Every singelton is bounded in a topological vector space

Question

I'm trying to prove that every singelton (one point set $\{x\}$) is bounded in a topological vector space.

I can't see it so easily. It is obvous that given a $V$ neighberhood of $0$, there is $t>0$ such that $x\in tV$ (since $X=\cup nV$), but why would $x\in sV$ for every $s>t$?

Thanks

P.S: I understood that the definition i'm working with isn't the standart. My definition of a set $X$ being bounded is that for every $V$ neighborhood of $0$, there is $t>0$ such that for all $s>t:\:X\subset sV$.

Answer 1

I figured it out:

Let $V$ be a neighborhood of $0$. Then there is $U\subset V$ balanced neighborhood of $0$.

Then $\exists t>0:\:\{x\}\subset tU$. then for any $s>t$ we have $t/s<1$ so $tU\subset sU\subset sV$.