TITLE: Proof that every metric space is homeomorphic to a bounded metric space
QUESTION [14 upvotes]: I have tried to show that every metric space $(X,d)$ is homeomorphic to a bounded metric space.  My book gives the hint to use a metric $d'(x,y)=\mbox{min}\{1,d(x,y)\}$.
If we can show that $d(x,y) \le c_1 \cdot d'(x,y)$ with $c_1$ some positive constant and $d'(x,y) \le c_2 \cdot d(x,y)$ for $c_2$ some positive constant, then the identity map $i:(X,d) \to (X,d')$ is continuous, and also obviously a bijection, thus showing that $(X,d)$ is homeomorphic to $(X,d')$, where $(X,d')$ is bounded, thus giving the desired result.
Suppose $d(x,y)<1$.  Then $d'(x,y)=d(x,y)$.  If $d(x,y)\ge 1$, then $d'(x,y) \le d(x,y)$.  Thus we can set $c_2 = 1$ and $d'(x,y) \le d(x,y)$ for all $x,y \in X$.
But when $d(x,y)>1$ why won't it always be the case that $c_1$ will depend on what $d(x,y)$ is?

REPLY [17 votes]: You’re working too hard: just show that $d$ and $d'$ generate the same open sets. Remember, a set $U$ is $d$-open if and only if for each $x\in U$ there is an $\epsilon_x>0$ such that $B_d(x,\epsilon_x)\subseteq U$. Once you have that $\epsilon_x$ that’s small enough, you can use any smaller positive $\epsilon$ just as well, so you might as well assume that $\epsilon_x<1$. Can you take it from there?