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?

  • 3
    $\begingroup$ Note: You are not trying to show that the two metrics are strongly equivalent; you are trying to show that they are topologically equivalent. In fact, the two metrics need not be strongly equivalent, since $d'$ is bounded but $d$ need not be. $\endgroup$ – Arturo Magidin Jun 10 '12 at 2:29
  • $\begingroup$ Requiring $d(x,y)\leq c_1\,d'(x,y)$ is for some $c_1$ and $d'(x,y)\leq c_1\,d(x,y)$ for some $c_2$ would show that $i$ and its inverse are Lipschitz continuous. This is a stronger condition than $i$ and its inverse being continuous (which is what is required). $\endgroup$ – matt Jun 10 '12 at 2:30

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?

  • $\begingroup$ If $\varepsilon_x < 1$ then we will also have $B_{d'}(x,\varepsilon_x) \subset U$, so that $U$ is also $d'$-open. Also if we let $\delta_y < 1$ be such that $B_{d'}(y,\delta_y) \subset V$, then $B_d(y,\delta_y) \subset V$ as well, so $V$ is $d$-open. Since all open sets in $(X,d)$ are open in $(X,d')$ and vice versa, $(X,d)$ is topologically equivalent to $(X,d')$, a bounded metric space. This works, right? $\endgroup$ – Alex Petzke Jun 10 '12 at 20:43
  • $\begingroup$ @Alex: Yes, that’s exactly right. $\endgroup$ – Brian M. Scott Jun 10 '12 at 20:49

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