Do boundedness in a metric and boundedness in a uniformity not coincide? A subset $A$ of a uniform space is said to be bounded if for each entourage $V$, $A$ is a subset of $V^n[F]$ for some natural number $n$ and some finite set $F$.  A subset of a metric space is said to be bounded if it is contained in some open ball.  Now if $U$ is the uniformity induced by a metric $d$, then the open balls with respect to $d$ are entourages in $U$, so clearly a set bounded with respect to $d$ is also bounded with respect to $U$.
But this journal paper says that the converse is not true:

In a metric space $(X,d)$ we have that each set that is bounded for the metric $d$ is bounded ... for the underlying uniformity, but the converse is in general not true.

So my question is, what is an example of a metric space $(X,d)$ where some sets bounded with respect to the the uniformity induced by $d$ are not bounded with respect to $d$?
 A: In a metric space $(X,d)$ bounded can mean three things: there are three bornologies we can talk about:


*

*$\mathcal{U}(d)$-bounded sets (in the Bourbaki sense).

*$\mathcal{U}(d)$-totally bounded sets (which Bourbaki calls precompact).

*$d$-bounded sets in the standard sense (being contained in a ball).


If $d$ and $d'$ are uniformly equivalent, of course 1 and 2 are the same but 3 can be different wrt $d$ or $d'$. The $\frac{d}{1+d}$ case is an example.
If $d$ and $d'$ are strongly equivalent, for all $i \in \{1,2,3\}$ $(X,d)$ and $(X,d')$ agree on boundedness notion $i$.
Concretely, take $\mathbb{R}$ in the metric $d(x,y) = \min(|x-y|,1)$. Then $A=\mathbb{R}$ is $d$-bounded but not "bounded" (from the uniformity, as in the paper).
A: I see the situation vice versa. Assume that a subset $A$ of a metric space $(X, d)$ is bounded with respect to the uniformity $\mathcal U(d)$ induced by $d$. Pick an arbitrary $\varepsilon>0$. Let $$V=\{(x,y)\in X\times X: d(x,y)<\varepsilon\}\in\mathcal U(d).$$ Therefore there exists a number $n$ and a finite subset $F$ of $X$ such that $A\subset V^n[F]$. That is 
for each point $y\in A$ there exists a point $x\in F$ such that $y\in V^n[F]$. The triangle inequality implies that $d(x,y)<n\varepsilon$. Pick any point $x\in F$. Then the triangle inequality implies that the set $A$ is contained in an open ball centered at $x$ with the radius $n\varepsilon+\max \{d(x,y):y\in F\}$, that is $A$ is bounded with respect to the metric $d$.
Conversely, let $X$ be an infinite set endowed with the metric $d(x,y)=0$, if $x=y$, and $d(x,y)=1$, otherwise for each $x,y$ in $X$. Then $X$ is contained in an open ball of radius $2$ centered at any point $x\in X$. Let $$V=\{(x,y)\in X\times X: d(x,y)<1\}\in\mathcal U(d).$$ Then $V$ is the diagonal of the set $X\times X$, so $V^n=V$ for each $n$. Therefore $V^n[F]=F$ for each (finite) subset $F$ of $X$, that is the space $X$ is not $\mathcal U(d)$-bounded.
