If $X$ s a vector space over the field $\Bbb F$ then the following notation well be used $$ x+A=\{x+a:a\in A\}\\ x-A:=\{x-a:a\in A\}\\ A+B:=\{a+b:a\in A\wedge b\in B\}\\ \lambda A:=\{\lambda a:a\in A\} $$ for any $A,B\subseteq X$, for any $x\in\ X$ and for any $\lambda\in\Bbb F$.
Definition
If $\tau$ is a topology on a vector space $X$ such that
- every point of $X$ is a closed set
- the vector space operation are continuous with respect to $\tau$
we say that $X$ is a topological vector space equipped with the vector topology $\tau$.
So we note that the condition $1$ imply that any topological vector space will be $T_1$ but many authors omitted the above condition from the definition of a topological vector space.
Definition
A subset $E$ of a topological vector space is said to be bounded if to every neighborhood $V$ of $0$ in $X$ corresponds a number $s>0$ such that $E\subseteq tV$ for every $t>s$.
Now a subset $S$ of a metric space $Y$ is bounded if there exist an open ball that contains it and this is equivalent to suppose that the supremum of the distances of the above set is finite. Now if $X$ is a normed space it is a vector topological space and a metric space so I ask to me if the definition two definition of bounded set agree in this particular case. So I observe that if $E$ is a bounded subset of normed space then for what above defined for any open ball $B(0,r)$ centered at $0$ there exist a $\sigma>0$ such that $$ E\subseteq\rho B(0,r) $$ for any $\rho\ge\sigma$ but $\rho B(0,r)$ is a bounded subset with the metric induced by the norm and so $$ E\subseteq\rho B(0,r)\subseteq B(0,r') $$ for some $r'>0$ so that $E$ is a bounded set with respect the metric induced by the norm. Now unfortunately I do not be able to prove the contrary, that is if $E$ is a bounded set with respect the metric induced by the norm then it is a bounded set in the normed space too so that I ask to prove this. So could someone help me, please?
If $E$ is a bounded subset (in the metric sense) then by the triangle inequality we may assume that $E$ is contained in some $B(0,r)$ for $r$ large enough. Now if $V$ is any neighborhood of $0$ it will contain some open ball $B(0,r')$ with $r'>0$. Now for any $\rho>\frac{r}{r'}$ we will have $E\subseteq B(0,r)\subseteq \rho B(0,r')$. So $E$ is bounded in the sense of topological vector spaces.
Edit: If $\rho>\frac{r}{r'}$ and $x\in B(0,r)$, then $x=\rho x'$ with $x'=\rho^{-1}x$ and $\left\lVert x'\right\rVert=\rho^{-1}\left\lVert x\right\rVert<\rho^{-1}r<\frac{r'}{r}r=r'$ so $x'\in B(0,r')$ and therefore $x\in \rho B(0,r')$.