$ and let $Q={\prod }_{j=1}^m[x_j-s\mathrm{/}2,x_j+s\mathrm{/}2]$. For every $y\in Q$, $\mathrm{\parallel }y-x{\mathrm{\parallel }}_2\le \sqrt{m}s\mathrm{/}2=r\mathrm{/}4<r$, so $Q\subseteq O$ and $\mathrm{vol}(Q)=s^m>0$.

Suppose $O$ were null. By [F1], with $\epsilon =s^m\mathrm{/}4$ there is a sequence of closed cubes $C_i$ covering $Q$ whose side lengths ${\mathrm{\ell }}_i\ge 0$ have ${\sum }_i{\mathrm{\ell }}_i^m\le s^m\mathrm{/}4$. For each $i\ge 0$, set ${\delta }_i=\min \{1,s^m{2}^{-i-5}\mathrm{/}(3^m(1+{\mathrm{\ell }}_i{)}^{m-1})\}$ and enlarge $C_i$ concentrically to an open cube $C_i^{\mathrm{\prime }}$ of side ${\mathrm{\ell }}_i+2{\delta }_i$. By [F6] and ${\delta }_i\le 1$, $0\le {\mathrm{\Delta }}_i:=({\mathrm{\ell }}_i+2{\delta }_i{)}^m-{\mathrm{\ell }}_i^m\le d_i:=s^m{2}^{-i-5}$. For every $N\ge 0$, induction using the finite-sum recursion gives ${\sum }_{i<N}{d}_i=s^m(1-2^{-N})\mathrm{/}16\le s^m\mathrm{/}16$. Thus every partial sum of the nonnegative enlarged-cost sequence $b_i=({\mathrm{\ell }}_i+2{\delta }_i{)}^m={\mathrm{\ell }}_i^m+{\mathrm{\Delta }}_i$ satisfies ${\sum }_{i<N}{b}_i\le s^m\mathrm{/}4+s^m\mathrm{/}16=5s^m\mathrm{/}16$, by [F4] and [F5]. A second application of [F5] gives a convergent total enlarged-cost series with sum at most $5s^m\mathrm{/}16<s^m\mathrm{/}2$.

The open cubes $C_i^{\mathrm{\prime }}$ cover the closed box $Q$, so [F2] gives a finite subcover $C_{i_1}^{\mathrm{\prime }},\dots ,C_{i_t}^{\mathrm{\prime }}$. Order its indices increasingly and take $N$ larger than the largest one. By [F4], the subcover's total cost is at most ${\sum }_{i<N}{b}_i$, which is at most $5s^m\mathrm{/}16<s^m\mathrm{/}2$ by step 2.1.

Write $\overline{C_i^{\mathrm{\prime }}}$ for the closed cube with the same center and side as $C_i^{\mathrm{\prime }}$, so $\mathrm{vol}(\overline{C_i^{\mathrm{\prime }}})=b_i$. Cut $Q$ and these finitely many closed covering cubes at all their coordinate endpoints. The positive-side grid cells inside $Q$ partition it; the interior of each cell lies in one covering cube because no covering-cube boundary crosses the cell. Assign each cell to the least index of a covering cube containing its interior. By finite cut-additivity from [F3], the assigned cells for each cube have total volume at most that closed cube's volume, while all cells sum to $\mathrm{vol}(Q)=s^m$. Thus $s^m\le {\sum }_{j=1}^t\mathrm{vol}(\overline{C_{i_j}^{\mathrm{\prime }}})<s^m\mathrm{/}2$, a contradiction. Therefore $O$ is not null.

Put $c=⟨v,u{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}$. For every $\mathrm{\mid }\lambda \mathrm{\mid }=1$, expand by [F2]: $\mathrm{\parallel }u-\lambda v{\mathrm{\parallel }}_2^2=2-\bar{\lambda }⟨u,v{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}-\lambda ⟨v,u{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}=2-2\mathrm{Re}(\lambda c)$.

