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I came across this during some reading: $|d(x,x_0)-d(y,x_0)|\le d(x,y)$. I can't seem to figure out why it holds. Here $d$ is a metric.

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3 Answers 3

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Considering the triangle inequality we know that:

$$ d(x,x_0) \leq d(x,y) + d(y,x_0) $$

Can you use this fact to achieve that $ |d(x,x_0) - d(y,x_0)| \leq |d(x,y)| $ ? Once you know this, recall that $d$ is a non-negative function...

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Combine the following two applications of the triangle inequality: \begin{align} d(x,x_0) &\le d(x,y) + d(y, x_0) \\ d(y, x_0) &\le d(x,y) + d(x, x_0) \end{align}

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HINT: Use the triangle inequality to show that $d(x,y)\ge d(x,x_0)-d(y,x_0)$ and $d(x,y)\ge d(y,x_0)-d(x,x_0)$.

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