Show that the metrics $\rho(x,y) = \sqrt{d(x,y)}$ and $\tau(x,y) = \min\{d(x,y),1\}$ are equivalent to $d(x,y)$. Exercise: Show that the metrics $\rho(x,y) = \sqrt{d(x,y)}$ and $\tau(x,y) = \min\{d(x,y),1\}$ are equivalent to $d(x,y)$.
What I've tried: I know that the metrics $d_0(x,y)$ and $d_1(x,y)$ are equivalent iff $x_n\stackrel{d_0}{\to}x$ implies $x_n\stackrel{d_1}{\to}x$. 
I think I've successfully proven the equivalence between $\rho$ and $d$. First suppose that $x_n\stackrel{d}{\to}x$. We have $\forall \epsilon>0, \exists N\geq 1: d(x_n,x)<\epsilon, \forall n\geq N$. Since $d(x_n,x) <\epsilon$ implies $\rho(x_n,x)<\epsilon$, we conclude that $x_n\stackrel{\rho}{\to}x$. Now suppose that $x_n\stackrel{\rho}{\to}x$. We know that $\forall\delta >0, \exists N\geq1:\rho(x_n,x)<\delta,\forall n\geq N.$ Since $\rho(x_n,x), \delta >0$ we know that $\rho(x_n,x)^2 < \delta^2$ and hence that $d(x_n,x) <\delta^2$. Pick $\delta = \sqrt{\epsilon}$ and we find that $d(x_n,x)\stackrel{d}{\to}x$. If we combine these results we find that $\rho$ and $d$ are equivalent.
I get into trouble when trying to prove the equivalence between $d$ and $\tau$. Suppose that $x_n\stackrel{d}{\to}x$. We have that $\forall\epsilon>0,\exists N\geq 1:d(x_n,x)<\epsilon, \forall n\geq N.$ Since $\tau(x_n,x) = \min\{d(x_n,x),1\}$ we have that $d(x_n,x)<\epsilon$ implies $\tau(x_n,x) <\epsilon$. Now suppose that $x_n\stackrel{\tau}{\to}x$. We have that $\forall \delta>0, \exists N\geq 1:\tau(x_n,x)<\delta,\forall n\geq N.$ If $\tau(x_n,x) = d(x_n,x)$ we have that $d(x_n,x) <\delta.$ If $\tau(x_n,x) = 1$ we have that $1 < \delta\Leftrightarrow d(x_n,x) < \delta d(x_n,x)\Leftrightarrow d(x_n,x)(1 - \delta) < 0 \Leftrightarrow d(x_n,x) < 0$. If this where possible I would be done, but since $d(x_n,x)$ is a metric it has to be larger than or equal to zero.
Question: How do I solve this exercise?
Thanks! 
 A: You seem to have done decently prior to the $x_n\stackrel{\tau}{\to}x\Longrightarrow x_n\stackrel{d}{\to}x$ part. Here is the general idea for the last part. Suppose $\tau(x_n,x)\to 0$. Then $\tau(x_n,x) = 1$ for only finitely many $n$. In addition, $\tau(x_n,x) < 1 \Longrightarrow \tau(x_n,x) = d(x_n,x)$. Since $d(x_n,x) = \tau(x_n,x)$ for all but finitely many $n$ and $\tau(x_n,x) \to 0$, we must have $d(x_n,x) \to 0$. 
Using your language, we can say if $x_n\stackrel{\tau}{\to}x$, then $\forall \epsilon >0$, $\exists N\in \Bbb N$ such that $\tau(x_n,x) < \epsilon$ whenever $n>N$. In particular, there must be $N_0 \in \Bbb N$ such that $\tau(x_n,x) < 1$ whenever $n > N_0$, which means $d(x_n,x) < 1$ whenever $n> N_0$. So, for $0 < \epsilon < 1$, $\exists N\in \Bbb N$ such that $d(x_n,x) = \tau(x_n,x) < \epsilon$ whenever $n> N$, and for $\epsilon \geq 1$, $N_0$ serves the same purpose. Thus, $\forall \epsilon >0$, there is some $N\in \Bbb N$ such that $d(x_n,x) < \epsilon$ whenever $n > N$, and we may conclude that $x_n\stackrel{d}{\to} x$.
