Finite Dimensional Normed Spaces and Riesz Lemma

By [L1], every $x\in X$ has exactly one coordinate list $a=(a_0,\dots ,a_{n-1})\in {\mathbb{K}}^n$ with $x={\sum }_{j<n}{a}_j{e}_j$. Therefore the displayed map $T$ is well defined, surjective, and injective.

$T$ is linear, because finite sums and scalar multiplication distribute over the coordinate formula: $T(\lambda a+b)={\sum }_{j<n}(\lambda a_j+b_j)e_j=\lambda {\sum }_{j<n}{a}_j{e}_j+{\sum }_{j<n}{b}_j{e}_j=\lambda T(a)+T(b)$.

In the complex case, [L5] makes both $P(a):=\mathrm{\parallel }a{\mathrm{\parallel }}_1$ and $N(a):=\mathrm{\parallel }T(a)\mathrm{\parallel }$ real norms on $({\mathbb{C}}^n{)}_{\mathbb{R}}$: for $N$, separation uses injectivity in step 1.1, and homogeneity and the triangle inequality use step 1.2 and the norm on $X$. The coordinate map $\mathrm{\Psi }(a)=(\mathrm{Re}{a}_0,\mathrm{Im}{a}_0,\dots ,\mathrm{Re}{a}_{n-1},\mathrm{Im}{a}_{n-1})$ is a real-linear bijection by [L4], so $P\circ {\mathrm{\Psi }}^{-1}$ and $N\circ {\mathrm{\Psi }}^{-1}$ are real norms on ${\mathbb{R}}^{2n}$. If $n\ge 1$, the pairwise equivalence clause of [L3] gives a positive real $c$ such that $cP(a)\le N(a)$ for every $a$. Thus $\mathrm{\parallel }{T}^{-1}x{\mathrm{\parallel }}_1\le c^{-1}\mathrm{\parallel }x\mathrm{\parallel }$. If $n=0$, both spaces consist only of zero by [L1], and the inverse is bounded with bound zero.

Put $C:={\sum }_{j<n}\mathrm{\parallel }{e}_j\mathrm{\parallel }$, a finite real. Then $\mathrm{\parallel }T(a)\mathrm{\parallel }=\parallel {\sum }_{j<n}{a}_j{e}_j\parallel \le {\sum }_{j<n}\mathrm{\mid }{a}_j\mathrm{\mid }\mathrm{\parallel }{e}_j\mathrm{\parallel }\le C{\sum }_{j<n}\mathrm{\mid }{a}_j\mathrm{\mid }=C\mathrm{\parallel }a{\mathrm{\parallel }}_1,$ so $T$ is bounded.

In the real case $\mathbb{K}=\mathbb{R}$, the coordinate $P(a)={\sum }_{j<n}\mathrm{\mid }{a}_j\mathrm{\mid }$ is a real norm: absolute-value definiteness, multiplicativity, and the triangle inequality give the three axioms by finite summation, including the empty list. The pullback $N(a):=\mathrm{\parallel }T(a)\mathrm{\parallel }$ is a norm on ${\mathbb{R}}^n$: definiteness uses step 1.1, and the triangle and homogeneity axioms come from the norm axioms on $X$ and the linearity of $T$. If $n=0$, then ${\mathbb{R}}^0=\{0\}$ and the inverse of $T$ is the zero map, hence bounded. If $n\ge 1$, [L3] gives $c>0$ with $c\mathrm{\parallel }a{\mathrm{\parallel }}_1\le N(a)=\mathrm{\parallel }T(a)\mathrm{\parallel }$ for every $a\in {\mathbb{R}}^n$, so $\mathrm{\parallel }{T}^{-1}x{\mathrm{\parallel }}_1\le c^{-1}\mathrm{\parallel }x\mathrm{\parallel }$ for every $x\in X$. Thus $T^{-1}$ is bounded in the real case.

Steps 1.1, 1.2, 2.1, 2.2, and 2.3 verify the three clauses of [L2]. Therefore $T$ is a topological isomorphism of normed spaces.