If $c=0$, the real part is zero for every unit $\lambda$; if $c\mathrm{\ne }0$, choose $\lambda =\bar{c}\mathrm{/}\mathrm{\mid }c\mathrm{\mid }$, giving $\mathrm{Re}(\lambda c)=\mathrm{\mid }c\mathrm{\mid }$. Since $\mathrm{Re}(\lambda c)\le \mathrm{\mid }c\mathrm{\mid }$ for every unit $\lambda$, this phase attains the infimum in [F1]; the value is $\sqrt{2-2\mathrm{\mid }c\mathrm{\mid }}$, a real nonnegative number by [F4].

Replacing $u$ and $v$ by unit scalar multiples changes $⟨u,v{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}$ by a unit scalar, so its modulus and the displayed value do not change. Since the infimum is attained, it is zero exactly when a minimizing phase satisfies $u=\lambda v$, which is exactly equality of the represented rays by [F3].

If $[z]\in U_j$, multiplying $z$ by any nonzero complex scalar leaves every ratio $z_i\mathrm{/}{z}_j$ unchanged, so ${\varphi }_j$ is well-defined on rays. The representative $z\mathrm{/}{z}_j$ has $j$th coordinate $1$ and all other coordinates equal to those ratios, so the stated inverse sends ${\varphi }_j([z])$ back to $[z]$.

Conversely, applying ${\varphi }_j$ to the inverse representative with $j$th coordinate $1$ returns each entry of $w$; hence the two maps are inverse bijections. Let $\pi$ be the quotient map and $O_j=\{z:z_j\mathrm{\ne }0\}={\pi }^{-1}(U_j)$. Since $O_j$ is open, $U_j$ is open. The restriction ${\pi }_j:O_j\to U_j$ is a quotient map: for $B\subseteq U_j$, if ${\pi }_j^{-1}(B)$ is open in $O_j$, it is open in ${\mathbb{C}}^D\setminus \{0\}$, so $B$ is open in the quotient and hence in $U_j$; the reverse implication follows from continuity of $\pi$. The composite ${\varphi }_j\circ {\pi }_j$ consists of continuous coordinate ratios on $O_j$, so ${\varphi }_j$ is continuous. The inverse is the quotient map composed with the continuous representative map $w\mapsto e_j+{\sum }_{i\mathrm{\ne }j}{w}_i{e}_i$, hence is continuous.

Let $\pi :{\mathbb{C}}^D\setminus \{0\}\to {\mathbb{C}\mathbb{P}}^{D-1}$ be the quotient map. Then ${\pi }^{-1}(U_j)=\{z:z_j\mathrm{\ne }0\}$ is Euclidean open, so $U_j$ is open by the quotient topology. On $U_j\cap U_k$ with $j\mathrm{\ne }k$, write $w={\varphi }_j([z])$. Then $w_k\mathrm{\ne }0$, the transition coordinates satisfy $w_j^{\mathrm{\prime }}=1\mathrm{/}{w}_k$ and $w_i^{\mathrm{\prime }}=w_i\mathrm{/}{w}_k$ for $i\mathrm{\ne }j,k$. Conversely, $w_k=1\mathrm{/}{w}_j^{\mathrm{\prime }}$ and $w_i=w_i^{\mathrm{\prime }}\mathrm{/}{w}_j^{\mathrm{\prime }}$ for $i\mathrm{\ne }j,k$. These rational maps are smooth as real maps on their open coordinate domains, so the charts are compatible and form a smooth atlas.

The patches cover every ray because a nonzero vector has at least one nonzero coordinate. If $D\ge 2$, choose $k\mathrm{\ne }j$; the ray of the basis vector $e_k$ has $j$th coordinate zero and is not in $U_j$, so each fixed standard patch is proper. Finally, ${\mathbb{C}}^{D-1}$ has $D-1$ complex, or $2D-2$ real, coordinates.

Put $m=D\mathrm{/}2$, $x=2\delta \in (0,1\mathrm{/}2)$ and $R=\lceil xm\rceil -1$. Greedily scan the finite set $\{0,1{\}}^m$ in lexicographic order: select the least remaining word and delete all words at Hamming distance at most $R$, repeating until none remain. Distinct selected words have distance $h\ge R+1\ge xm=\delta D$, and each selection deletes at most $B={\sum }_{j=0}^R\left(\genfrac{}{}{0px}{}{m}{j}\right)$ words, so the selected code has at least $2^m\mathrm{/}B$ words.

