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Let $d$ be a metric on $X$ and $(X,d)$ a non-compact metric space i.e. $X = \mathbb{R}$ or $\mathbb{Q}$ For $x,y \in X$ define $$ \tilde{d}(x,y) := \begin{cases} d(x,y), & \text{if }d(x,y) < 1, \\ 1, & \text{else.} \end{cases} $$ Show that $\tilde{d}$ is metric on $X$.

My attempts

Because $d$ is a metric, $\tilde{d}(x,y) \ge 0$ for all $x,y \in X$. If $x = y$, we have $d(x,y) = 0 < 1$ and therefore $\tilde{d}(x,y) = 0$ If $\tilde{d}(x,y) = 0 \neq 1$, we have $\tilde{d}(x,y) = d(x,y) = 0$ and therefore somit $x = y$.

This metric is trivially symmetric, since $d$ is.

Let $x,y,z \in X$.

Case 1: $d(x,z) < 1$. Then we have $$ \tilde{d}(x,z) = d(x,z) \le d(x,y) + d(y,z). $$

Case 2: $d(x,z) \ge 1$. Then, we have $$ 1 = \tilde{d}(x,z) \le d(x,z) \le d(x,y) + d(y,z). $$

But I know know how to continue from here, I know that we have $\tilde{d}(x,y) \le d(x,y)$ für alle $x,y \in X$, but I don't know how to use it.

Is my approach for the positive definiteness and symmetry correct? How can I prove the triangle inequality?

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marked as duplicate by Viktor Glombik, YuiTo Cheng, José Carlos Santos general-topology Sep 8 at 10:35

This question was marked as an exact duplicate of an existing question.

  • $\begingroup$ yes symmetry and positive definite is correct. For the triangle inequality you can again distinguish between the cases $d(x,y)\leq 1$ and $d(x,y)>1$ and the same for $d(y,z)$. $\endgroup$ – Pink Panther Mar 12 at 0:00
  • $\begingroup$ Your first case for the inequality is correct, for case 2 assuming the inequality does not hold true leads to a fairly simple contradiction. $\endgroup$ – babemcnuggets Mar 12 at 0:21
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Proof of triangle inequality: let $x,y,z \in X$. If $d(x,z) \geq 1$ then $\tilde {d} (x,z) =1$ so $\tilde {d} (x,y)\leq 1 \leq \tilde {d} (x,z)+\tilde {d} (y,z)$. Similarly, if $d(y,z) \geq 1$ then $\tilde{d}(x,y) \leq \tilde {d} (x,z)+\tilde {d} (y,z)$. Finally if $d(x,z) < 1$ and $d(y,z) < 1$ the $\tilde {d} (x,y) \leq d(x,y) \leq d(x,z)+d(y,z) = \tilde {d} (x,z)+\tilde {d} (y,z)$.

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Alternatively, one could rewrite the metric as $$ \tilde{d}(x,y) := \min(1, d(x,y)). $$ Then, the triangle inequality becomes easy to verify: \begin{align} \tilde{d}(x,z) & = \min(1, d(x,z)) \le \min(1, d(x,y) + d(y,z)) \\ & \le \min(1, d(x,y)) + \min(1,d(y,z)) = \tilde{d}(x,y) + \tilde{d}(y,z). \end{align}

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For all $x \in X$ and every $r > 0$, we have $B(x,r) \subset \bar B(x,r)$, where $\bar B(x,r)$ is the ball with the $\bar d$ metric.

For some positive $s < \min(r,1)$ we have $\bar B(x,s) \subset B(x,r)$, which suffices to show equivalence.

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