Let $T:{\mathbb{C}}^n\to X$ be the common algebraic basis map $T(a_0,\dots ,a_{n-1})={\sum }_{j<n}{a}_j{e}_j$. Applied to the norm $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_a$, [L1] gives constants $A,B>0$ such that $\mathrm{\parallel }T(a){\mathrm{\parallel }}_a\le A\mathrm{\parallel }a{\mathrm{\parallel }}_1\text{and}\mathrm{\parallel }a{\mathrm{\parallel }}_1\le B\mathrm{\parallel }T(a){\mathrm{\parallel }}_a$ for every $a\in {\mathbb{C}}^n$.

Applied to the norm $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_b$, [L1] gives constants $A^{\mathrm{\prime }},B^{\mathrm{\prime }}>0$ such that $\mathrm{\parallel }T(a){\mathrm{\parallel }}_b\le A^{\mathrm{\prime }}\mathrm{\parallel }a{\mathrm{\parallel }}_1\text{and}\mathrm{\parallel }a{\mathrm{\parallel }}_1\le B^{\mathrm{\prime }}\mathrm{\parallel }T(a){\mathrm{\parallel }}_b$ for every $a\in {\mathbb{C}}^n$.

Let $x\in X$ and write $x=T(a)$, which is possible and unique by [L1]. Then $\mathrm{\parallel }x{\mathrm{\parallel }}_b=\mathrm{\parallel }T(a){\mathrm{\parallel }}_b\le A^{\mathrm{\prime }}\mathrm{\parallel }a{\mathrm{\parallel }}_1\le A^{\mathrm{\prime }}B\mathrm{\parallel }T(a){\mathrm{\parallel }}_a=A^{\mathrm{\prime }}B\mathrm{\parallel }x{\mathrm{\parallel }}_a\mathrm{.}$ Interchanging $a$ and $b$ gives $\mathrm{\parallel }x{\mathrm{\parallel }}_a\le A{B}^{\mathrm{\prime }}\mathrm{\parallel }x{\mathrm{\parallel }}_b\mathrm{.}$

Taking $c=(A{B}^{\mathrm{\prime }}{)}^{-1}>0$ and $C=A^{\mathrm{\prime }}B>0$, step 2.1 is exactly the two-sided estimate of [L2], so the two norms are equivalent.

By [L1], it is enough to prove that ${\mathbb{K}}^n$ is complete for the coordinate ${\mathrm{\ell }}^1$ norm, because a bounded bijection with bounded inverse preserves Cauchy sequences and their limits.

In the real case $\mathbb{K}=\mathbb{R}$, if $n=0$ then ${\mathbb{R}}^0=\{0\}$ is complete trivially. If $n\ge 1$, [L2] applies directly to the ${\mathrm{\ell }}^1$ norm on ${\mathbb{R}}^n$, so ${\mathbb{R}}^n$ is complete.

In the complex case, if $n=0$ the coordinate space is a singleton and complete. If $n\ge 1$, let $\mathrm{\Psi }:{\mathbb{C}}^n\to {\mathbb{R}}^{2n}$ send each coordinate to its real and imaginary parts. By [L3] it is a real-linear bijection, and $Q(u):=\mathrm{\parallel }{\mathrm{\Psi }}^{-1}u{\mathrm{\parallel }}_1$ is a real norm on ${\mathbb{R}}^{2n}$. It is complete by [L2]. Since $Q(\mathrm{\Psi }a-\mathrm{\Psi }b)=\mathrm{\parallel }a-b{\mathrm{\parallel }}_1$, the bijection $\mathrm{\Psi }$ transports Cauchy sequences and limits exactly; thus ${\mathbb{C}}^n$ is complete in its complex coordinate norm metric.

Let $(x_m)$ be a Cauchy sequence in $X$, and write $a_m:=T^{-1}{x}_m$. Since $T^{-1}$ is bounded, $(a_m)$ is Cauchy in ${\mathbb{K}}^n$; by steps 1.2 and 1.3 it converges to some $a\in {\mathbb{K}}^n$. Since $T$ is bounded, $x_m=T(a_m)\to T(a)$ in $X$. Thus every Cauchy sequence in $X$ converges in $X$.

By [L4], step 2.1 says exactly that $X$ is Banach.

