Let $X=R$ and $d_1$ be the normal norm induced metric, and $d_2=\frac{d_1}{d_1+1}$. I want to show that $d_1$ and $d_2$ determine the same topology.

Now, I know that the metric topologies induced by each of $d_1$ and $d_2$, have basis consisting of open balls in $\mathbb{R}$. So if I want to show that the metrics determine the same topology, I then want to show $B_{d_1}(x_1,r_1)\subseteq B_{d_2}(x_2,r_2)$ and $B_{d_2}(x_2,r_2)\subseteq B_{d_1}(x_1,r_1)$ for some $x_1, x_2\in \mathbb{R}$ and $r_1, r_2>0$

First I pick $y\in\mathbb{R}, \epsilon>0$ and consider

\begin{align*} B_{d_1}(y,\epsilon) &= \{z\ |\ d_1(y,z)<\epsilon\} \\ &\supseteq \{z\ |\ d_1(y,x)+d_1(x,z)<\epsilon\} \\ &= \{z\ |\ d_1(x,z)<\epsilon-d_1(y,x)\} \\ &= \{z\ |\ \frac{d_1(x,z)}{d_1(x,z)+1}<\frac{\epsilon-d_1(y,x)}{d_1(x,z)+1}\} \\ &= \{z\ |\ d_2(x,z)<\frac{\epsilon-d_1(y,x)}{d_1(x,z)+1}\}\\ &= B_{d_2}(x,\frac{\epsilon-d_1(y,x)}{d_1(x,z)+1}) \end{align*}

So we have one of the inclusions. Now I'm not sure if this is even allowed? My doubts come from the fact the the ball $B_{d_2}(x,\frac{\epsilon-d_1(y,x)}{d_1(x,z)+1})$ have radius dependent of $z$ which is not fixed.

  • $\begingroup$ Show that there are constants $C_1, C_2 $ such that $$C_1d_1(x,y)\le d_2(x,y)\le C_2d_1(x,y).$$ Then $d_1$ and $d_2$ are said to be equivalent and they induces the same topology. $\endgroup$
    – Bumblebee
    Nov 29, 2017 at 17:38
  • 1
    $\begingroup$ The two metrics are not equivalent since there is no $c > 0$ with $ d_1(x,y) \leq c \cdot d_2(x,y)$ as the inequality $n \leq c \frac{n}{n + 1}$ does not hold for $n > c$. $\endgroup$
    – njlieta
    Nov 29, 2017 at 17:51

1 Answer 1


Your arguments use some weird claims and manipulations with $x$, $y$ and $z$, whereas the deal is simple. We have to show the following. For each $x\in X$ and for each $r>0$ there exist $r_1,r_2>0$ such that $B_{d_1}(x,r_1)\subset B_{d_2}(x,r)$ and $B_{d_2}(x,r_2)\subset B_{d_1}(x,r)$. Since $d_2(x,y)\le d_1(x,y)$ for each $y\in X$, it suffices to put $r_1=r$. It suffices to put $r_2=\frac r{1+r}$, because if $d_2(x,y)= \frac {d_1(x,y)}{1+ d_1(x,y)}<\frac r{1+r}$ then $d_1(x,y)<r$, because the function $\frac r{1+r}=1-\frac 1{1+r}$ is monotically increasing.


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