
Equivalent Metrics
Define $\displaystyle \rho$ on $\displaystyle X \times X$ by
$\displaystyle \rho(x,y) = min(1,d(x,y)), \ \ \ \ \ \ \ \ \ \ x,y \in X$
Show that $\displaystyle \rho$ is a metric that is equivalent to d
My solution:
Two metrics are equivalent if and only if the convergent sequences in $\displaystyle (X,d)$ are the same as the convergent sequences in $\displaystyle (X,\rho)$
Let $\displaystyle (x_{n})^\infty_{n=1}$ be convergent in $\displaystyle \rho$ i.e.,
$\displaystyle \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 $\displaystyle 0 \ < \epsilon \ < 1$. Then $\displaystyle \exists \ N_{\epsilon}$ such that $\displaystyle \forall\ n \ \geq N_{\epsilon}: min (1,d(x_n,x)) < \epsilon < 1$.
This implies $\displaystyle min (1,d(x_n,x))=d(x_n,x)\ <\ \epsilon$.
What if $\displaystyle \epsilon > 1$?

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

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

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 :p

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

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

(Headbang) (Headbang) (Headbang) (Headbang) (Headbang)
Oops (Speechless)