1
$\begingroup$

This excerpt from a journal paper says if X is a uniform space and $A$ is a subset of $X$, then $A$ is said to be bounded if for each entourage $V$, "there exists a finite set $F$ and a positive integer $m$ such that $A\subseteq V^m[F]$". My question is, what does $V^m[F]$ mean?

The $V^m$ part is clear enough; you just compose $V$ with itself $m$ times. But I haven't encountered a set in brackets after an entourage. If it said $V^m[x]$ where $x\in X$, that would make complete sense. How are the two notations related? If $F=\{x_1,...,x_n\}$, then does $V^m[F]=\cup_{i=1}^n V^m[x_i]$ or something?

$\endgroup$
1
$\begingroup$

If $R$ is a relation on $X$ (so $R \subseteq X \times X$) then $R[F]$ is just the "functional image" for a subset $F$ of $X$:

$$R[F]=\{y \in X: \exists x \in F: (x,y) \in R\}$$

So yes, this does commute with unions as you suggested. This is the same definition in essence as $f[F]$ for a function $f: X \to Y$ and $F \subseteq X$; note that a function is just a special case of a relation.

$\endgroup$
  • $\begingroup$ OK thanks. Also, why is it necessary to talk about finite sets in the definition rather than just a single point? The definitions of boundedness for metric spaces and topological vector spaces are about sets centered at a single point, so why are things different in the case of uniform spaces? $\endgroup$ – Keshav Srinivasan Dec 19 '18 at 7:13
  • $\begingroup$ @KeshavSrinivasan One thing is to make clear the relation to compactness, I think. And for a finite space in the trivial uniformity we really need the finite set definition! One point won't do. $\endgroup$ – Henno Brandsma Dec 19 '18 at 7:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.