The following question is taken from Arrows, Structures and Functors the categorical imperative by Arbib and Manes, and Theory of Mathematical Structures by Jiří Adamek.
$\color{Green}{Background:}$
[From Arbib and Mane]
$\textbf{(1) Definition:}$ The $\textbf{Free monoid on the set X of generators}$ is the set $X^*$ of all finite sequences of elements from $X$ (including the "empty" sequence $\Lambda$ of length $0$) with associative multiplication of $\textbf{concatenation}$
$$(x_1\ldots x_m)\cdot ({x_1}^{'}\ldots {x_n}^{'})=(x_1\ldots x_m, {x_1}^{'}\ldots {x_n}^{'})$$
for which $\Lambda$ is clearly the identity: $\Lambda\cdot w=w=w\cdot \Lambda$ for all $w\in X^*.$ Note that an element $x$ yields a string $(x)$ in $X^*$ of length one.
A more extensive definition from Adamek's text.
$\textbf{(2) Definition:}$ 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 $m\in M)$
$\textbf{(3)}$ 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 $M=n.$
$\textbf{(4)}$ Proposition: Let $(X,\alpha)$ be a free object over $M\subset X.$ Then $M$ is a set of generators of $(X,\alpha).$
$\textbf{(5)}$ 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 $X.$
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 $f:(X,\alpha)\to (X,2\alpha)$
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 $\textbf{Met}_k.$
$\textbf{(6) Exercise:}$ Let $U:\textbf{Met}\to \textbf{Set}$ be the underlying set functor. Show that there exists a free metric space over $B$ with respect to $U$ if and only if $B$ has at most one element.
$\color{Red}{Questions:}$
I understand the example in $\textbf{(5)}$ is the answer to the exercise in $\textbf{(6)}.$ I don't understand what it means for vanilla metric space is only free if it has at most one element. The notion of a free object involves the notion of "generators." So what does it mean for a metric space being free, having only one element to be a generators. I know that like topological space, being a free object is a discrete space. But translating to metric space, having one element means that the metric space has one element, so what does that generate.
Thank you in advance.