A homogeneous distance over a topological vector space E verify $d(\alpha x,\alpha y)=|\alpha|d(x,y)$.
Show that if the structure of the topological vector space is generated by a homogeneous distance, then the notions of dimensioning in the metric sense and in the direction of the topological vector spaces coincide.
I understand the question in the following way:
- Definition of bounded set in a metric space: $A\subset X$ is bounded if $diam(A)<+\infty$
- Definition of bounded set in a topological vector space: $A\subset X$ is bounded if for every neighborhood U of 0, $\exists s>0$, s.t $A \subset tU, \forall t>s$
Let $A\subset E$, prove that
$diam(A)<+\infty \iff $ for every neighbourhood U in $0$, $\exists s>0$ such that $A\subset tU, \forall t>s$.
My half solution:
$\Leftarrow)$ U is a neighbourhood in 0, then $\exists$ a ball $B(0,r)\subset U$.
Further, as $A\subset tU\Rightarrow diam(A)\le diam(tU)$
$diam(tU)=sup\{d(w,z)|w,z\in tU\}$ $=sup\{d(tx,ty)|x,y\in U\}$
$=sup\{|t|d(x,y)|x,y\in U\}$
$=|t|sup\{d(x,y)|x,y\in U\}$
$\le |t|sup\{d(x,0)+d(0,y)|x,y\in U\}$
$\le |t|[sup\{d(x,0)|x\in U\}+sup\{d(0,y)|y\in U\}]$
$< |t|(r+r)=2r|t|$.
Then, $diam(A)\le diam(tU)< 2r|t|$.
Therefore, $A$ is bounded.
Can you give me some idea or suggestion for the ($\Rightarrow$) ?
Let $k=\text{diam}\,(A)<\infty$. This means that, for any $a,b\in A$, $d(a,b)\leq k$. Fix $a_0\in A$. Then $$ |d(a,0)-d(a_0,0)|\leq d(a,a_0)\leq k. $$ So, for every $a\in A$, $$d(a,0)\leq d(a_0,0)+k.$$ That is, $A$ is inside the ball of radius $k$ around the origin $B_k(0)$. Now let $U$ be open with $0\in U$. Then there exists $r>0$ such that $B_r(0)\subset U$. Let $s=k/r$. Then $$A\subset B_k(0)=B_{sr}(0)=sB_r(0)\subset sU,$$and the inclusion will hold for any $t\geq s$.