The bounded functions on a set form a complete metric space in the supremum metric, and the coding map of the two-sign shift space is an isometry

Let $(f_j{)}_{j\in \mathbb{N}}$ be a Cauchy sequence in $(\mathcal{B}(S),d_{\mathrm{\infty }})$ and fix $s\in S$. For all $j,k$ one has $\mathrm{\mid }{f}_j(s)-f_k(s)\mathrm{\mid }\in \{\mathrm{\mid }{f}_j(s^{\mathrm{\prime }})-f_k(s^{\mathrm{\prime }})\mathrm{\mid }:s^{\mathrm{\prime }}\in S\}$ and therefore $\mathrm{\mid }{f}_j(s)-f_k(s)\mathrm{\mid }\le d_{\mathrm{\infty }}(f_j,f_k)$ by [F1] and [F6]. Hence $(f_j(s){)}_j$ is a Cauchy sequence of reals, so it converges; define $f(s):={\lim }_j{f}_j(s)\in \mathbb{R}$. This defines a function $f:S\to \mathbb{R}$.

Isometry property: for $s,t\in X_{\pm }$ put $\mathrm{\Phi }(s,t):=d_2(\phi (s),\phi (t))$. Then $\mathrm{\Phi }(s,t)=0$ if and only if $\phi (s)=\phi (t)$ (by [F2], [F6] and definiteness of $d_2$), if and only if $s=t$ since $\phi$ is injective ([F2]). Symmetry holds because $d_2$ is symmetric, and the triangle inequality follows from that of $d_2$: $D_{\pm }(s,u)=d_2(\phi s,\phi u)\le d_2(\phi s,\phi t)+d_2(\phi t,\phi u)=D_{\pm }(s,t)+D_{\pm }(t,u)$ ([F2], [F6]). Finally $D_{\pm }(s,t)=d_2(\phi s,\phi t)$ says exactly that the bijection $\phi$ preserves distances, so it is an isometry onto $(X_2,d_2)$.

Boundedness of $f$: since $(f_j)$ is Cauchy, there is $N$ with $d_{\mathrm{\infty }}(f_j,f_N)\le 1$ for all $j\ge N$; then $\mathrm{\mid }{f}_j(s)\mathrm{\mid }\le \mathrm{\mid }{f}_N(s)\mathrm{\mid }+1\le {\sup }_{s^{\mathrm{\prime }}}\mathrm{\mid }{f}_N(s^{\mathrm{\prime }})\mathrm{\mid }+1=:M$ for all $j\ge N$ and all $s$, where $M<\mathrm{\infty }$ because $f_N$ is bounded ([F1], [F6]). For fixed $s$, the bounds $-M\le f_j(s)\le M$ pass to the limit by [F5], giving $-M\le f(s)\le M$, hence $\mathrm{\mid }f(s)\mathrm{\mid }\le M$ by [F6]. Thus $f\in \mathcal{B}(S)$.

Uniform error estimate: given $\epsilon >0$, choose $N$ with $d_{\mathrm{\infty }}(f_j,f_k)\le \epsilon$ for all $j,k\ge N$. Fix $j\ge N$ and $s\in S$; then $\mathrm{\mid }{f}_k(s)-f_j(s)\mathrm{\mid }\le d_{\mathrm{\infty }}(f_k,f_j)\le \epsilon$ for every $k\ge N$ ([F1], [F6]), thus $f_j(s)-\epsilon \le f_k(s)\le f_j(s)+\epsilon$. Passing these inequalities to the coordinate limit with [F5] gives $f_j(s)-\epsilon \le f(s)\le f_j(s)+\epsilon$, or $\mathrm{\mid }f(s)-f_j(s)\mathrm{\mid }\le \epsilon$ by [F6]. Taking the supremum over $s$ in [F1] gives $d_{\mathrm{\infty }}(f,f_j)\le \epsilon$ for every $j\ge N$, so $f_j\to f$ in $(\mathcal{B}(S),d_{\mathrm{\infty }})$.

The space $(B_R(S),d_{\mathrm{\infty }})$ is complete: if $(f_j)$ is Cauchy in $(B_R(S),d_{\mathrm{\infty }})$, then by steps 2.1 and 3.1 its pointwise limit $f$ lies in $\mathcal{B}(S)$ and $f_j\to f$; for each $s$ one has $-R\le f_j(s)\le R$ for all $j$, so [F5] gives $-R\le f(s)\le R$, hence $\mathrm{\mid }f(s)\mathrm{\mid }\le R$ by [F6]; hence $f\in B_R(S)$ and the Cauchy sequence converges inside $B_R(S)$.

Conclusion. Clause 1 is proved by steps 1.1, 2.1 and 3.1: the pointwise limit of a Cauchy sequence of bounded functions is bounded (step 2.1) and is the uniform limit (step 3.1). Clause 2 is step 4.1, and clause 3 is step 1.2. For clause 4, take $S=\mathbb{Z}$: by the definitions of The two-sign bilateral shift space, its coding map, and the ball of bounded bilateral real sequences the space $\mathcal{B}(\mathbb{Z})$ is exactly ${\mathrm{\ell }}^{\mathrm{\infty }}(\mathbb{Z})$ and $d_{\mathrm{\infty }}(f,g)={\sup }_k\mathrm{\mid }{f}_k-g_k\mathrm{\mid }=\mathrm{\parallel }f-g{\mathrm{\parallel }}_{\mathrm{\infty }}$, so clauses 1 and 2 give completeness of ${\mathrm{\ell }}^{\mathrm{\infty }}(\mathbb{Z})$ and of each $B_R$, $R\ge 0$; with $S=\mathbb{N}$ the same identifications give the one-sided statement, the argument being identical since it never uses any structure of the index set beyond its nonemptiness.