Questions about how to construct Free Metric Space?

Question

The following are from: "Arrows, Structures and Functors the categorical imperative" by Arbib and Manes, "Theory of Mathematical Structures" by Jiří Adamek, "Lectures and Exercises on Functional Analysis" By A. Ya. Helemskii, Translated by S. Akbarov, the article: "Free objects in the theory of Categories" by Z. Semadeni (Poznan)

$\color{Green}{Background:}$

[From Adamek]

$\textbf{(1a) Definition:}$ An object $(X,\alpha)$ is said to be $\textit{free over a set}$ $M\subset X$ provided that for each object $(Y,\beta)$ and each map $f_0:M\to Y$ there exists a unique morphism $f:(X,\alpha)\to (Y,\beta)$ extending $f_0$ (i.e., with $f(m)=f_0(m)$ for all $m\in M)$

$\textbf{(1b) Definition:}$ A construct is said to have $\textit{free object}$ if for each cardinal number $n$ there exists a free object on $n$ generators.

$\textbf{(2)}$ Terminology. If $(X,\alpha)$ is a free object over $M$ then $M$ is called a $\textit{set of free generators}$ We also say that $(X,\alpha)$ is a $\textit{free object on n generators}$ if card $M=n.$

$\textbf{(3)}$ Proposition: Let $(X,\alpha)$ be a free object over $M\subset X.$ Then $M$ is a set of generators of $(X,\alpha).$

$\textbf{(4)}$ Generalizing the situation in $\textbf{Top}$ and $\textbf{Pos},$ we call an object $(X,\alpha)$ $\textit{discrete}$ if for each object $(Y,\beta),$ all maps $f:X\to Y$ are morphisms $f:(X,\alpha)\to (Y,\beta).$ Equivalently, an object $(X,\alpha)$ is discrete iff it is free over all of $X.$

[From Arbib and Manes]

$\textbf{(5a)}$ $\textit{The Category}$ $\textbf{Met}.$ Let $(X,d), (Y,e)$ be metric spaces. A $\textbf{contraction}$ from $(X,d)$ $\textit{to}$ $(Y,e)$ is a function $f:X\to Y$ satisfying $e(fx_1,fx_2)\leq d(x_1,x_2)$ It is clear that a contraction is Lipschitz (take $\lambda=1$). By adapting the proof for Lipschitz maps above it is clear that $g\cdot f$ is a contraction when $f$ and $g$ are and, of course, $\text{id}_X:(X,d)\to (X,d)$ is a contraction. We obtain the category $\textbf{Met}$ of metric spaces and contractions.

[From Adamek]

$\textbf{(5b)}$$\textbf{Examples:}$ The construct $\textbf{Met}$ does not have free objects on two or more generators. If card $M>1$ and if $(X,\alpha)$ is a free metric space over $M\subset X,$ consider the space $(X,2\alpha):$ the inclusion $f_0=v:M\to X$ has, of course, no extension to a contraction $f:(X,\alpha)\to (X,2\alpha)$

$\textbf{(5c)}$$\textbf{Examples:}$ For each real number $k>0$ denote by

$$\textbf{Met}_k$$

the full subconstruct of $\textbf{Met},$ the objects of which are the metric spaces $(X,\alpha)$ with diameter at most $k,$ i.e., such that

$\alpha(x_1,x_2)\leq k\quad$ for all $x_1,x_2\in X.$

Then $\textbf{Met}_k$ has free objects, in fact,discrete objects: for each set $X$ define a metric $\alpha$ by

$\begin{equation*} \alpha(x_1,x_2)=\begin{cases} k \quad &\text{if } \, x_1\neq x_2 \quad\{x_1,x_2\in X\} \\ 0 \quad &\text{if } \, x_1= x_2 \\ \end{cases} \end{equation*}$

Then $(X,\alpha)$ is discrete in $\textbf{Met}_k.$

$\textbf{(6)}$[from: A. Ya. Helemskii, Translated by S. Akbarov, pp 50-51]

"Free objects in $\textbf{Gr}$ and in $\textbf{Ab}$ are precisely what is called in free groups and free abelian groups (indicate the basis). Finally free objects in $\textbf{Top}$ are discrete topological spaces, and in $\textbf{Met},$ discrete metric spaces; in these cases the basis coincide with the underlying set."

$\textbf{(7)}$ [From: Z. Semadeni (Poznan), pg 1]

Let us consider the category of metric spaces of diameter less than or equal to 1, the morphisms being contractions (a $\textit{contraction}$ means a transformation which satisfies the Lipschitz condition with the constant 1). Then an isomorphism is just an isometry. A $\textit{free join}$ of a family $\{A_t\}_{t\in T}$ of objects in this category of all sets) metrized as follows: $c(a,b)=c_t(a,b)$ if both $a$ and $b$ belong to the same space $A_t,$ and $c(a,b)=1$ if $a\in A_t, b\in A_u, t\neq u.$ A $\textit{direct join}$ is the Cartesian product $\prod {A_t}$ with the uniform metric

$c(\{a_t\},\{b_t\})=\text{sup}\{c_t(a_t,b_t):t\in T\}.$

Clearly a one-point set is a basic free object in this category and hence a "free metric space" is a set $A$ with the trivial distance $c(a,b)=1$ for $a\neq b.$

$\color{Red}{Questions:}$

I have several questions I like to ask about how to construct a free metric space.

