Two definitions of a bounded set in topological vector spaces

Question

Let $X$ be a topological vector space. A subset $E$ is bounded if to every open set $V$ containing $0$ in $X$ there corresponds a number $s>0$ such that $E\subseteq tV$ for every $t > s$. Would the content of this definition be altered if it were required merely that to every open set $V$ containing $0$ there corresponds some $t>0$ such that $E\subseteq tV$?

Answer 1

The two definitions are equivalent.

The key point is that every neighbourhood of $0$ contains a neighbourhood $V$ of $0$ which is also balanced, i.e. $\alpha V\subset V$ for every scalar $\alpha$ with $\vert\alpha\vert\leq 1$ (see e.g. chapter 1 of Rudin's Functional Analysis). It follows that in any of the two definitions, one can restrict oneself to balanced neighbourhoods of $0$. Now, if $V$ is balanced and $t>0$, then $tV\subset t'V$ for every $t'\geq t$ (because $tV=\alpha\, t'V$ with $\alpha=t/t'\leq 1$ and $t'V$ is also balanced). This shows that the second definition implies the first, and so is equivalent to it.