An open ball contains another open ball Consider $\mathbb R^n$ with the distance induced by the Euclidean norm $\|\;\|_2$. For any $x\in\mathbb{R}^n$ and $r>0$ define $B(x;r)=\{y\in\mathbb{R}^n: \|x-y\|_2<r\}$. If
$$B(a, r_1)\subset B(b, r_2),$$ how do I prove that $$r_1\leq r_2?$$ It seems extremely obvious but I can't efficiently construct a concise and rigorous proof.
My attempt so far involves circular reasoning. I tried to prove by contradiction.
Suppose $r_1>r_2$, then $$B(a,r_1)\cap B(b,r_2)\subset B(b,r_2).$$
By the property of proper sets, $\exists p\in B(a,r_1)$ such that $p\notin B(b,r_2)$. But by definition, $B(a,r_1)\subset B(b,r_2)$, which is a contradiction.
 A: Check that in any metric space, if $A\subseteq B$, then $\mathrm{diam}(A)\le\mathrm{diam}(B)$, where the diameter of a set $E$ is defined as
$$
\mathrm{diam}(E) = \sup\{d(x,y):x,y\in E\}.
$$
In $\mathbb R^n$ with the usual metric, a brief calculation shows the diameter of a ball $B(a,r)$ is $2r$ (consider any unit vector $u$ and the points $a + tu$ with $-r<t<r$). Hence if $B(a,r_1)\subseteq B(b,r_2)$, we have $2r_1\le 2r_2$.
A: Depends on the space. For example, on $X=\mathbb{N}$ with the metric inherited as a subspace of the real line,
$B(1;3/2)=\{1,2\}=B(1;2)$.
Consider now the subspace $Y=\{1,2, 5\}$ of $\mathbb{N}$.
$B_Y(1;4)=\{1,2\}\subset \{1,2,5\}\subset B_Y(2;7/2)$ and yet $7/2<4$
In linear spaces that have a translation invariant metric where open balls are convex (normed spaces for example) the statement holds.
Edit: Th OP now specified that the space in question is $\mathbb{R}^n$ with the Euclidean norm. Here is another proof that the statement holds.
Suppose $a\neq b$. Notice that open balls in the Euclidean space (actually any normed space) are convex. Take the line $\{b+t(b-a): t\in\mathbb{R}\}$
let $t^*=\inf\{t>0: b+t(b-a)\in B(b;r_2)\}$.  THen $1<t^*\leq \frac{r_2}{\|b-a\|}$
$a+s(b-a)\in B(a;r_1)\subset B(b;r_2)$ for all $0<s<\frac{r_1}{\|b-a\|}$; hence
$$\frac{r_1}{\|b-a\|}\leq t^*\leq \frac{r_2}{\|b-a\|}$$
