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 $d$?

  • $\begingroup$ The paper quotes "Topologie Génerale" as a source for the definition. I looked (it is defined in an exercise, not in the main text) but there is no mention of a comparison with the metric case there. Sloppy of that paper to state this without reference. Have you asked the author? $\endgroup$ – Henno Brandsma Dec 22 '18 at 22:13
  • $\begingroup$ The final part of said exercise suggests that the converse does hold for connected spaces, so a counterexample has to be disconnected I think. $\endgroup$ – Henno Brandsma Dec 22 '18 at 22:34
  • $\begingroup$ Maybe the author means that $X$ can have an equivalent uniformity (ie yields same topology) and be bounded in that uniformity but not in $d$? I think one can prove that if we use the uniformity induced by $d$ then Bourbaki-bounded sets are $d$-bounded. $\endgroup$ – Henno Brandsma Dec 22 '18 at 23:13
  • $\begingroup$ @HennoBrandsma I just emailed the author, let's see what he says. By the way, do you know whether two metrics which induce the same uniformity have the same bounded sets? $\endgroup$ – Keshav Srinivasan Dec 22 '18 at 23:44
  • $\begingroup$ @HennoBrandsma I'm starting to question my whole understanding of uniformities. This book claims that for any metric $d$, the metric $\frac{d}{1+d}$ is uniformly equivalent to $d$. But if two metrics are uniformly equivalent then don't they induce the same uniformity? I'm really confused. $\endgroup$ – Keshav Srinivasan Dec 23 '18 at 1:45

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.


In a metric space $(X,d)$ bounded can mean three things: there are three bornologies we can talk about:

  1. $\mathcal{U}(d)$-bounded sets (in the Bourbaki sense).
  2. $\mathcal{U}(d)$-totally bounded sets (which Bourbaki calls precompact).
  3. $d$-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).

  • $\begingroup$ Wait a minute, doesn’t the paper say all $d$-bounded sets are $U(d)$-bounded, but the converse need not be true? So how are you giving an example of a set which is $d$-bounded but not $U(d)$-bounded? $\endgroup$ – Keshav Srinivasan Dec 23 '18 at 14:10
  • $\begingroup$ What am I missing? $\endgroup$ – Keshav Srinivasan Dec 23 '18 at 21:25

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