I know that for free monoids, from its definition, if we have a set $A$ equipped with binary operations of monoid homomorphism and an identity element. With those binary operations and elements from $A$, along with the operation of concatenation, we can form different combinations of words etc to generate the set $A^*$ and hence $A^*$ is the set of list strings from set $A.$ Here the elements from $A$ are used to generate the set $A^*.$ Also the construction of a free monoid does not resemble $\textbf{(1a)}$ definition, in the sense from how is describe in most texts about free monoids, there is never a mention of the unique morphism $f$ extending each map $f_0$.

I don't understand in the example referred to $\textbf{(5b)},$ how the morphism $f:(X,\alpha)\to (X,2\alpha)$ shows that $\textbf{Met}$ does not have free objects. I also don't understand what it means for a metric space to be free. I understand from the definition $\textbf{1a},$ and $\textbf{1b},$ and the example in $\textbf{5b},$ that in order for a metric space in the category $\textbf{Met}$ to be considered free, it has to have only one element, but then in what sense is that a generator.

For a concrete example along the lines of $\textbf{(5c)}$ Say we have a set $X=\mathbb{N}$ and we have a subset $M=\{2\}\subset X.$ Let $k=1,$ and we define a metric on $X$ such that for any $x\in X$

$\begin{equation*} d(x,2)=\begin{cases} 1 \quad &\text{if } \, x\neq 2 \quad\{x\in X\} \\ 0 \quad &\text{if } \, x=2 \\ \end{cases} \end{equation*}$

We can repeat the above construction for different values of $k$ and letting the subset $M={x_0},$ for some $x_0\in X$. The general example in $\textbf{(5c)},$ the subset $M$ consisting of a single element is never specified. After such construction, according to the example in $\textbf{(5c)},$ we have $(X,d)$ a discrete metric space in the category of $\textbf{Met}_k$

But the example in $\textbf{(5c)}$ never specific that $(X,\alpha)$ has one generator, and also, having one generator, what set is it suppose to be generating. According to the quoted passage from $\textbf{(6)},$ for a free metric space, it has the same basis as its underlying set, which I also don't know what that means and how it can be.

I also did some further research on google for more information about what a free metric space is or its definition. The only thing I found are the quoted passages from $\textbf{(6)}$ and $\textbf{(7)}.$

Thank you in advance

Answer 1

You write:

"I don't completely understand how in the case of metric space, what it means to be a free metric space with one generator. [S]pecifically what is it suppose to generate?"

The whole idea of this abstract definition is to shift away from the idea of (say) a free group being built out of its generators and towards that same free group having homomorphisms out of it being determined by, and without restrictions on, their actions on those generators. That is, we're shifting from

The elements of $F_2$ are the things that can be built out of $a$ and $b$

to

To check whether some group $G$ is "$F_2$ in disguise," we see whether there are elements $a,b\in G$ such that for every group $H$ and every map of sets $f:\{a,b\}\rightarrow H$ there is exactly one group homomorphism $\hat{f}:G\rightarrow H$ extending $f$.

The point is that this latter "map-theoretic" definition can be applied directly to things like metric spaces where there is no obvious notion of "generation" matching our usual intuitions. You'll notice that Definition 1a makes no reference to generation at all.

(Also, note that the starting $f$ is not required to be, and usually will not be, a morphism in the relevant category; it's just a map-of-sets.)

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A test problem

To get a feel for this, let's check the following:

• Claim: $\mathbb{R}$ is not free over $\{0,1\}$.

Here $\mathbb{R}$ is being viewed as a metric space with the usual metric.

• Proof: Well, I need to push back against either existence or uniqueness. In fact, existence already breaks down, so let's attack that.

  Let $M$ be the two-point metric space with points $a,b$ and $d(a,b)=1$. Then the map-of-sets $$f:\{0,1\}\rightarrow M: 0\mapsto a,\,\, 1\mapsto b$$ cannot be extended to a contraction $\mathbb{R}\rightarrow M$. Indeed, there is no nonconstant contraction map from $\mathbb{R}$ to $M$ at all, do you see why?

Example 5b in the text shows that this is in fact overkill: we could show in a "uniform" way that no metric space is every free over a set with more than one element (and as above this only leans on the existence part of the definition).

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Another test problem

It may help to see a "geometry-flavored" situation where (somewhat) interesting free objects do exist.

Let's let ${\bf Met'}$ be the category whose objects are metric spaces and whose morphisms are continuous maps (not just contractions). In ${\bf Met'}$, we do in fact have lots of free objects:

If $(X,d)$ is discrete, then $(X,d)$ is free over $X$ in the sense of ${\bf Met'}$.

Proof: Suppose $(Y,e)$ is a metric space and $f:X\rightarrow Y$ is a set function. Since $(X,d)$ is discrete, $f$ is in fact already continuous and so is a morphism in ${\bf Met'}$. $\quad\Box$

This example much more closely matches our intuition for "building spaces out of points," and so might be easier to think about at first.

Exercise: given a metric space $(X,d)$ and a subset $M\subseteq X$, when is $(X,d)$ free in the sense of ${\bf Met'}$ over $M$?