I tried $|d(x, z) - d(y, z)| \leq d(x, z) + d(y, z)$, but the the triangle equation goes the wrong side. Is there an analogue to the reverse triangle inequality with metrics? The reason I want to know this is to proof that $d(x, A)$ is continious where $A$ is a set. (Let $|x - y| < \delta$, then after some algebra $|d(x, A) - d(y, A)| = \inf(\{|d(x, a) - d(y, a)| \ | \ a \in A \})$.
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2Yes, this is really just the triangle inequality. – Angina Seng Nov 25 '17 at 12:41
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@LordSharktheUnknown the triangle inequality is $d(x, y) \leq d(x, z) + d(y, z)$ right? How does this imply the statement from the title? – user388557 Nov 25 '17 at 12:43
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@LordSharktheUnknown nvm, I saw B.Mehta's answer, I'm facepalming so hard now. – user388557 Nov 25 '17 at 12:44
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In particular for $z=0$ we get the useful formula $\big||x|-|y|\big|\le|x\pm y|\le|x|+|y|$. – zwim Nov 25 '17 at 13:33
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$d(x,y) + d(y, z) \geq d(x, z)$ so $d(x,y) \geq d(x,z) - d(y,z)$. Similarly, $d(y,x) + d(x,z) \geq d(y,z)$ so $d(x,y) \geq d(y,z) - d(x,z)$. Hence, $d(x,y) \geq |d(y,z) - d(x,z)|$.
B. Mehta
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