No structure in ##(d,x)## in ##\textbf{Met*}(X)## is admissible

[elias001]

TL;DR

I want to know how to show no structure $(d,x)$ in $\textbf{Met*}(X)$ such that each $\pi_i:(X,d,x)\to (\textbf{R}_+, |x-y|,0)$ is admissible

The following question are taken from $\textit{Arrows, Structures and Functors the categorical imperative}$ by Arbib and Manes, and from $\textit{Algebraic Theories}$ by Manes

$\color{blue}{Question}\color{blue}{/difficulties:}$

I am having a lot of difficulties doing the $\color{green}{Exercise:}$ from the end of the second quoted passage below for a few reasons. The $\textbf{difficulties}$ I have are as follows:

I don't know how to describe the mapping $\pi_i:(X,d,x)\to (\textbf{R}_+, |x-y|,0)$ written out in explicit functional elements notation. From $X=\textbf{R}^I$ with $I$ being infinite, the domain of $\pi_i$ would involved some sort of product of metrics, and $\pi_i$ is suppose to serve as some sort of projection map. So we would have $d_1(a_1,b_1), d_2(a_2,b_2), \ldots, d_i(a_i,b_i), i\in I,$ and $\pi_jd_i(a,b)=d_j(a_j,b_j),$ where each $d_i(a_i,b_i),$ is a metric in $\textbf{Met*}(X)$ for all $i\in I,$ $d_j(a_j,b_j)$ is a metric in $\textbf{Met*}(\textbf{R}_+)$ and both $a=(a_i)_{i\in I},b=(b_i)_{i\in I},$ are elements of $X=\textbf{R}.$ I am not sure if this interpretation is correct.

In that exercise question, the author asks the reader to show that there is no structure in $(d,x)$ that would make each $\pi_i$ admissible. I think I am suppose to do it by contradiction and assume such added structure exists, then either show that $\pi_i$ is either not a contraction or is not an isometry. I am not sure if this would be a feasible approach.

$\color{purple}{Background}$ $\color{purple}{Information}$

(From Manes)

$\textbf{Categories of}$ $\mathscr{K-}$$\textbf{Objects with Structure.}$ Let $\mathscr{K}$ be a (fixed base) category. A $\textit{literal category,}$ $\mathscr{C,}$ of $\mathscr{K-}$$\textit{objects with structure}$ is defined by the following two data and two axioms:

$\mathscr{C}$ assigns to each object $K$ of $\mathscr{K}$ a class $\mathscr{C}(K)$ of $\mathscr{C-}$$\textit{structures on}$ $K.$ A $\mathscr{C-}$$\textit{structure}$ is a pair $(K,s)$ with $s\in \mathscr{C}(K).$

For each ordered pair $(K,S;L,t)$ of $\mathscr{C-}\textit{structures,} \mathscr{C,}$ assigns a subset $\mathscr{C}(s,t)$ of $\mathscr{K}(K,L)$ of $\mathscr{C}-\textit{admissible } \mathscr{K-}\textit{morphisms from } (K,s) \textit{to } (L,t);$ to denote that $f:K\rightarrow L$ is in $\mathscr{C}(s,t)$ we will write $f(K,s)\rightarrow (L,t)$ of $f:s\rightarrow t$ (if necessary, imposing additional decoration should more than one $\mathscr{C}$ be in the picture).

The two axioms are:

$\textit{Axiom of Composition.}$ If $f:s\rightarrow t$ and $g:t\rightarrow u$ then $f\circ g:t\rightarrow u.$

$\textit{Structure is Abstract.}$ If $f:K\rightarrow L$ is an isomorphism in $\mathscr{K}$ then for all $t\in \mathscr{C}(L)$ there exists unique $s\in \mathscr{C}(K)$ such that $f:s\rightarrow t$ and $f^{-1}:t\rightarrow s.$ optimal; that is, specifically, if whenever $(K',s')$ is a $\mathscr{C-}$structure and $g:K\rightarrow K'$ is a $\mathscr{K-}$morphism such that (please see the following image)


$f_i\circ g$ is admissible for all $i$ then $g$ is also admissible.

