# 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 $$d$$?

• 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? – Henno Brandsma Dec 22 '18 at 22:13
• The final part of said exercise suggests that the converse does hold for connected spaces, so a counterexample has to be disconnected I think. – Henno Brandsma Dec 22 '18 at 22:34
• 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. – Henno Brandsma Dec 22 '18 at 23:13
• @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? – Keshav Srinivasan Dec 22 '18 at 23:44
• @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. – 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 $$d(x,y). Pick any point $$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).

• 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? – Keshav Srinivasan Dec 23 '18 at 14:10
• What am I missing? – Keshav Srinivasan Dec 23 '18 at 21:25