2
$\begingroup$

I was trying to prove the following obvious fact , but despite having the intuition on why this is true I could not prove it If , $ B(x,r)\cap B(x',r') = \emptyset $ then , $ d(x,x') \geqslant r + r'$

I tried forming the following equations $ \forall X \in B(x,r) $

1.)$ d(x,X) < r $

2.) $ d(x',X) >r' $

Similarly , $\forall Y \in B(x',r') $

  1. $d(x',Y) < r'$

  2. $ d(x,Y)> r $

Then I tried to add the equations to see if I can use the properties of the metric function to get the desired inequality but I could not,any help , hint would be appreciated .

  • 4
    Intuition can be a misleading thing in metric spaces. It is not true. Let $X=\{0,1\}$ and $x=0,x'=1, r=1$. The $B(x,r), B(x',r)$ are disjoint, but $d(x,x') = 1 < r+r$. – 2017-02-11
  • 0
    @copper.hat Maybe I should specify that the d metric being used is the euclidean norm , and that the domian in the statement is being defined is the $R^n$ domain . I am not quite sure how norms , and metrics works on set like {0,1} – 2017-02-11
  • 0
    It is true in normed spaces, there is a point on a straight line joining $x,x'$ that is in neither ball, and you can use this to show the desired result (essentially reducing the space to the real line). – 2017-02-11
  • 0
    @copper.hat But then , I would have to show that their exists such a point , how do I go about doing that ? – 2017-02-11
  • 0
    Consider the (connected) segment $[x,x']$. Since the balls are disjoint and open, there must be a point on the segment that does not lie in either ball (otherwise this would contradict connectedness). – 2017-02-11
  • 0
    @copper.hat Could you please define what you mean by connectedness , I am sure that I would have read about it but by some other name . – 2017-02-11
  • 0
    https://en.wikipedia.org/wiki/Connected_space#Formal_definition – 2017-02-11
  • 1
    @copper.hat Ahh , I see I didn't know about connectedness after all. I am a Masters in Econ student , and I have only taken one course in pure mathematics ( Intro. to Mathematical analysis ) which primarily dealt with real analysis . Thus I did not know this much topology . This was one of the subpoints made by professor in the ( Intro. to Mathematical Analysis 2 ) which deals primarily with introducing lebesgue integration . – 2017-02-11

2 Answers 2

3

Show the contrapositive instead. Define $s = r + r'$ and $\lambda = r/s$. Then $r = \lambda s,r' = (1-\lambda) s$. If $d(x,x') < r + r'$, define $y := (1-\lambda) x + \lambda x'$. Then $$ d(y,x) = \|\lambda x' - \lambda x\| = \lambda \|x' - x\| < \lambda s = r $$ so $y \in \mathrm B(x,r)$. A symmetric argument shows $y \in \mathrm B(x',r')$, so the balls are not disjoint.

$y$ is a convex combination of $x$ and $x'$.

1

Here is how you can establish the result in a normed space without using connectedness directly (this follows the usual proof that an interval is connected).

Let $p(x) = x+t(x'-x)$, note that $p(0) \in B(x,r), p(1) \in B(x',r')$ and $B(x,r), B(x',r')$ are disjoint.

Let $t^* = \sup \{ t \in [0,1] | p(t) \in B(x,r) \}$. Note that $t^* >0$ since $B(x,r)$ is open and $t^* < 1$ since $B(x',r')$ is open. Also note that $\|p(t_1)-p(t_2)\| = \|(t_1-t_2)(x'-x)\| = |t_1-t_2| \|x'-x\|$.

Note that $t^* \notin B(x,r)$ since $B(x,r)$ is open (otherwise this would contradict the definition of $t^*$) and $t^* \notin B(x',r')$ since $B(x',r')$ is open and the two balls are disjoint.

Hence $p(t^*) \notin B(x,r) \cup B(x',r')$.

Note that $p(1)-p(0) = p(1)-p(t^*) + p(t^*)-p(0)$ and $\|p(1)-p(0) \| = \|p(1)-p(t^*)| + \|p(t^*)-p(0)\|$ (that is, we have equality) and hence $\|x'-x\| = \|x'-p(t^*)\| + \|p(t^*)-x\| \ge r' + r$.