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After looking a bit at uniform spaces, as their general definition seems relevant to the study of topological vector spaces, it seems that they provide just enough structure to define the notion of total boundedness:

Say a set $B$ is totally bounded if for any entourage $V$, there are finitely many points $x_n$ such that $B \subseteq \bigcup_n V[x_n]$.

I haven't seen this definition explicitly, but I have seen a special case of it applied in the context of TVS, though not named as such. However, I can't seem to get a similar definition for boundedness proper. Specifying that $B$ be contained in some $V[x]$ just won't work, as the $V$ could be made arbitrarily large.

Is it possible to get such a definition of boundedness in uniform spaces?

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  • $\begingroup$ Correct me if I'm wrong, but if $d$ denotes the euclidean distance on $\mathbb{R}$, don't $d$ and $d \wedge 1$ generate the same set of entourages, but different notions of boundedness? $\endgroup$ – Dominik Sep 1 '15 at 8:56
  • $\begingroup$ I don't understand what you mean by d∧1... $\endgroup$ – Sir Jective Sep 1 '15 at 8:57
  • $\begingroup$ $\wedge$ denotes the minimum, so i mean the metric $\min\{d(\cdot, \cdot), 1\}$. $\endgroup$ – Dominik Sep 1 '15 at 8:59
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    $\begingroup$ A reference is Boundedness in uniform spaces, topological groups and homogeneous spaces by C.J. Atkin, according the the OP of the questions I linked above. $\endgroup$ – Stefan Hamcke Sep 1 '15 at 9:30
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    $\begingroup$ One may define a set $A$ in a uniform space to be bounded if for any symmetric entourage $V$ there is a natural number $n$ and a finite set $F$ such that $$ A \subseteq V^n(F) $$ Assume $d$ is a metric on $X$ inducing the uniform structure $\cal U$. If $A$ is bounded in $(X,\cal U)$, then for any $ε>0$ there exists a number $n$ and a finite set $F$ such that $A⊆V_ε^n(F)$, where $V_ε$ is entourage containing all pairs $(x,y)$ such that $d(x,y)<ε$. By the triangle inequality, $A$ is then contained in $$ V_{nε}(F)⊆V_{nε+\max_{x\in F}(d(y,x))}(y) $$ so $A$ is bounded in $(X,d)$ $\endgroup$ – Stefan Hamcke Sep 1 '15 at 9:49
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The answer is no.

Let $d_1$ be the euclidean metric on the real numbers and $d_2 = \min\{d_1(\cdot, \cdot), 1\}$. It is easy to check that $d_2$ satisfies all conditions of a metric and that all subsets of $\mathbb{R}$ are bounded. Therefore $d_1$ and $d_2$ induce different notions of boundedness in $\mathbb{R}$.

however, both metrics induce the same entourages. Consider for a metric $d$ the set $U_\alpha^d = \{(x, y) \mid d(x, y) < \alpha \}$. Then the set $\Phi^d = \{B \subset \mathbb{R}^2 \mid U_\alpha^d \subset B \text{ for some } \alpha > 0\}$ is an entourage for any metric $d$. Now if a set $M$ is in $\Phi^{d_1}$, then there is an $\alpha > 0$ with $U_\alpha^{d_1} \subset M$. We can wlog assume $\alpha < 1$ [by replacing $\alpha$ with $\min\{\alpha, \frac{1}{2}\}$] and therefore $U_\alpha^{d_1} = U_\alpha^{d_2}$, so $M \in \Phi^{d_2}$. In a similar fashion, the inclusion $\Phi^{d_2} \subset \Phi^{d_1}$ can be shown.

This means that $d_1$ and $d_2$ induce the same entourages on $\mathbb{R}$, but different notions of boundedness.

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  • $\begingroup$ Thank you for your help. I'm going to try to sleep on this now, and if later the logic seems airtight, I'll accept it. Currently my brain is under the fog of insomnia. $\endgroup$ – Sir Jective Sep 1 '15 at 9:38
  • $\begingroup$ Wait a second... the $B$ are subsets of $\mathbb R$, but the $U_\alpha^d$ are collections of ordered pairs. How can the latter be contained in the former? $\endgroup$ – Sir Jective Sep 1 '15 at 17:40
  • $\begingroup$ My bad, the $B$ should be subsets of $\mathbb{R}^2$. I've fixed the mistake now. $\endgroup$ – Dominik Sep 1 '15 at 17:45
  • $\begingroup$ Also, the $\Phi^d$ is a system of entourages, right? $\endgroup$ – Sir Jective Sep 1 '15 at 17:52
  • $\begingroup$ Yes, it's the system that is induced by $d$. $\endgroup$ – Dominik Sep 1 '15 at 17:57

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