If $n=0$, then $W=\{0\}$ by [L2]. Every $x\in V\setminus \{0\}$ has the open ball of radius $\mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}2$ disjoint from $W$, so its complement is open and $W$ is closed. It remains to treat $n>0$.

For $n>0$, in the real case put $m=n$ and $e_j=b_j$. In the complex case put $m=2n$, $e_{2j}=b_j$, and $e_{2j+1}=i{b}_j$, and use real scalars. Splitting complex coefficients proves real spanning; grouping a real relation into a complex relation among the $b_j$ proves independence. Thus $(e_j{)}_{j<m}$ is a supplied ordered real basis of $W$, and its norm is still the restricted ambient norm. By [L1], $W$ is complete for this metric.

Define maps $E_j:\mathbb{N}\to {\mathbb{Q}}^j$ for $j\le m$ as follows: $E_0(s)$ is the empty tuple, and $E_{j+1}(p(s,t))=(E_j(s),q(t))\mathrm{.}$ The unique inverse pair of each natural makes this a total definition. Finite recursion, or [L5] on states consisting of a stage and such a map, produces the maps uniformly; after stage $m$ keep the state fixed. By induction each $E_j$ is surjective: given a tuple of length $j+1$, its initial tuple has a preimage $s$ under $E_j$, its final coordinate has a preimage $t$ under $q$, and $p(s,t)$ maps to the entire tuple. This uses two witnesses for one tuple, not a family of selected enumerations. Define $d(s)={\sum }_{j<m}(E_m(s){)}_j{e}_j\mathrm{.}$ Its image is exactly the rational span of this finite basis.

This image is dense in $W$. For $w={\sum }_{j<m}{a}_j{e}_j$ and $\epsilon >0$, let $t_j$ be the least index satisfying $\mathrm{\mid }{a}_j-q(t_j)\mathrm{\mid }<\delta ,\delta =\frac{\epsilon }{2m(1+{\sum }_{j<m}\mathrm{\parallel }{e}_j\mathrm{\parallel })}>0.$ Existence follows from rational density and [L5] gives the least index. Surjectivity of $E_m$ gives a value of $d$ equal to $v={\sum }_{j<m}q(t_j)e_j$, and $\mathrm{\parallel }w-v\mathrm{\parallel }\le \delta {\sum }_{j<m}\mathrm{\parallel }{e}_j\mathrm{\parallel }<\epsilon \mathrm{.}$ Hence every ball in $W$ meets $d[\mathbb{N}]$.

Let $x$ be an ambient adherent point of $W$, meaning every ball about $x$ meets $W$. For each $k\in \mathbb{N}$, first such adherence and then step 3.1 give a point $w\in W$ within $1\mathrm{/}(2(k+1))$ of $x$ and a value $d(s)$ within $1\mathrm{/}(2(k+1))$ of $w$. Thus the set $A_k=\{s\in \mathbb{N}:\mathrm{\parallel }x-d(s)\mathrm{\parallel }<1\mathrm{/}(k+1)\}$ is nonempty. Set $s_k=\min A_k$ and $u_k=d(s_k)$. This defines the sequence outright by a unique-index rule. By [L6] the error tends to zero, so $u_k\to x$ in $V$, and $\mathrm{\parallel }{u}_k-u_l\mathrm{\parallel }\le \mathrm{\parallel }{u}_k-x\mathrm{\parallel }+\mathrm{\parallel }x-u_l\mathrm{\parallel }$ makes it Cauchy in the restricted metric of $W$.

Completeness from step 1.2 supplies $y\in W$ with $u_k\to y$ in the restricted norm. Since the norm agrees with the ambient one, $\mathrm{\parallel }x-y\mathrm{\parallel }\le \mathrm{\parallel }x-u_k\mathrm{\parallel }+\mathrm{\parallel }{u}_k-y\mathrm{\parallel }⟶0,$ so $x=y\in W$. Every ambient adherent point therefore belongs to $W$. A point outside $W$ consequently has a ball disjoint from $W$, making the complement open and $W$ closed. Together with step 1.1 this proves all cases, without manufacturing a sequence in an arbitrary subset or using countable choice.