(From Arbib and Manes):

A $\textbf{category, C, of sets with structure}$ is given by the following two data and two axioms:

$\textbf{C}$ assigns to each set $X$ a set $\textbf{C}(X)$ of $\textbf{C-structures on } X$. $\textbf{C-structure,}$ then, is a pair $(X,s)$ with $s$ in $\textbf{C}(X).$ For each pair of sets $(X,Y),\textbf{C}$ assigns a function

$$\textbf{C}(X)\times \textbf{C}(Y)\rightarrow\textbf{P}(Y^{X}):s,t\mapsto \textbf{C}(s,t)$$

where $Y^{X}$ is, recall, the set of functions from $X$ to $Y$ and $\textbf{P}(Y^{X})$ is the set of its subsets. We write "$f:(X,s)\rightarrow (Y,t)$" and say "$f$ is $\textbf{admissible}$ in $\textbf{C}$ from $s$ to $t$" just in case $f$ is in $\textbf{C}(s,t).$ The axioms are:

$\textbf{Admissible maps to composable:}$ If $f:(X,s)\rightarrow (Y,t)$ and $g:(Y,t)\rightarrow (Z,u)$ then $g\circ f:(X,s)\rightarrow (Z,u).$

$\textbf{Structure is abstract:}$ If $f:X\rightarrow Y$ is a bijection (i.e. an isomorphism of sets) and if $t$ is in $\textbf{C}(Y)$ there exists a unique $s$ in $\textbf{C}(X)$ with $f:(X,s)\rightarrow (Y,t)$ and $f^{-1}:(Y,t)\rightarrow (X,s).$

$\textit{Some mappings between metric spaces.}$ Let $(X,d), (Y,e)$ be metric spaces. A function $f:X\to Y$ is an $\textbf{isometry}$ $\textit{from}$ $(X,d)$ $\textit{into}$ $(Y,e)$ if for all $x_1,x_2\in X, e(fx_1,fx_2)=d(x_1,x_2),$ that is, if $f$ preserves distances. An isometry is automatically one-to-one, for if $x_1\neq x_2$ then $e(fx_1,fx_2)=d(x_1,x_2)\neq 0,$ which implies $fx_1\neq fx_2. f$ is an $\textbf{isomorphism}$ $(X,d)\to (Y,e)$ if $f$ is an isometry and $f$ is onto. In this case $f^{-1}$ is also an isomorphism, since $d(f^{-1}y_1, f^{-1}y_2)=e(ff^{-1}y_1, ff^{-1}y_2)=e(y_1,y_2).$
$f:(X,d)\to (Y,e)$ is a $\textit{Lipschitz map}$ if there exists $\lambda >0$ such that $e(fx_1,fx_2)\leq \lambda d(x_1,x_2)$ for all $x_1,x_2\in X.$

$\textit{The categories}$ $\textbf{Met1}$ $\textit{and}$ $\textbf{Met*}.$ Let $\textbf{Met1}$ be the category whose objects are metric spaces $(X,d)$ $\textit{of diameter}\leq 1$ (that is, $d(x,y)\leq 1$ for all $x,y$) and whose morphisms are contractions. Let $\textbf{Met*}$ be the category whose objects are $\textit{metric spaces with base point}$
$(X,d,x)$ (meaning $(X,d)$ is a metric space and $x$ "the base point" is an arbitrary element of $X$) and whose morphisms $f:(X,d,x)\to (Y,e,y)$ are contractions $f:(X,d)\to (Y,e)$ such that $fx=y.$ It is clear that $\textbf{Met*}$ is a category since if $fx=y$ and $gy=z$ then $gfx=gy=z.$

$\textit{Every}$ family in $\textbf{Met*}$ has a product, although it need not be built on the cartesian product set! Given $(X_i,d_i,x_i)$ let $X$ be the subset of all $I-$tuples $(a_i)$ with each $a_i\in X$ such that the $I-$tuple of numbers $d_i(a_i,x_i)$ has an upper bound, and define $d((a_i),(b_i))=\text{Sup }\{d_i(a_i,b_i)\mid i\in I\}.$ Then $x=(x_i)$ is in $X.$ Since there exist, by definition of $X,$ numbers $s$ and $t$ such that $d_i(a_i,x_i)\leq t$ for all $i$ and $d_i(b_i,x_i)\leq s$ for all $i$ then $$d_i(a_i,b_i)\leq d_i(a_i,x_i)+d_i(x,b_i)=d_i(a_i,x_i)+d_i(b_i,x_i)\leq t+s$$

which proves that $d((a_i),(b_i))$ is a well -defined number. Let $p_i:(X,d_,x)\to (X_i,d_i,x_i)$ be the usual coordinate projections.


$\color{green}{Exercise:}$ Consider $(\textbf{R}_+,|x-y|,0)$ in $\textbf{Met*}.$ Show that, if $X=\textbf{R}^I$ with $I$ infinite, there is no structure $(d,x)$ in $\textbf{Met*}(X)$ such that each $\pi_i:(X,d,x)\to (\textbf{R}_+, |x-y|,0)$ is admissible.

Thank you in advance.