# Show this induced metric is a metric [duplicate]

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

• 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 $d(y,z)$. – Pink Panther Mar 12 at 0:00
• Your first case for the inequality is correct, for case 2 assuming the inequality does not hold true leads to a fairly simple contradiction. – babemcnuggets Mar 12 at 0:21

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 $$\tilde {d} (x,y) \leq d(x,y) \leq d(x,z)+d(y,z) = \tilde {d} (x,z)+\tilde {d} (y,z)$$.
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}
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.