Equivalence for bounded sets Let $(X,d)$ be a metric space and $A \subset X$. Show that $A$ is bounded if and only if there exists a constant $\lambda > 0$ such that $A \subset \lambda \mathcal{B}_{1}(0)$ where $\mathcal{B}_{1}(0)$ is the unit ball centered at $0$.
Now for my proof, I've two definitions: $A$ is bounded if the diameter $d(A) \leq M < \infty$ or $A$ is bounded if there exist $r>0$ and $a \in A$ such that $A \subset \mathcal{B}_{r}(a)$. With the second definition, I can see why the statement holds; if $A \subset \mathcal{B}_{r}(a)$ I can "move" the center of the ball to $0$ and dilate the radius as necessary. What I should prove is that if $x \in A$ then $x = \lambda y$ with $d(y,0) < 1$ for some $\lambda >0$, but how can I relate these concepts to give a formal proof? Thanks
Edit: the diameter of a set is $d(A) = \sup_{x,y \in A} d(x,y)$
 A: For this theorem to be true, some relation between scalar multiplication and the metric has to be assumed, otherwise it may be false. For example, with the discrete metric, every subset is metrically bounded but only $\{0\}$ is topologically bounded. Usually the following axiom is assumed: $$d(\lambda x,\lambda y)=|\lambda|d(x,y)$$ In what follows, the slighly weaker axiom $d(\lambda x,\lambda y)\le c\lambda\, d(x,y)$ for $\lambda>0$ is taken.
Let $A$ be metrically bounded, $\mathrm{diam}(A)\le M$. Fix a point $x_0\in A$. Then for any $x\in A$, $$d(x,0)\le d(x,x_0)+d(x_0,0)\le M+d(x_0,0)=:r$$ Let $\lambda>cr$ and $x':=x/\lambda$; then $$d(x',0)=d(x/\lambda,0)\le \frac{c}{\lambda}d(x,0)<1$$ so $A\subseteq\lambda B_1(0)$.
Let $A$ be topologically bounded, $A\subseteq \lambda B_1(0)$. Then for any $x,y\in A$, $x=\lambda x'$, $y=\lambda y'$ with $x',y'\in B_1(0)$. $$d(x,y)=d(\lambda x',\lambda y')\le c\lambda d(x',y')\le c\lambda(d(x',0)+d(0,y'))<2c\lambda$$ so $\mathrm{diam}(A)\le2c\lambda$.
