Is a set bounded in every metric for a uniformity bounded in the uniformity?

Question

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?

Answer 1

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 $V_1\in\mathcal U$ such that for each finite subset $F$ of $X$ and each natural number $n$, $A\not\subset V^n_1[F]$.

Choose a base $\{V_i\}$, $n\ge 2$ of the uniformity $\mathcal U$ consisting of symmetric entourages such that $V^3_{i+1}\subset V_i$ for each $i\ge 1$. For each $n\le 0$ 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 $\{V_i:i\in\Bbb Z\}$ 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 $x\in X$ 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 $\{(x,y):\rho(x,y)<1/2^i\}\subset V_i\subset \{(x,y):\rho(x,y)\le 1/2^i\}.$

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 $[A]=\{[x]\in V: [x]\cap A\ne\varnothing\}$. Define a function $f: \widehat V \to\Bbb N$ such that $f\equiv 1$, if $[A]$ is finite, and $f|[A]$ is unbounded, otherwise.

At last, for each $x,y\in X$ 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 $x,y\in X$, the metric $\rho’$ induces the uniformity $\mathcal U$ on the set $X$.

Let $a\in X$ be any element. If $[A]$ is finite, there exists a class $[x]\in \widehat V$ such that $A\cap [x]\not\subset V^n_1[p[x]]$ 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.

Answer 2

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 $A\not\subseteq V^n[F].$ 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 $a\sim b$ if there is a path $a=x_0,x_1,\dots,x_n=b$ with $d(x_i,d_{i+1})<\epsilon$ for each $0\leq i<n.$ 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

• $d'(a,b)=\inf\left\{\sum_{i=0}^nd(x_i,x_{i+1})\mid x_0=a, x_n=b, d(x_i,x_{i+1})<\epsilon\right\}$ if $a\sim b$
• $d'(a,b)=\infty$ 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 $C\in X/\sim$ (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 $f:(X/\sim)\to\mathbb N$ 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 $a\sim b$
• $d''(a,b)=d'(a,t_C)+\max(1,|f(C)-f(C')|)+d'(t_{C'},b)$ if $a\in C$ and $b\in C'$ 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 $a\in A.$ For large enough $i$ we have $i>r+f([x])$ where $[x]$ is the equivalence class of $x.$ There exists $a\in C_i\cap A,$ 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 $a\sim b$
• $d''(a,b)=d'(a,t_C)+1+d'(t_{C'},b)$ if $a\in C$ and $b\in C'$ 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 $C\in X/\sim$ such that for all $n,F$ we have $A\cap C\not\subseteq V^n[F].$ (Suppose not; for each $C$ intersecting $A$ there are $n_C,F_C$ with $A\cap C\subseteq V^{n_C}[F_C],$ but then $A\subseteq V^{\max n_C}[\bigcup F_C]$ which contradicts the definition of $V.$)

Suppose $A\cap C$ is contained in the $d''$-ball of radius $r$ around $a\in X.$ If $a\notin C,$ replace it by $t_C$ - the ball will still contain $A\cap C$ 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 $A\cap C\not\subseteq V^N[\{x\}],$ which implies there is a point $b\in (A\cap C)\setminus V^N[\{x\}].$

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

$$(n-1)\epsilon\leq\sum_{i=0}^{n-2}(d(x_{i},x_{i+1})+d(x_{i+1},x_{i+2}))<2r\tag{*}$$

so $n<2r/\epsilon+1<N.$ But that implies $b\in A\cap C\setminus V^N[\{x\}],$ 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 $$\{(a,b)\mid d(a,b)<\alpha\}=\{(a,b)\mid d'(a,b)<\alpha\}=\{(a,b)\mid d''(a,b)<\alpha\}.$$