Let $T:{\mathbb{K}}^n\to X$ be the basis map from [L1]. Since $T^{-1}$ is bounded, there is $B>0$ such that $\mathrm{\parallel }{T}^{-1}x{\mathrm{\parallel }}_1\le B\mathrm{\parallel }x\mathrm{\parallel }(x\in X)\mathrm{.}$

Put $M:={\sum }_{j<n}\mathrm{\parallel }S(e_j)\mathrm{\parallel }$, a finite real. If $x={\sum }_{j<n}{a}_j{e}_j=T(a_0,\dots ,a_{n-1})$, then by [L3] $Sx={\sum }_{j<n}{a}_{j}S(e_j),$ so $\mathrm{\parallel }Sx\mathrm{\parallel }\le {\sum }_{j<n}\mathrm{\mid }{a}_j\mathrm{\mid }\mathrm{\parallel }S(e_j)\mathrm{\parallel }\le M{\sum }_{j<n}\mathrm{\mid }{a}_j\mathrm{\mid }=M\mathrm{\parallel }{T}^{-1}x{\mathrm{\parallel }}_1\mathrm{.}$

Combining steps 1.1 and 1.2 gives $\mathrm{\parallel }Sx\mathrm{\parallel }\le MB\mathrm{\parallel }x\mathrm{\parallel }(x\in X)\mathrm{.}$ Therefore $S$ is bounded, and with [L3] this makes $S$ a bounded linear operator by [L2].

By [A1], choose $y\in X\setminus M$ and put $d:={\inf }_{m\in M}\mathrm{\parallel }y-m\mathrm{\parallel }$. Because $M$ is closed, its complement is open, so there is $\rho >0$ with $B(y,\rho )\subseteq X\setminus M$. Hence $\mathrm{\parallel }y-m\mathrm{\parallel }\ge \rho$ for every $m\in M$, which gives $d\ge \rho >0$.

Since $\alpha <1$, one has $d<d\mathrm{/}\alpha$. By definition of the infimum, choose $m\in M$ with $d\le \mathrm{\parallel }y-m\mathrm{\parallel }<\frac{d}{\alpha }\mathrm{.}$ Put $x:=\frac{y-m}{\mathrm{\parallel }y-m\mathrm{\parallel }}\mathrm{.}$ Then $\mathrm{\parallel }x\mathrm{\parallel }=1$.

Let $m^{\mathrm{\prime }}\in M$. Because $m+\mathrm{\parallel }y-m\mathrm{\parallel }{m}^{\mathrm{\prime }}\in M$, the definition of $d$ gives $d\le \mathrm{\parallel }y-(m+\mathrm{\parallel }y-m\mathrm{\parallel }{m}^{\mathrm{\prime }})\mathrm{\parallel }=\mathrm{\parallel }y-m-\mathrm{\parallel }y-m\mathrm{\parallel }{m}^{\mathrm{\prime }}\mathrm{\parallel }=\mathrm{\parallel }y-m\mathrm{\parallel }\mathrm{\parallel }x-m^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{.}$ Therefore $\mathrm{\parallel }x-m^{\mathrm{\prime }}\mathrm{\parallel }\ge \frac{d}{\mathrm{\parallel }y-m\mathrm{\parallel }}>\alpha ,$ the last inequality by step 2.1. Since $m^{\mathrm{\prime }}\in M$ was arbitrary, $\mathrm{dist}(x,M)>\alpha$.

Step 3.1 produces a unit vector $x$ whose distance from $M$ exceeds $\alpha$, as required.

Assume $X$ admits an ordered basis of length $n$. If $\mathbb{K}=\mathbb{R}$, the coordinate space ${\mathbb{R}}^n$ is locally compact by [L4] when $n\ge 1$, and ${\mathbb{R}}^0=\{0\}$ is compact, hence locally compact. If $\mathbb{K}=\mathbb{C}$, then ${\mathbb{C}}^n$ identifies with ${\mathbb{R}}^{2n}$ by [L5], so the same conclusion holds there. By [L1], $X$ is homeomorphic to that coordinate space, hence locally compact.

Assume now that $X$ is locally compact. By [L6], applied to the norm metric, $0$ has a compact neighbourhood $K$ containing some open ball $B(0,r)$ with $r>0$. The closed ball $\bar{B}(0,\frac{r}{2}):=\{x\in X:\mathrm{\parallel }x\mathrm{\parallel }\le \frac{r}{2}\}$ is a closed subset of $K$ and is therefore compact.

