# Proof idea: Let $(X,d)$ be a metric space, and $\rho$ be bounded metric, show that they will generate the same topology

Let $(X,d)$ be a metric space, $d$ generates the metric topology $\mathcal{T}$ via metric ball $B_\epsilon(x)$. Show that bounded metrics:

1. $\rho_1(x,y) = \dfrac{d(x,y)}{1+d(x,y)}$ with metric ball $B_\epsilon^1(x)$
2. $\rho_2(x,y) = \min\{1, d(x,y)\}$ with poke ball $B_\epsilon^2(x)$

generates the same metric topology $\mathcal{T}$

Is there anything else to this question other than noticing that we can pick $\epsilon > \epsilon'> \epsilon'', \epsilon' \in (0,1)$ such that $B_{\epsilon''}(x) \subseteq B_{\epsilon'}^i(x) \subseteq B_\epsilon(x)$ will hold for $i \in \{1,2\}$?

• Presumably poke ball is metric ball. – Batman Jul 15 '16 at 22:14

You just have to prove every open ball around any point $x$ in the first topology contains an open ball in the second topology around $x$ and vice-versa. (Do you see why)?