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 $\mathbb{K}=\mathbb{C}$, write $\mathrm{\Psi }:{\mathbb{C}}^n\to {\mathbb{R}}^{2n}$ for the coordinatewise real-imaginary-part map from [L4]. Define $N_{\mathbb{R}}(u,v):=\mathrm{\parallel }T(u_0+i{v}_0,\dots ,u_{n-1}+i{v}_{n-1})\mathrm{\parallel }\mathrm{.}$ By [L4] and [L5] this is a real norm on ${\mathbb{R}}^{2n}$. If $n=0$ the inverse is again bounded trivially. If $n\ge 1$, [L3] applied to ${\mathbb{R}}^{2n}$ gives $c>0$ with $c\mathrm{\parallel }(u,v){\mathrm{\parallel }}_2\le N_{\mathbb{R}}(u,v)$ for every $(u,v)\in {\mathbb{R}}^{2n}$. Also $\mathrm{\mid }{u}_j+i{v}_j\mathrm{\mid }\le \mathrm{\mid }{u}_j\mathrm{\mid }+\mathrm{\mid }{v}_j\mathrm{\mid }\le \sqrt{2}(u_j^2+v_j^2{)}^{1\mathrm{/}2}$, so $\mathrm{\parallel }(u_0+i{v}_0,\dots ,u_{n-1}+i{v}_{n-1}){\mathrm{\parallel }}_1\le \sqrt{2n}\mathrm{\parallel }(u,v){\mathrm{\parallel }}_2\mathrm{.}$ Hence $\mathrm{\parallel }a{\mathrm{\parallel }}_1\le \frac{\sqrt{2n}}{c}\mathrm{\parallel }T(a)\mathrm{\parallel }(a\in {\mathbb{C}}^n),$ which is the boundedness of $T^{-1}$.

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 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, 1.3, 2.1, and 2.2 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 $\mathrm{\parallel }x{\mathrm{\parallel }}_a\le A{B}^{\mathrm{\prime }}\mathrm{\parallel }x{\mathrm{\parallel }}_b\mathrm{.}$

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 $\mathbb{K}=\mathbb{C}$, if $n=0$ the same trivial argument applies. If $n\ge 1$, [L3] identifies ${\mathbb{C}}^n$ with ${\mathbb{R}}^{2n}$ by real and imaginary parts, and the proof of A chosen algebraic basis identifies a finite-dimensional normed space with a coordinate space already shows that the complex coordinate ${\mathrm{\ell }}^1$ norm is equivalent to a real norm on ${\mathbb{R}}^{2n}$. By [L2], that real norm is complete, hence so is ${\mathbb{C}}^n$ with the complex coordinate ${\mathrm{\ell }}^1$ norm.

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.

By [L3], the hypothesis makes $W$ a normed space in its own right, with an ordered basis of finite length. Therefore [L1] makes $W$ Banach, hence complete for its restricted norm.

Applying [L2] to that complete normed subspace shows that $W$ is closed in $V$.

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 $X$.

Let $Q_{\mathbb{K}}:=\mathbb{Q}$ in the real case and $Q_{\mathbb{K}}:=\{a+ib:a,b\in \mathbb{Q}\}$ in the complex case. In either case $Q_{\mathbb{K}}$ is countable by [L3], and it is dense in $\mathbb{K}$. Let $D\subseteq X$ be the set of all finite $Q_{\mathbb{K}}$-linear combinations of the basis vectors $b_n$. Coding a finite combination by the finite list of its indices together with its coefficient list gives an explicit surjection from a countable set onto $D$, so $D$ is countable. Also $0\in D$, so $D$ is nonempty.

For $N\in \mathbb{N}$ put $F_N:={\mathrm{span}}_{\mathbb{R}}\{b_0,\dots ,b_N\}$ in the real case, and $F_N:={\mathrm{span}}_{\mathbb{R}}\{b_0,i{b}_0,\dots ,b_N,i{b}_N\}$ in the complex case. In either case $F_N$ is finite-dimensional over $\mathbb{R}$. It is proper: in the real case $b_{N+1}\mathrm{\notin }{F}_N$ by linear independence of the basis image, and in the complex case a real-linear relation expressing $b_{N+1}$ in terms of $b_0,i{b}_0,\dots ,b_N,i{b}_N$ would be the same as a complex-linear relation expressing $b_{N+1}$ in terms of $b_0,\dots ,b_N$. Thus [L2] makes every $F_N$ closed. Also $X={\bigcup }_{N\in \mathbb{N}}{F}_N$, because every vector uses only finitely many basis vectors, and in the complex case every complex coefficient splits into real and imaginary parts.

Every proper linear subspace of a normed space has empty interior. Indeed, if $W⊊X$ were a linear subspace containing some ball $B(x,r)$, then $B(0,r)\subseteq W$ because $W$ is closed under subtraction, and for any $y\in X\setminus \{0\}$ the vector $(r\mathrm{/}(2\mathrm{\parallel }y\mathrm{\parallel }))y$ would lie in $B(0,r)\subseteq W$, forcing $y=(2\mathrm{\parallel }y\mathrm{\parallel }\mathrm{/}r)(r\mathrm{/}(2\mathrm{\parallel }y\mathrm{\parallel }))y\in W$; also $0\in W$. So $W=X$, contradiction.

