# Show two metrics are equivalent

The definition I'm given is that two metrics $\tau, \rho$ on a set X are equivalent if there exist two numbers $c_1, c_2$ such that $c_1\tau(u,v)\leq\rho(u,v)\leq c_2\tau(u,v)$.

I'm supposed to show if they are equivalent, then there exists some c such that ${1 \over c} \tau(u,v)\leq\rho(u,v)\leq c\tau(u,v)$.

I'm using the fact that $c_1\tau(u,v)\leq c_2\tau(u,v)$ to show that $\tau(u,v)\leq {c_2 \over c_1 }\tau(u,v)$ and ${c_1 \over c_2 }\tau(u,v)\leq \tau(u,v)$. Doing some substitution gives me $c_1\tau(u,v) \leq c_1 {c_2 \over c_1 }\tau(u,v) \leq\rho(u,v)\leq c_2{c_1 \over c_2 }\tau(u,v)\leq c_2\tau(u,v)$. This seems to me to suggest $c_1\tau(u,v) \leq c_2 \tau(u,v) \leq\rho(u,v)\leq c_1\tau(u,v)$ and I keep ending up with $c={1 \over c}=1$ and $c_1\tau=c_2\tau$ which seems wrong. Am I completely off track?

If you pick $c>c_2$ large enough, then $1/c < c_1$.
• How did you get $c_1 \frac{c_2}{c_1} \tau(u,v) \leq \rho(u,v)$ in the second line of the third paragraph? – Catalin Zara Mar 15 '16 at 1:30
• $\tau(u,v)\leq {c_2 \over c_1 }\tau(u,v)$ so $c_1 \tau(u,v)\leq c_1 {c_2 \over c_1 }\tau(u,v)$ – chris Mar 15 '16 at 1:35
• Sure, but if $c_1\tau(u,v) \leq c_1 \frac{c_2}{c_1}\tau(u,v)$ and $c_1\tau(u,v) \leq \rho(u,v)$, it doesn't follow that $c_1\frac{c_2}{c_1}\tau(u,v) \leq \rho(u,v)$. [$A\leq B$ and $A\leq C$ do not imply $B \leq C$.] – Catalin Zara Mar 15 '16 at 1:38