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

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?

• Let $\rho$ be any metric which induces the uniformity $U$. Put $d=\frac{\rho}{1+\rho}$. Then $d$ also induces the uniformity $U$ and $X$ (and so each its subset) is bounded with respect to $\rho$, right? – Alex Ravsky Dec 25 '18 at 3:09
• @AlexRavsky Yes, that’s right. I posted a question about that here: math.stackexchange.com/q/3050017/71829 – Keshav Srinivasan Dec 25 '18 at 4:28
• @AlexRavsky By the way, I changed this question to reflect your answer to my other question. – Keshav Srinivasan Dec 25 '18 at 4:39
• @AlexRavsky Yes, I’d be interested in doing that. – Keshav Srinivasan Jan 18 at 8:48
• @AlexRavsky: I don't think I can help but feel free to use any ideas or material from my answer if you think it would improve your paper – Dap Jan 18 at 12:28

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.

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 $$0\leq i The basic idea of this argument (see the argument around (*) below) is that $$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 $$d''(a,x) for all $$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 $$\sum_{i=0}^nd(x_i,x_{i+1}) If any two consecutive distances $$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 $$n<2r/\epsilon+1 But that implies $$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 $$\{(a,b)\mid d(a,b)<\alpha\}=\{(a,b)\mid d'(a,b)<\alpha\}=\{(a,b)\mid d''(a,b)<\alpha\}.$$

• A bounded subset of a uniform space is not the same as a totally bounded, see a definition at the beginning of the question. Each ball of a normed space (over $\Bbb R$) is bounded, so $A$ is not a counterexample. – Alex Ravsky Jan 17 at 9:20
• @AlexRavsky: thanks, I had misread. I've now tried to answer the question using the correct definition of boundedness in a uniform space. – Dap Jan 17 at 20:31
• @Dap May I ask what part of your mathematical career (if any) you are in? – mathworker21 Jan 18 at 4:41
• @AlexRavsky: thanks for the comments, I agree those steps were missing – Dap Jan 18 at 11:51