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Problem: Given a metric space $(X, d)$, prove that $\mid d(x,z) - d(y,u)\mid \leq d(x,y) + d(z,u) , (x,y,z,u \in X)$.

The only thing I could think to use was the triangle inequality for each of $d(x,z)$ and $d(y,u)$, which gave me $d(x,z) + d(y,u) \leq d(x,y) + d(z,u) + 2d(y,z)$. I tried a few different things from that point, nothing worked however. I'm probably missing something fairly straightforward, can anyone help?

Franz
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  • $d(x,z)\leq d(x,y)+d(y,z)\leq d(x,y)+d(y,u)+d(u,z)$. This proves one part. Similarly, you can prove the other part also. – Krish Aug 28 '17 at 07:41

2 Answers2

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Note that $d(x,z) \leq d(x,y) + d(y,u) + d(u,z)$ by the triangle inequality. Hence, $d(x,z) - d(y,u) \leq d(x,y) + d(u,z)$.

Similarly, $d(y,u) \leq d(y,x) + d(x,z) + d(z,u)$ by the triangle inequality. Hence, $d(y,u) - d(x,z) \leq d(x,y) + d(u,z)$.

From the above two statements, it follows that $|d(y,u) - d(x,z)| \leq d(x,y) + d(u,z)$.

Sarvesh Ravichandran Iyer
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Suppose $d(x,z) \geq d(y,u)$:

$$d(x,z) - d(y,u) \leq d(x,y) + d(z,u) = d(x,y) + d(u,z) \Rightarrow d(x,z) \leq d(x,y) + d(y,u) + d(u,z)$$ we know that $d(x,u) \leq d(x,y) + d(y,u)$ (triangle inequality):

$$d(x,u) + d(u,z) \leq d(x,y) + d(y,u) + d(u,z)$$ And we know $d(x,z) \leq d(x,u) + d(u,z)$ (triangle inequality). So, $$d(x,z) \leq d(x,y) + d(y,u) + d(u,z)$$

You can do this for the other case ($d(x,z) < d(y,u)$).

OmG
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