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Given a set $X$, define a metric by $d_1=\frac{d(x,y)}{1+d(x,y)}$ for all $x,y\in X$. I want to show that $d_1$ is a metric equivalent with $d$, a given metric.

Here is my attempt so far:

Given a set $U\subseteq X$ which is open in the $d$ metric and $x\in U$, there exists an $\varepsilon>0$ such that $B_{d}(x,\varepsilon)\subseteq U$. We want to show that there exists a $\delta>0$ such that $B_{d_1}(x,\delta)\subseteq U$, and one way to do that is to show that there exists a $\delta>0$ such that $B_{d_1}(x,\delta)\subseteq B_{d}(x,\varepsilon).$

Given any $y\in B_{d}(x,\varepsilon)$, we have $d(x,y)<\varepsilon$. Since $d_1(x,y)\leq d(x,y)$, we know $d_1(x,y)< \varepsilon$, so that $y\in B_{d_1}(x,\varepsilon)$. Therefore $B_{d}(x,\varepsilon)\subseteq B_{d_1}(x,\varepsilon)$. But this is the opposite of what I want, isn't it?

Can someone help me complete this proof and help me understand what I am doing wrong? Thank you.

Anne Bauval
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user284178
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  • why not to use the definition https://en.wikipedia.org/wiki/Metric_%28mathematics%29#Definition to prove that $d_1$ is a metric ? – Nizar Oct 26 '15 at 20:55
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    The function $x\to \frac{x}{1+x}$ is strickly increasing on $[0,\infty)$. Use this to show that $B_{d}(x,r)$ is contained in $B_{d_1}\left(x, \frac{r}{1+r}\right)$ and that $B_{d_1}(x,r)$ is contained in $B_{d}\left(x,\frac{r}{1-r}\right)$. – Winther Oct 26 '15 at 20:58
  • I am not trying to show $d_1$ is a metric in this case, just that it is equivalent to the metric $d$. – user284178 Oct 26 '15 at 20:59

1 Answers1

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Above answer by @Whyka contains an error (albeit subtle one), so I've decided to post a correct answer.

Since $d_1\leq d$, we have $B_d(x, r)\subseteq B_{d_1}(x, r)$, and so for all $\varepsilon > 0$ and $x\in X$ we can find $\delta > 0$ such that $B_d(x, \delta)\subseteq B_{d_1}(x, \varepsilon)$, indeed just take $\delta = \varepsilon$.

In the other direction, given $\varepsilon > 0$ and $x\in X$ we need to find $\delta > 0$ such that $B_{d_1}(x, \delta)\subseteq B_d(x, \varepsilon)$. In other words, $d_1(x, y) < \delta$ should imply $d(x, y) < \varepsilon$.

Notice that $d_1(x, y) = \frac{d(x, y)}{1+d(x, y)} < \delta$ is equivalent to $d(x, y) < \delta + \delta\cdot d(x, y)$ which is equivalent to $d(x, y) < \frac{\delta}{1-\delta}$ for $0 < \delta < 1$. Now in order for $\frac{\delta}{1-\delta}\leq \varepsilon$ we need to have $\delta\leq \varepsilon - \delta\cdot \varepsilon$ which is equivalent to $\delta \leq \frac{\varepsilon}{1+\varepsilon}$. And indeed we see that if we take $\delta = \frac{\varepsilon}{1+\varepsilon}$ then $0 < \delta < 1$ and so $d_1(x, y) < \delta$ implies $d(x, y) < \varepsilon$. This shows $B_{d_1}(x, \delta)\subseteq B_d(x, \varepsilon)$ for this choice of $\delta > 0$.

Hence the two metrics $d, d_1$ are equivalent.

In fact the argument above shows that $d, d_1$ are uniformly equivalent so that they induce the same uniform structures on $X$.

Additional remarks. The choice of metric $d_1 = \frac{d}{1+d}$ instead of, say, $d_2 = \min(d, 1)$ is convenient because the map $x\mapsto \frac{x}{1+x}$ for $x \geq 0$ is smooth and strictly increasing. This means in particular that it's more useful for arguments where we care about differentiable maps.

Jakobian
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