Boundedness in uniform spaces? 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?

 A: 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.
