Proving inequality with metrics

Question

I was trying show that $|D(x,B) - D(y,B)| \le d(x,y)$ with $D(x,B) = \inf_{b \in B} d(x,b)$ and $(X,d)$ is a metric space.

My try: $d(x,y) \ge d(x,b) -d(y,b) \ge \inf_{b\in B}d(x,B) - d(y,b)$ forall $b \in B$.

Then:

$\inf_{b\in B}d(x,B) - d(y,b) = K - d(y,b) \ge \sup_{b\in B} (K - d(y,b)) \stackrel{(*)}{=} K - \inf_{b \in B} d(y,b) = D(x,B) - D(y,B)$.

Is the equality $(*)$ holds? Thank you for help!!

yes, the equality holds, but the inequality before it does not — cats, Feb 18 '13 at 09:16

score 2 · Answer 1 · answered Feb 18 '13 at 09:18

Here is the classic argument. For any $b\in B,\, d(x,y) + d(y,b) \ge d(x,b) \ge \displaystyle\inf_{b\in B} d(x,b) = d(x,B).$ Since this holds for all $b\in B,$ $d(x,y) + \displaystyle\inf_{b\in B} d(y,b) \ge d(x,B)$ (this is an easy exercise in the definition of infimum). Hence, $d(x,y) \ge d(x,B) - d(y,B).$ Similarly, $d(x,y) \ge d(y,B) - d(x,B),$ from which the result follows.

score 1 · Accepted Answer · answered Feb 18 '13 at 09:20

I would write your first step as

$$d(x,y)\ge D(x,B)-d(y,b)\text{ for each }b\in B\;,\tag{1}$$

from which it follows immediately that

$$d(x,y)\ge D(x,B)-\inf_{b\in B}d(y,b)=D(x,B)-D(y,B)\;.$$

If that step isn’t quite clear, rewrite $(1)$ as

$$d(y,b)\ge D(x,B)-d(x,y)\text{ for each }b\in B\;;$$

then $D(x,B)-d(x,y)$ is clearly a lower bound for $\{d(y,b):b\in B\}$, and hence

$$\inf_{b\in B}d(y,b)\ge D(x,B)-d(x,y)\;.$$

Then repeat the argument with $x$ and $y$ interchanged to get

$$d(x,y)\ge D(y,B)-D(x,B)\;,$$

and you’re done.

Proving inequality with metrics

2 Answers2