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Suppose $(X,P)$ is a metric space and $x_0$ is fixed point of $X$.

Define $f:X \longrightarrow \mathbb{R}$ ($x\in X$) by $x\mapsto P(x,x_0)$, which means that $f(x)=P(x,x_0)=|x-x_0|$.

Show that f is continuous on $X$.

I know that for $f$ to be continuous on $X$, $f$ is continuous at each point of $X$.

This is what I did Show $f$ is continuous on $X$

$\forall \epsilon>0 \exists \delta>0$ such that $P(x,x_0)<\delta$ then $P(f(x),f(x_0)<\epsilon=|x-x_0|+|x_0-x_0|$

But this show it is continuous on $x_0$ and I am a bit stuck.

Another User
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Fernando Martinez
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    $x,x_0\in X$, so how do you define $|x-x_0|$? In other words, what is subtraction? How do you have the triangle inequality? I think you should leave it as $P(x,x_0)$ and work with this instead. – Clayton Oct 29 '15 at 15:42
  • I thought because the metric for R is $|x-x_0|$ maybe I am wrong I am begginer at metric spaces. – Fernando Martinez Oct 29 '15 at 15:47
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    The standard metric in $\Bbb R$ is given by $|x-x_0|$ (where $x,x_0\in\Bbb R$), but how do you define absolute values for elements in $X$? – Clayton Oct 29 '15 at 15:51

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Show that $|f(x) - f(y)| = |P(x_0,x) - P(x_0,y)| \le P(x,y)$ using the triangle inequality.

Henno Brandsma
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