For the successor step, let $M_m:=\mathrm{span}\{x_0,\dots ,x_m\}$. The finite list $x_0,\dots ,x_m$ generates $M_m$, so by deleting dependent terms if necessary one gets an ordered basis of finite length for $M_m$. Thus $M_m⊊X$ by the contradiction hypothesis, and A finite-dimensional normed subspace is closed makes $M_m$ closed. Applying [L2] with $\alpha =1\mathrm{/}2$ yields a unit vector $x_{m+1}$ with $\mathrm{dist}(x_{m+1},M_m)>1\mathrm{/}2$. In particular $\mathrm{\parallel }{x}_{m+1}-x_j\mathrm{\parallel }>1\mathrm{/}2$ for every $j\le m$. [L2, A finite-dimensional normed subspace is closed, choose]

Suppose for contradiction that $X$ admits no ordered basis of finite length. We recursively build, for each $m\ge 0$, unit vectors $x_0,\dots ,x_m\in X$ such that $\mathrm{\parallel }{x}_j-x_k\mathrm{\parallel }>\frac{1}{2}(j\mathrm{\ne }k)\mathrm{.}$ For $m=0$ choose any nonzero $x_0$ and normalize it.

Every $x_j$ lies in the closed ball $\bar{B}(0,r\mathrm{/}2)$ after rescaling by $r\mathrm{/}2$: namely $y_j:=(r\mathrm{/}2)x_j$ satisfies $\mathrm{\parallel }{y}_j\mathrm{\parallel }=r\mathrm{/}2$, so $y_j$ lies in that compact set. Also $\mathrm{\parallel }{y}_j-y_k\mathrm{\parallel }=\frac{r}{2}\mathrm{\parallel }{x}_j-x_k\mathrm{\parallel }>\frac{r}{4}(j\mathrm{\ne }k)\mathrm{.}$

By [L3], the compact metric space $\bar{B}(0,r\mathrm{/}2)$ is totally bounded. Taking $\epsilon =r\mathrm{/}8$, it admits a finite $\epsilon$-net. But one $\epsilon$-ball can contain at most one of the points $y_j$, since distinct ones are more than $r\mathrm{/}4=2\epsilon$ apart. Therefore a finite $\epsilon$-net cannot cover arbitrarily large finite sets $\{y_0,\dots ,y_m\}$, contradiction.

The contradiction in step 4.1 shows that $X$ must admit an ordered basis of finite length. Together with step 1.1, this proves the equivalence.

Let $S$ be the set of all finite lists $s=(u_0,\dots ,u_m)$ of unit vectors in $X$ such that $\mathrm{\parallel }{u}_j-u_k\mathrm{\parallel }>\alpha$ for all $j\mathrm{\ne }k$. The set $S$ is nonempty because any single unit vector lies in it: choose any nonzero $u\in X$ and normalize it.

Define a relation $R$ on $S$ by $sRt$ when $t$ is obtained from $s$ by appending one more unit vector $u$ with $\mathrm{dist}(u,\mathrm{span}(s))>\alpha$. If $s=(u_0,\dots ,u_m)\in S$, then the finite span $M:=\mathrm{span}(s)$ is generated by a finite list, so by deleting dependent terms one obtains an ordered basis of finite length for $M$. Since $X$ is not the span of any finite list, $M⊊X$; by [L2] it is closed. Now [L3] applies to $M$ and produces a unit vector $u$ with $\mathrm{dist}(u,M)>\alpha$, so $s$ has an $R$-successor. Thus $R$ is entire on $S$.

By [L1], there is a sequence $(s_n{)}_{n\in \mathbb{N}}$ in $S$ with $s_{n}R{s}_{n+1}$ for every $n$. Because each successor appends exactly one new term, the first entries stabilize: if $x_n$ denotes the last entry appended when passing from $s_n$ to $s_{n+1}$, then every earlier $x_j$ remains in all later lists.

