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Prove that $|d(a,b) - d(a_{1},b_{1})| \leq d(a,a_{1}) + d(b,b_{1})$

Granted their are two cases to this. I will save one to do independently, but I wanted to see if my proof for the other case is correct.

Case: $d(a,b) \leq d(a_{1},b_{1})$.

Then $d(a_{1}, b_{1}) - d(a,b) \leq d(a,a_{1}) + d(b,b_{1})$

$d(a_{1},b_{1}) \leq d(a,b) + d(a,a_{1}) + d(b,b_{1}) \geq d(a,b_{1}) + d(b,b_{1}) \geq d(a,b)$

Note that these are metric spaces.

user121895
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1 Answers1

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On the one hand:
$d(a,b)\leq d(a,a')+d(a',b')+d(b',b) \Rightarrow d(a,b)-d(a',b')\leq d(a,a')+d(b',b)$
On the other hand:
$d(a',b')\leq d(a',a)+d(a,b)+d(b,b') \Rightarrow d(a',b')-d(a,b)\leq d(a',a)+d(b,b')$
Both together:
$\vert d(a,b)-d(a',b')\vert\leq d(a,a')+d(b,b')$

C-star-W-star
  • 16,275
  • I am a little unsure I follow your logic. For $d(a,b) \geq d(a_{1},b_{1})$ I have that $d(a,b) - d(a_{1},b_{1}) \leq d(a,a_{1}) + d(b,b_{1})$. Then $d(a,b) \leq d(a_{1},b_{1}) + d(a,a_{1}) + d(b,b_{1}) \geq d(a,b) + d(b,b_{1}) \geq d(a,b)$. – user121895 Jan 23 '14 at 02:55
  • I think your prob is: $\lvert x\rvert\leq c$ First note that for $c<0$ the statement $0\leq\lvert x\rvert\leq c<0$ is certainly wrong. For $c\geq 0$, there's no need to consider the cases on $x$ separately: Checking that $x\leq c$ as well as $-x\leq c$ gives $\lvert x\rvert\leq c$. This is the same as checking both cases separately: On the one hand if $x\geq 0$ then check that $+x=\lvert x\rvert\leq c$ while $-x=-\lvert x\rvert\leq 0\leq c$ becomes trivially true; on the other hand if $x<0$ then check that $-x=\lvert x\rvert\leq c$ while $+x=-\lvert x\rvert\leq 0\leq c$ becomes trivially true. – C-star-W-star Jan 23 '14 at 22:19