# Clarifying the definition of Boundedness in Uniform Spaces

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?

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.

• 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? – Keshav Srinivasan Dec 19 '18 at 7:13
• @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. – Henno Brandsma Dec 19 '18 at 7:16