For $j\le R<xm$, the ratio $x\mathrm{/}(1-x)<1$ gives $x^j(1-x{)}^{m-j}\ge x^{xm}(1-x{)}^{(1-x)m}$. The binomial identity [F6], $1={\sum }_{j=0}^m\iota \left(\genfrac{}{}{0px}{}{m}{j}\right)x^j(1-x{)}^{m-j}$ therefore yields $\iota B\le [x^{-x}(1-x{)}^{-(1-x)}{]}^m=2^{m{H}_2(x)}$. Put $h(t)=-[t\ln t+(1-t)\ln (1-t)]\mathrm{/}\ln 2$ for $0<t<1$, which is $H_2(t)$ by [F4]. The product and chain rules [F5] give $h^{\mathrm{\prime }}(t)=(\ln (1-t)-\ln t)\mathrm{/}\ln 2=\ln ((1-t)\mathrm{/}t)\mathrm{/}\ln 2$. For $0<t<1\mathrm{/}2$, the ratio exceeds $1$, so [F4] gives $h^{\mathrm{\prime }}(t)>0$. The same rules show $h$ is differentiable on $(0,1)$, and [F5] makes it continuous there; in particular it is continuous on $[x,1\mathrm{/}2]$. By [F4], ${\log }_2(1\mathrm{/}2)=-1$, so $h(1\mathrm{/}2)=1$. The mean value theorem [F3] now gives $1-h(x)=h^{\mathrm{\prime }}(\xi )(1\mathrm{/}2-x)>0$ for some $\xi \in (x,1\mathrm{/}2)$. Thus $H_2(x)<1$ and the code size is at least $2^{m(1-H_2(x))}=\exp (c_{\delta }D)$.

Map each selected bit word $b$ to the sign word $s_i=(-1{)}^{b_i}$; this is a bijection preserving Hamming distance. Form $u_s=D^{-1\mathrm{/}2}(1,\dots ,1,s_1,\dots ,s_m)$. It is unit because it has $D$ entries of modulus $D^{-1\mathrm{/}2}$. If two selected words differ in $h$ places, then $⟨u_s,u_t{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}=1-2h\mathrm{/}D$. Since $h\le m=D\mathrm{/}2$ and $h\ge \delta D$, this lies in $[0,1-2\delta ]$, so [F1] gives $d_{\mathrm{c}\mathrm{h}}([u_s],[u_t])=2\sqrt{h\mathrm{/}D}\ge 2\sqrt{\delta }$. The first half of each vector is fixed and nonzero, so scalar-proportional unit vectors must have scalar $1$ and identical last-half signs; distinct code words therefore give distinct rays.

A gate of arity $r\le k$ has at most $W^r$ ordered wire tuples. Summing over possible arities $0\le r\le k$ gives at most ${\sum }_{r=0}^k{W}^r\le (k+1)W^k$ tuples because $W\ge 1$; including its label gives at most $C_{k}A{W}^k$ choices for one occurrence.

A length-$g$ serial description is an ordered list of $g$ such choices, so there are at most $(C_{k}A{W}^k{)}^g$ descriptions. For two gates on disjoint supports, reorder the tensor factors as the first support, the second support, and the remaining wires. Their embeddings are then $G\otimes I\otimes I$ and $I\otimes H\otimes I$. By [F2], both compositions act as $Gx\otimes Hy\otimes z$ on elementary tensors, which span the register; thus they commute. By [F3], serially listing each layer therefore preserves its operator; listing the layers in circuit order preserves the full denotation and adds no occurrence.