For $m<n$, the vector $x_n$ was chosen with $\mathrm{dist}(x_n,\mathrm{span}(x_0,\dots ,x_{n-1}))>\alpha$, and $x_m$ lies in that span. Hence $\mathrm{\parallel }{x}_n-x_m\mathrm{\parallel }>\alpha$. Every $x_n$ is a unit vector by the definition of $R$. Therefore $(x_n{)}_{n\in \mathbb{N}}$ is the required separated sequence.

Assume $X$ admits an ordered basis of length $n$, and let $T:{\mathbb{K}}^n\to X$ be the coordinate isomorphism from [L1]. Then $T^{-1}[{\bar{B}}_X]$ is closed in ${\mathbb{K}}^n$, because $T$ is continuous, and bounded in the coordinate ${\mathrm{\ell }}^1$ norm, because $T^{-1}$ is bounded.

Assume conversely that ${\bar{B}}_X$ is compact. Then [L4] makes it totally bounded. Suppose for contradiction that $X$ admits no ordered basis of finite length.

Let $F\subseteq {\bar{B}}_X$ be a finite $1\mathrm{/}2$-net, and put $M:=\mathrm{span}(F)$. The finite set $F$ generates $M$, so by deleting dependent terms one gets an ordered basis of finite length for $M$; thus [L3] makes $M$ closed. Since $X$ is not finitely generated, $M⊊X$.

In the real case $\mathbb{K}=\mathbb{R}$, if $n=0$ then ${\bar{B}}_X=\{0\}$ is compact. If $n\ge 1$, step 1.1 and [L5] show that $T^{-1}[{\bar{B}}_X]$ is compact in ${\mathbb{R}}^n$, hence ${\bar{B}}_X$ is compact as its homeomorphic image.

In the complex case $\mathbb{K}=\mathbb{C}$, [L6] identifies ${\mathbb{C}}^n$ with ${\mathbb{R}}^{2n}$. Under that identification the coordinate ${\mathrm{\ell }}^1$ norm is equivalent to a real norm on ${\mathbb{R}}^{2n}$, so the bounded closed set $T^{-1}[{\bar{B}}_X]$ is also closed and bounded in Euclidean space. If $n=0$ it is a singleton; if $n\ge 1$, [L5] makes it compact in ${\mathbb{R}}^{2n}$, hence compact in ${\mathbb{C}}^n$ and therefore in $X$.

Applying [L2] with $\alpha =1\mathrm{/}2$ gives a unit vector $x\in X$ with $\mathrm{dist}(x,M)>1\mathrm{/}2$. Since $F$ is a $1\mathrm{/}2$-net in the unit ball and $x\in {\bar{B}}_X$, some $y\in F$ satisfies $\mathrm{\parallel }x-y\mathrm{\parallel }<1\mathrm{/}2$. But $y\in M$, which contradicts $\mathrm{dist}(x,M)>1\mathrm{/}2$.

Therefore $X$ must admit an ordered basis of finite length. Together with steps 2.1 and 2.2, this proves the equivalence.

If ${\bar{B}}_X$ were compact, [L1] would force $X$ to admit an ordered basis of finite length.

That contradicts the hypothesis, so ${\bar{B}}_X$ is not compact.

The identity map is linear and satisfies $\mathrm{\parallel }{I}_{X}x\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$ for every $x\in X$, so [L1] makes it a bounded linear operator.

One has $I_X[{\bar{B}}_X]={\bar{B}}_X$. By [L2], that set is not compact. Therefore the identity operator does not send the closed unit ball to a compact subset of $X$.

Fix $p$ and $q$ from [L3, L4]. Encode a finite word $(a_0,\dots ,a_{\mathrm{\ell }-1})$ of naturals by $p(\mathrm{\ell },c_{\mathrm{\ell }})$, where $c_0=0$ and $c_{j+1}=p(c_j,a_j)$. To decode any natural, invert its outer pair to obtain $(\mathrm{\ell },c)$, and invert the inner pair $\mathrm{\ell }$ times, recovering the symbols in reverse order. If the final residue is not zero, return the empty word; otherwise return the recovered word. These are uniform finite recursions: one may retain the remaining length, current residue and recovered finite list as the state and stop changing the state when the remaining length is zero. Thus [L5] defines the decoder, and decoding an encoded word returns that word by induction and injectivity of $p$. We have a total surjection from $\mathbb{N}$ onto all finite words, without choosing separate enumerations for each length.

