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This is a follow-up to my question here. 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 this answer shows that if $U$ is the uniformity induced by a metric $d$, then a set bounded with respect to $U$ is also bounded with respect to $d$, but the converse need not be true.

But I’m interested in whether something weaker is true. Suppose that $(X,U)$ is a metrizable uniform space, and $A$ is a subset of $X$ which is bounded with respect to every metric which induces $U$. Then is $A$ bounded with respect to $U$?

To put it another way, is the collection of bounded sets with respect to a metrizable uniformity equal to the intersection of the collections of bounded sets with respect to each of the metrics for the uniformity?

Yesterday I lost Internet connection, so I wrote my answer offline and didn’t see similer Dap’s answer.

The answer is affirmative. Assume that $A$ is unbounded. Then there exists a symmertic entourage $V_1inmathcal U$ such that for each finite subset $F$ of $X$ and each natural number $n$, $Anotsubset V^n_1[F]$.

Choose a base ${V_i}$, $nge 2$ of the uniformity $mathcal U$ consisting of symmetric entourages such that $V^3_{i+1}subset V_i$ for each $ige 1$. For each $nle 0$ put $V_i=V_1^{3^{1-i}}$.

To construct a metric $rho$ in which $A$ is not contained in any ball we formulate an unbounded counterpart of a fundamental Theorem 8.1.10 from Engelking’s “General topology” (2nd edn.).

Lemma. For every sequence ${V_i:iinBbb Z}$ of symmetric members of a uniformity $mathcal U$ on a set $X$, where $V^3_{i+1}subset V_i$ for each $i$ there exists a function $rho$ on the set $V=bigcup V_i$ such that

(i) For each $xin X$ we have $(x,x)in V$ and $rho(x,x)=0$.

(ii) For each $(x,y)in V$ we have $(y,x)in V$ and $rho(x,y)=rho(y,x)$.

(iii) For each $(x,y),(y,z)in V$ we have $(x,z)in V$, and $rho(x,z)le rho(x,y)+ rho(y,z)$.

(iv) For each $i$ we have ${(x,y):rho(x,y)<1/2^i}subset V_isubset {(x,y):rho(x,y)le 1/2^i}.$

The proof of Lemma is almost the same as that of Theorem 8.1.10, so we skip it.

Remark that conditions (i)-(iii) imply that $V$ is an equivalence relation. Let $widehat V$ be the set of classes of the relation $V$. For each class $[x]in widehat V$ pick a point $p[x]in [x]$. Let $[A]={[x]in V: [x]cap Anevarnothing}$. Define a function $f: widehat V toBbb N$ such that $fequiv 1$, if $[A]$ is finite, and $f|[A]$ is unbounded, otherwise.

At last, for each $x,yin X$ put
$$rho’(x,y)=cases{rho(x,y), mbox{ if }(x,y)in V,\
1+|f([x])- f([y])|+rho(x, p[x])+ rho(y,p[y]), mbox{ otherwise}.}$$

It is easy to check that $rho’$ is a metric on $X$. Since and $r(x,y)le 1/2$ iff $r’(x,y)le 1/2$ for each $x,yin X$, the metric $rho’$ induces the uniformity $mathcal U$ on the set $X$.

Let $ain X$ be any element. If $[A]$ is finite, there exists a class $[x]in widehat V$ such that $Acap [x]notsubset V^n_1[p[x]]$ for each natural number $n$. Condition (iv) of Lemma imply that a set $rho(A,p[x])$ is unbounded, so a set $rho’(A,a)$ is unbounded too. If $[A]$ is infinite then $f|[A]$ is unbounded, so a set $rho’(A,a)$ is unbounded too.

Fix a set $A$ and an entourage $V$ witnessing that $A$ is not bounded with respect to the uniformity. So for all $n,F$ we have $Anotsubseteq V^n[F].$ We need to construct a metric for the uniformity in which $A$ is not bounded.