Let $K\subseteq {\mathbb{R}}^m$ be compact and null, and let $\eta >0$. If $K=\mathrm{\varnothing }$, the empty family covers it. Otherwise [F1] gives a countable closed-cube cover $(C_j)$ with total volume at most $\eta \mathrm{/}8$. Write $C_j={\prod }_{r<m}[a_{jr},a_{jr}+{\mathrm{\ell }}_j]$ and let $x_{jr}=a_{jr}+{\mathrm{\ell }}_j\mathrm{/}2$ be its center coordinates. Set ${\delta }_j=\min \{1,\eta 2^{-j-5}\mathrm{/}(3^m(1+{\mathrm{\ell }}_j{)}^{m-1})\}$ and let $C_j^{\mathrm{\prime }}={\prod }_{r<m}(x_{jr}-({\mathrm{\ell }}_j+2{\delta }_j)\mathrm{/}2,x_{jr}+({\mathrm{\ell }}_j+2{\delta }_j)\mathrm{/}2)$, an open concentric enlargement. By [F11], $({\mathrm{\ell }}_j+2{\delta }_j{)}^m-{\mathrm{\ell }}_j^m={\sum }_{r=1}^m\iota \left(\genfrac{}{}{0px}{}{m}{r}\right){\mathrm{\ell }}_j^{m-r}(2{\delta }_j{)}^r\le {\delta }_j(1+{\mathrm{\ell }}_j{)}^{m-1}{\sum }_{r=1}^m\iota \left(\genfrac{}{}{0px}{}{m}{r}\right)2^r<3^m{\delta }_j(1+{\mathrm{\ell }}_j{)}^{m-1}\le \eta 2^{-j-5}$, because ${\delta }_j^r\le {\delta }_j$, ${\mathrm{\ell }}_j^{m-r}\le (1+{\mathrm{\ell }}_j{)}^{m-1}$ and ${\sum }_{r=1}^m\iota \left(\genfrac{}{}{0px}{}{m}{r}\right)2^r=3^m-1$. Define the covering cost of an open cube of side $s$ to be $s^m$; this agrees with [F5] for its closed counterpart. Put $d_j=\eta 2^{-j-5}$ and $b_j=({\mathrm{\ell }}_j+2{\delta }_j{)}^m={\mathrm{\ell }}_j^m+{\mathrm{\Delta }}_j$, where $0\le {\mathrm{\Delta }}_j\le d_j$. For every $N\ge 0$, induction using [F9] gives ${\sum }_{j<N}{d}_j=\eta (1-2^{-N})\mathrm{/}16\le \eta \mathrm{/}16$: the formula is $0$ at $N=0$, and the successor step adds $\eta 2^{-N-5}$. By [F7], the original nonnegative volume series has every partial sum at most $\eta \mathrm{/}8$. Finite-sum additivity and monotonicity [F10] therefore give ${\sum }_{j<N}{b}_j\le \eta \mathrm{/}8+{\sum }_{j<N}{d}_j\le 3\eta \mathrm{/}16$ for every $N$. Since $b_j\ge 0$, [F7] gives a convergent total covering-cost series with sum at most $3\eta \mathrm{/}16<\eta \mathrm{/}4$. The open cubes cover $K$, and compactness [F2] gives a finite subcover.

Order the finite subcover by its original indices and write $L_r$ and $x_r$ for the side and center of its $r$th open cube. Its finite total cost is at most the partial sum of $(b_j)$ through its largest original index, hence at most $3\eta \mathrm{/}16<\eta \mathrm{/}4$ by [F7, F10, step 1.1]. The finite polynomial sum ${\sum }_r(L_r+\zeta {)}^m$ is continuous at $\zeta =0$, so choose $\zeta >0$ for which it is less than ${\sum }_r{L}_r^m+\eta \mathrm{/}8<3\eta \mathrm{/}8$. By [F4], choose rational $q_r$ with $L_r<q_r<L_r+\zeta$ and a rational center $y_r\in {\mathbb{Q}}^m$ whose coordinates differ from $x_r$ by less than $(q_r-L_r)\mathrm{/}4$. The closed rational cube of side $q_r$ centered at $y_r$ contains the open cube of side $L_r$ centered at $x_r$, since each coordinate displacement is less than $L_r\mathrm{/}2+(q_r-L_r)\mathrm{/}4<q_r\mathrm{/}2$. These finitely many rational cubes cover $K$, and their total volume is less than $3\eta \mathrm{/}8<\eta \mathrm{/}2$ by [F5].

