1. ## Equivalent Metrics

Define $\rho$ on $X \times X$ by

$\rho(x,y) = min(1,d(x,y)), \ \ \ \ \ \ \ \ \ \ x,y \in X$

Show that $\rho$ is a metric that is equivalent to d

My solution:
Two metrics are equivalent if and only if the convergent sequences in $(X,d)$ are the same as the convergent sequences in $(X,\rho)$

Let $(x_{n})^\infty_{n=1}$ be convergent in $\rho$ i.e.,

$\forall \ \epsilon \ \exists \ N_{\epsilon}\ \mbox{such that}\ \forall\ n \geq N_{\epsilon}\ \rho(x_n,x)={min(1,d(x_{n},x))}}<\epsilon$

Choose $0 \ < \epsilon \ < 1$. Then $\exists \ N_{\epsilon}$ such that $\forall\ n \ \geq N_{\epsilon}: min (1,d(x_n,x)) < \epsilon < 1$.

This implies $min (1,d(x_n,x))=d(x_n,x)\ <\ \epsilon$.

What if $\epsilon > 1$?

2. Irrelevant. The definition of "convergent" says for any $\epsilon> 0$ but is is small $\epsilon$ that is important.

3. And if you still wish to show it for $\epsilon>1$, you can simply note that $min(x,y) \le x$ & $min(x,y) \le y \ \forall x,y \in \mathbb{R}$

4. Defunkt, how would that be relevant ? oO

I think it's possible to show it for e>1 if you consider the contrapositive. But as Hallsofivy says, it's not useful

5. $min(1, d(x_n,x)) \le 1 < \epsilon$

6. But what we want to show is that $d(x_n,x)<\epsilon$. We already know that $\min(1,d(x_n,x))<\epsilon$

7. Oops