An open ball contained within another open ball Suppose $$B(a, r_1)\subset B(b, r_2).$$ 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 $r>s$, 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.

In case $a\ne b$, then the statement is false. I will give a couterexample but will keep the case $a=b=v$ below in case someone in the future needed it.
Let $X=\{0,3,6\}$ be endowed with the regular distance on $\mathbb{R}$
then,
$$B(0,5)=\{0,3\} \text{ and } B(3,4)=\{0,3,6\}$$
$$\Rightarrow B(0,5)\subset B(3,4)\text{ but } 5\not\le 4$$
I am assuming by $\subset$ means a strict subset or else the property is not true.
If
$$B(v,r_1)\subset B(v,r_2)$$
Then there is an $x\in B(v,r_2)$ such that $x\notin B(v,r_1)$
$$\Rightarrow r_1\le d(v,x)\text{ and } d(v,x)<r_2$$
$$\Rightarrow r_1<r_2$$
