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Is the following just a matter of showing the 3 properties that make up a metric??

Define d on $\Bbb R\times\Bbb R$ by $d(x,y)=\min \{1,|x-y|\}$. Show that $d$ is a metric on $\Bbb R$

  1. $d(x,y)=0$ if $x=y$
  2. $d(x,y)=d(y,x)$ for every $x,y \in X$
  3. $d(x,y)\le d(x,z)+d(z,y)$ for every $x,y,z \in X$
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  • $\begingroup$ Yes. Note that for every metric $d'$ the construction $d(x,y)=\min\{1,d'(x,y)\}$ yields a metric. $\endgroup$ – Stefan Hamcke Oct 20 '13 at 22:32
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    $\begingroup$ Which of the three conditions are you having trouble showing? $\endgroup$ – Brian M. Scott Oct 20 '13 at 22:36
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    $\begingroup$ well the triangle inequality always seems to be the more difficult one to show. $\endgroup$ – cele Oct 20 '13 at 22:37
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    $\begingroup$ You're missing the property that $d(x,y)>0$ if $x\neq y$. $\endgroup$ – Stefan Smith Oct 20 '13 at 22:41
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That's correct (though your first condition should read "if and only if $x=y$"). This is what is known sometimes as the standard bounded metric induced by the metric $\rho(x,y)=|x-y|$. More generally, if $\langle X,\rho\rangle$ is any metric space, then the function $$d(x,y)=\min\{1,\rho(x,y)\}$$ is again a metric on $X$, called the standard bounded metric induced by $\rho$, and induces the same metric topology.

To prove that it is in fact a metric, the first two properties are relatively straightforward. For the triangle inequality, note that if $\rho(x,y)\ge1$ or $\rho(y,z)\ge 1,$ then $$d(x,y)+d(y,z)\ge1\ge d(x,z).$$ Otherwise, we have $$d(x,y)+d(y,z)=\rho(x,y)+\rho(y,z)\ge\rho(x,z)\ge d(x,z)$$ by definition of $d,$ since $\rho$ is a metric.

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  • $\begingroup$ It even induces the same uniform structure :-) $\endgroup$ – Stefan Hamcke Oct 20 '13 at 22:37
  • $\begingroup$ @StefanH : what's "uniform structure"? $\endgroup$ – Stefan Smith Oct 20 '13 at 22:42
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    $\begingroup$ @StefanSmith: It is a structure between metric and topology where you have a kind of neighborhoods which are uniform over the whole space, they are called entourages. So you can take an entourage $U$ and put it around a point $x$ to get the neighborhood $U(x)$, the same way as you can choose an $ϵ$ and then put the ball $B_ϵ$ around $x$ to get the neighborhood $B_ϵ(x)$ in a metric space. So a metric induces a uniform structure, and a uniform structure induces a topology. $\endgroup$ – Stefan Hamcke Oct 20 '13 at 22:47
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    $\begingroup$ @StefanSmith: My first comment may be understood as: The identity map is uniformly continuous. So it is not only continuous (preserves the topology) but uniformly continuous (preserves the uniform structure). It is, however, not an isometry, as it does not maintain the distances. $\endgroup$ – Stefan Hamcke Oct 20 '13 at 22:50

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