We are given some metric $d$ for the uniformity, and we can assume that $V={(a,b)mid d(a,b)<epsilon}$ for some $epsilon>0.$
Define $asim b$ if there is a path $a=x_0,x_1,dots,x_n=b$ with $d(x_i,d_{i+1})<epsilon$ for each $0leq i<n.$
The basic idea of this argument (see the argument around (*) below) is that $A$ is not contained in any finite union of balls of the extended metric $d'$ defined as a path metric by

• 
  $d'(a,b)=infleft{sum_{i=0}^nd(x_i,x_{i+1})mid x_0=a, x_n=b, d(x_i,x_{i+1})<epsilonright}$ if $asim b$
  


• 
  $d'(a,b)=infty$ otherwise.

The problem is that $d'$ may take infinite values so fail to be a metric.

Pick an element $t_C$ in each equivalence class $Cin X/sim$ (using the axiom of choice).

Case 1. $A$ intersects infinitely many classes in $X/sim.$

By the axiom of choice there is a sequence $C_1,C_2,dots$ of distinct equivalence classes intersecting $A.$ Define $f:(X/sim)tomathbb N$ such that $f(C_i)=i$ and $f(C)=1$ if $C$ is not equal to any $C_i.$ Define a metric $d''$ by:

• 
  $d''(a,b)=d'(a,b)$ if $asim b$
  


• 
  $d''(a,b)=d'(a,t_C)+max(1,|f(C)-f(C')|)+d'(t_{C'},b)$ if $ain C$ and $bin C'$ where $C,C'in X/sim$ are disjoint equivalence classes

I claim that $d''$ is a metric for the uniformity in which $A$ is not bounded. Suppose not, so there exists $x,r$ such that $d''(a,x)<r$ for all $ain A.$ For large enough $i$ we have $i>r+f([x])$ where $[x]$ is the equivalence class of $x.$ There exists $ain C_icap A,$ but then $d''(a,x)>r$ which contradicts the choice of $r.$

Case 2. $A$ intersects finitely many $sim$-equivalence classes.

Define $d''$ in the same way but with $f$ constant, so

• 
  $d''(a,b)=d'(a,b)$ if $asim b$
  


• 
  $d''(a,b)=d'(a,t_C)+1+d'(t_{C'},b)$ if $ain C$ and $bin C'$ where $C,C'in X/sim$ are disjoint equivalence classes

I claim that $d''$ is a metric for the uniformity in which $A$ is not bounded.

There must be some class $Cin X/sim$ such that for all $n,F$ we have $Acap Cnotsubseteq V^n[F].$ (Suppose not; for each $C$ intersecting $A$ there are $n_C,F_C$ with $Acap Csubseteq V^{n_C}[F_C],$ but then $Asubseteq V^{max n_C}[bigcup F_C]$ which contradicts the definition of $V.$)

Suppose $Acap C$ is contained in the $d''$-ball of radius $r$ around $ain X.$ If $anotin C,$ replace it by $t_C$ - the ball will still contain $Acap C$ since the distance from any point in $C$ to $t_C$ is less than its distance to any point not in $C.$ Pick an integer $N>2r/epsilon+1.$ We know $Acap Cnotsubseteq V^N[{x}],$ which implies there is a point $bin (Acap C)setminus V^N[{x}].$

Consider a list $a=x_0,x_1,dots,x_n=b$ with each $d(x_i,x_{i+1})<epsilon$ and $sum_{i=0}^nd(x_i,x_{i+1})<r.$ If any two consecutive distances $d(x_i,x_{i+1}),d(x_{i+1},x_{i+2})$ sum to less than $epsilon$ we can delete the middle element $x_{i+1}$ to get a shorter list with the same properties. Eventually we get a list where every two consecutive distances sum to at least $epsilon.$ Therefore

$$(n-1)epsilonleqsum_{i=0}^{n-2}(d(x_{i},x_{i+1})+d(x_{i+1},x_{i+2}))<2rtag{*}$$

so $n<2r/epsilon+1<N.$ But that implies $bin Acap Csetminus V^N[{x}],$ contradicting the choice of $b.$

Finally note that $d,d',$ and $d''$ (for either case) all define the same uniformity since for $alpha<min(1,epsilon)$ we have
$${(a,b)mid d(a,b)<alpha}={(a,b)mid d'(a,b)<alpha}={(a,b)mid d''(a,b)<alpha}.$$