By [L5], fix a surjection $d:\mathbb{N}\to D$.

$D$ is dense in $X$. Indeed, let $x={\sum }_{j=0}^m{\lambda }_j{b}_{n_j}\in X$ and let $\epsilon >0$. Choose $q_j\in Q_{\mathbb{K}}$ with $\mathrm{\mid }{\lambda }_j-q_j\mathrm{\mid }<\epsilon \mathrm{/}(2(m+1)(1+{\sum }_{j=0}^m\mathrm{\parallel }{b}_{n_j}\mathrm{\parallel }))$. Then $\parallel x-{\sum }_{j=0}^m{q}_j{b}_{n_j}\parallel \le {\sum }_{j=0}^m\mathrm{\mid }{\lambda }_j-q_j\mathrm{\mid }\mathrm{\parallel }{b}_{n_j}\mathrm{\parallel }<\epsilon \mathrm{.}$ So every vector of $X$ lies in the closure of $D$.

We now run the separable-complete Baire argument inside the open unit ball $U_0:=\{x:\mathrm{\parallel }x\mathrm{\parallel }<1\}$. Because $F_0$ is closed with empty interior, the set $U_0\setminus F_0$ is nonempty and open. By step 2.2, the set $A_0:=\{m\in \mathbb{N}:d(m)\in U_0\setminus F_0\}$ is nonempty, so [L5] gives its least element $m_0$. Put $x_0:=d(m_0)$. Since $U_0\setminus F_0$ is open at $x_0$, the set $B_0:=\{k\in {\mathbb{N}}_{\ge 1}:\bar{B}(x_0,\frac{1}{k})\subseteq U_0\setminus F_0\}$ is nonempty; let $k_0$ be its least element and set $r_0:=1\mathrm{/}{k}_0$. Then $\bar{B}(x_0,r_0)\subseteq U_0\setminus F_0$.

Inductively, if closed balls $\bar{B}(x_n,r_n)\subseteq U_0\setminus F_n$ have been chosen with $\bar{B}(x_n,r_n)\subseteq B(x_{n-1},r_{n-1}\mathrm{/}2)$ for $n\ge 1$, then $B(x_n,r_n\mathrm{/}2)\setminus F_{n+1}$ is a nonempty open set. By step 2.2, the set $A_{n+1}:=\{m\in \mathbb{N}:d(m)\in B(x_n,r_n\mathrm{/}2)\setminus F_{n+1}\}$ is nonempty, so [L5] gives its least element $m_{n+1}$. Put $x_{n+1}:=d(m_{n+1})$. Since $B(x_n,r_n\mathrm{/}2)\setminus F_{n+1}$ is open at $x_{n+1}$, the set $B_{n+1}:=\{k\in {\mathbb{N}}_{\ge 1}:\frac{1}{k}<\frac{r_n}{2}\text{ and }\bar{B}(x_{n+1},\frac{1}{k})\subseteq B(x_n,r_n\mathrm{/}2)\setminus F_{n+1}\}$ is nonempty; let $k_{n+1}$ be its least element and set $r_{n+1}:=1\mathrm{/}{k}_{n+1}$. Then $\bar{B}(x_{n+1},r_{n+1})\subseteq B(x_n,r_n\mathrm{/}2)\setminus F_{n+1}\subseteq \bar{B}(x_n,r_n),$ and $r_{n+1}<r_n\mathrm{/}2$. Hence $r_n\le r_0\mathrm{/}{2}^n$ for every $n$, so $r_n\to 0$.

For $m>n$, the inclusion from step 3.2 gives $x_m\in B(x_n,r_n\mathrm{/}2)$, so $\mathrm{\parallel }{x}_m-x_n\mathrm{\parallel }<r_n\mathrm{/}2$. Hence $(x_n)$ is Cauchy. Since $X$ is Banach, [L1] gives $x_n\to x$ for some $x\in X$. Each $\bar{B}(x_n,r_n)$ is closed and contains all later $x_m$, so it contains the limit $x$; therefore $x\in \bar{B}(x_n,r_n)\subseteq X\setminus F_n$ for every $n$.

Step 4.1 contradicts $X={\bigcup }_N{F}_N$ from step 1.2. Therefore no countably infinite Hamel basis exists. The foundational point recorded in [L4] is that the proof used only a fixed countable dense set, with both the recurring point selections and the ball radii chosen canonically from $\mathbb{N}$, and not the unrestricted complete-metric Baire theorem.

By [L1], the map $K_o$ is well defined as a map into $C_b(M)$. For any $x,y,z\in M$, the triangle inequality gives $d(x,z)\le d(x,y)+d(y,z)\text{and}d(y,z)\le d(y,x)+d(x,z),$ so $\mathrm{\mid }{K}_o(x)(z)-K_o(y)(z)\mathrm{\mid }=\mathrm{\mid }d(x,z)-d(y,z)\mathrm{\mid }\le d(x,y)\mathrm{.}$ Taking the supremum over $z$ yields $\mathrm{\parallel }{K}_o(x)-K_o(y){\mathrm{\parallel }}_{\mathrm{\infty }}\le d(x,y)$.