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I need to solve this question:

Pick $b\notin B[a,r]$ show that there exists $s>0$ such that $B[a,r]\cap B[b,s]$ is empty

My idea is to suppose a point $p$ in the intersection. Then

$$d(a,b)\le d(a,p)+d(p,b)$$

Note that $d(a,p) = r$ and $d(p,b) = s$. So we get: $d(a,b)\le r+s$.

Now, I showed:

$p$ in the intersection of balls $B[a,r]$ and $B[b,s]$ $\implies$ $d(a,b)\leq r+s$, so $d(a,b)> r+s \implies$ $p$ not in the intersection, therefore, $B[a,r]\cap B[b,s]$ is empty. Given $a,b,r$ I can always pick $s$ such that $d(a,b)> r+s$, because $d(a,b)- r$ is positive. Why? Because in order for $b$ to be out of the closed ball, we must have $d(a,b)> r$ which is the ball`s radius.

Is my reasoning correct? There is another exercise after this one, that asks me to prove:

Let $(M,d)$ be a metric space, $a,b\in M$, $r,s\in > \mathbb{R}_{+}^{*}$. Show that if $r+s<d(a,b)$ then the balls $B[a,r]$ and $B[b,s]$ are disjoint

Doesn`t my proof cover both questions?

Andrés E. Caicedo
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1 Answers1

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Pick $b\notin B[a,r]$ then $d(b,a)\gt r$ , let $s= dist (b, B[a,r])=inf \{d(b,x): x \in B[a,r]\}$

Now, consider the ball $B[b,\frac s2]$

Claim:$B[a,r] \cap B[b,\frac s2]=\phi$

If not let $x\in B[a,r] \cap B[b,\frac s2]$

$\Rightarrow d(x,b)\le \frac s2$ where $x\in B[a,r]$

$\Rightarrow dist (b, B[a,r])\le \frac s2$

$\Rightarrow s\le \frac s2$

which is a contradiction.(as $s \gt 0$)

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