In the real case let $e_j=b_j$. In the complex case let $e_{2j}=b_j$ and $e_{2j+1}=i{b}_j$, and regard $X$ as a real normed space. In either case $(e_j)$ is a real Hamel basis: in the complex case a finite real relation groups into a complex relation among the $b_j$, forcing both real coefficients for each $b_j$ to vanish, and spanning follows by splitting each complex coefficient into real and imaginary parts. Completeness is unchanged.

Interpret a decoded symbol $p(j,t)$ as the term $q(t)e_j$, and interpret a decoded word as the sum of its terms, the empty word giving zero. Step 1.1 now defines a total function $d:\mathbb{N}\to X$ whose image $D$ is exactly the finite rational span of $(e_j)$. In particular $0\in D$. For each fixed $N$, the same decoder, with symbols having $j>N$ contributing zero, defines $d_N:\mathbb{N}\to F_N$, where $F_N={\mathrm{span}}_{\mathbb{R}}\{e_0,\dots ,e_N\}$; its image is exactly the rational span of this finite list.

For any finite expression $v={\sum }_{j<\mathrm{\ell }}{\lambda }_j{e}_{n_j}$ and $\epsilon >0$, if $\mathrm{\ell }=0$ then $v=0\in D$. Otherwise put $\delta =\epsilon \mathrm{/}(2\mathrm{\ell }(1+{\sum }_{j<\mathrm{\ell }}\mathrm{\parallel }{e}_{n_j}\mathrm{\parallel }))>0$ and let $t_j$ be the least natural with $\mathrm{\mid }{\lambda }_j-q(t_j)\mathrm{\mid }<\delta$. Such an index exists by [L3]; [L5] determines it. The triangle inequality gives $\parallel v-{\sum }_{j<\mathrm{\ell }}q(t_j)e_{n_j}\parallel \le \delta {\sum }_{j<\mathrm{\ell }}\mathrm{\parallel }{e}_{n_j}\mathrm{\parallel }<\epsilon \mathrm{.}$ Consequently $D$ is dense in $X$, and using only indices at most $N$ shows that $d_N[\mathbb{N}]$ is dense in $F_N$. Only the displayed finite representation of the fixed vector is used, not a selection of representations for a family of vectors.

Each $F_N$ is proper since $e_{N+1}\mathrm{\notin }{F}_N$ by real independence. Also $X={\bigcup }_N{F}_N$, because every vector has a finite expression in the $e_j$ and a finite list of natural indices has an upper bound.

Fix $N$ and a point $z$ in the ambient closure of $F_N$, meaning every ball around $z$ meets $F_N$. For each $k\in \mathbb{N}$, some $u\in F_N$ satisfies $\mathrm{\parallel }z-u\mathrm{\parallel }<1\mathrm{/}(2(k+1))$, and density from step 3.1 gives some $d_N(j)$ with $\mathrm{\parallel }u-d_N(j)\mathrm{\parallel }<1\mathrm{/}(2(k+1))$. Hence the set of $j$ with $\mathrm{\parallel }z-d_N(j)\mathrm{\parallel }<1\mathrm{/}(k+1)$ is nonempty. Define $j(k)$ to be its least member and $u_k=d_N(j(k))$. This defines one sequence outright by [L5]; no countable choice is used. The reciprocal estimate [L6] and the triangle inequality give $u_k\to z$ in $X$ and show $(u_k)$ is Cauchy in $F_N$.

The displayed list $(e_0,\dots ,e_N)$ is an ordered real basis of $F_N$ by step 1.2, so [L2] supplies a limit $y\in F_N$ of this already constructed Cauchy sequence. Then $\mathrm{\parallel }z-y\mathrm{\parallel }\le \mathrm{\parallel }z-u_k\mathrm{\parallel }+\mathrm{\parallel }{u}_k-y\mathrm{\parallel }\to 0$, whence $z=y\in F_N$. Therefore $F_N$ is closed: if a point outside it had no disjoint ball, it would belong to its closure, contrary to what we just proved. This argument uses completeness only on the specific sequence from step 4.1.