closure of bounded set in uniform space

Question

I have stumbled upon the following statement but I could not see why it is true. Let $(X,V)$ be a uniform space and $A$ bounded in it. Bounded means that for each $W\in V $ there is a finite set $F\subseteq X$ and an integer m, s.t. $A\subseteq W^m[F]$. So the statement says that in such a case, $\overline A\subseteq W^{m+1}[F]$. Can someone explain why that is true?

Answer 1

If $y\in\overline A$, then there is some $x\in A\cap W^{-1}(y)\subseteq W^m[F]\cap W^{-1}(y)$, thus $(x,y)\in W$, and there is an $f\in F$ and $f=x_0,x_1,\dots,x_m=x$ such that $(x_i,x_{i+1})\in W$ for all $i$. It then follows that $(f,y)\in W^{m+1}$.