Fix a bijection $\nu :\mathbb{N}\to \mathbb{Q}$ and a pairing bijection $\pi :{\mathbb{N}}^2\to \mathbb{N}$ from [F3]. A list of $t$ rational cubes is determined by $(m+1)t$ rationals: its side and its $m$ center coordinates for each cube. Encode a tuple $(a_0,\dots ,a_{r-1})\in {\mathbb{N}}^r$ recursively by $c_0=0$ and $c_{r+1}(a_0,\dots ,a_r)=\pi (a_0,c_r(a_1,\dots ,a_r))$. Since $\pi$ and $\nu$ are injective, pairing the list length $t$ with the code of its $(m+1)t$ rational coordinates gives an injection from all finite rational cube lists into $\mathbb{N}$. For each natural code, use its unique list preimage if it has one, and otherwise use the empty list; this defines an enumeration $({\mathcal{C}}_s{)}_{s\ge 0}$ that includes every list. By step 2.1, every compact null $K$ has at least one enumerated list covering it with total volume less than $\eta \mathrm{/}2$. The set of qualifying codes is nonempty, so it has a least member by [F6]. The least-code rule assigns a unique cover to each pair $(K,\eta )$ without choosing from an arbitrary family.

Fix $\epsilon >0$. For each $i\ge 0$, apply the unique rule of step 3.1 to $(K_i,{\eta }_i)$ with ${\eta }_i=\epsilon 2^{-i-1}$. Let $t_i$ be the finite length of its selected list, whose total volume is less than $\epsilon 2^{-i-2}$. Define $N_0=0$ and $N_{i+1}=N_i+t_i$ recursively as in [F8]. For each $n\in \mathbb{N}$, if $N_i\le n<N_{i+1}$ for some $i$, take the cube at position $n-N_i$ in the $i$th list; otherwise take the degenerate zero-side cube at the origin. The half-open index intervals are disjoint: if $i<j$, then $N_{i+1}\le N_j$, so no $n$ lies in both. Each cube of the $i$th list, at position $j<t_i$, occurs at index $N_i+j$; thus every selected list occurs in the sequence.

Any finite initial segment meets only finitely many blocks. If it meets none, its total volume is zero. Otherwise, its finite set of block indices has a largest member $r$: scan the finite list and retain the larger at each comparison using the linear order in [F8]. Each partial block costs at most its full-list bound from step 4.1. By [F9], the budget sum $B_r={\sum }_{i=0}^r\epsilon 2^{-i-2}$ satisfies $B_0=\epsilon \mathrm{/}4=\frac{\epsilon }{2}(1-2^{-1})$ and $B_{r+1}=B_r+\epsilon 2^{-r-3}=\frac{\epsilon }{2}(1-2^{-r-1})+\epsilon 2^{-r-3}=\frac{\epsilon }{2}(1-2^{-r-2})$. So induction gives $B_r=\frac{\epsilon }{2}(1-2^{-r-1})<\epsilon \mathrm{/}2$. Thus every partial sum of the nonnegative volume series is at most $\epsilon \mathrm{/}2$, and [F7] gives convergence with total at most $\epsilon \mathrm{/}2<\epsilon$. The sequence covers ${\bigcup }_i{K}_i$, so [F1] shows that this union is null. Since every selected list is assigned by the unique least-code rule of step 3.1, the proof uses no choice principle.

Fix one gate topology, one choice of its gate-family parameter charts, and one standard target patch $U_j$ from [F1]. The product $\mathrm{\Omega }$ of the finitely many selected open parameter charts is open in ${\mathbb{R}}^q$. By [F8] and [F9], each local embedding has matrix entries that are fixed finite linear combinations of the local gate entries, and the circuit matrix is a finite ordered product of those embeddings. Its entries are therefore finite sums of products of smooth gate-matrix entries. The projected data vector $v(\theta )$ from [F6] depends smoothly on $\theta \in \mathrm{\Omega }$. The set ${\mathrm{\Omega }}_j=\{\theta \in \mathrm{\Omega }:v_j(\theta )\mathrm{\ne }0\}$ is open. Under the fixed real/imaginary identification ${\mathbb{C}}^{D-1}\cong {\mathbb{R}}^{2D-2}$, the chart map $f_j(\theta )=(v_i(\theta )\mathrm{/}{v}_j(\theta ){)}_{i\mathrm{\ne }j}$ is $C^{\mathrm{\infty }}$ on ${\mathrm{\Omega }}_j$.

The parameter dimension $q$ is at most $pg<2D-2$, so the derivative of $f_j$ has rank at most $q$ at every point; every point is critical. With target dimension $m=2D-2$ and $r=1$, the hypothesis $r>\max \{q-m,0\}=0$ holds, and [F2] makes $f_j({\mathrm{\Omega }}_j)$ null.