# Show that two metrics known not to be strongly equivalent actually induce the same topology.

Suppose on $\mathbb{R}$, we have the usual Euclidean metric, $\rho_{1}(x,y) = \Vert x-y \Vert$, and also the metric $\rho_{2}(x,y) = \displaystyle \frac{\rho_{1}(x,y)}{1+\rho_{1}(x,y)}$.

I need to show that the topologies that $\rho_{1}$ and $\rho_{2}$ respectively induce are equal to each other (i.e., if $\rho_{1}$ induces $\tau_{1}$ and $\rho_{2}$ induces $\tau_{2}$, then $\tau_{1} = \tau_{2}$).

I understand that $\rho_{1}$ and $\rho_{2}$ are not strongly equivalent metrics on $\mathbb{R}$, so I am looking to use them as an example of two metrics that induce the same topology but are not strongly equivalent.

The only problem is I don't know how to show that $\rho_{1}$ and $\rho_{2}$ induce the same topology...Could someone please let me know how to do this? Thanks. I would prefer something very elementary rather than something very technical.

I am putting a bounty on this question for a fully worked solution.

• @BrianM.Scott, I edited the question to reflect this.
– user100463
May 3 '16 at 21:52
• each open set in $\tau_1$ will be open in $\tau_2$ and conversely May 3 '16 at 21:54

In fact something more general is true: if $\langle X,\rho\rangle$ is any metric space, and we define

$$\rho_1:X\times X\to\Bbb R:\langle x,y\rangle\mapsto\frac{\rho(x,y)}{1+\rho(x,y)}\;,$$

then $\rho_1$ is a metric on $X$ and is equivalent to $\rho$ (i.e., generates the same topology).

HINT: To show the equivalence, it suffices to show that for each $\epsilon>0$ and $x\in X$,

• if $y\in B_\rho(x,\epsilon)$, then there is $\delta>0$ such that $B_{\rho_1}(x,\delta)\subseteq B_\rho(x,\epsilon)$, and
• if $y\in B_{\rho_1}(x,\epsilon)$, then there is $\delta>0$ such that $B_\rho(x,\delta)\subseteq B_{\rho_1}(x,\epsilon)$.

If $\tau$ and $\tau_1$ are the topologies generated by $\rho$ and $\rho_1$, respectively, the first of these points shows that $\tau\subseteq\tau_1$, and the second shows that $\tau_1\subseteq\tau$.

For the first, you can begin by using the triangle inequality to show that if $\alpha=\epsilon-\rho(x,y)$, then $B_\rho(y,\alpha)\subseteq B_\rho(x,\epsilon)$. Then you just need to find a $\delta>0$ such that $B_{\rho_1}(y,\delta)\subseteq B_\rho(y,\alpha)$. The second point can be approached similarly.

It may be helpful to notice that if

$$b=\frac{a}{1+a}\;,$$

then

$$a=\frac{b}{1-b}\;.$$

• actually, I just started a bounty on this question. If you could answer every part of the question in detail, I will award you the +50.
– user100463
May 15 '16 at 16:56
• or at least more of the question in detail - like maybe the first bullet point with all the triangle inequalities done out. That should be helpful.
– user100463
May 16 '16 at 0:28

Theorem: Let $(X,\rho)$ be a metric space and consider the function $d:X\times X\to \mathbb{R}$, defined by $d(x,y):=\rho(x,y)/(1+\rho(x,y))$. Then $d$ is a metric and induces on $X$ the same topology than $\rho$.

Proof: Consider the function $f:[0,\infty)\to\mathbb{R}$, $f(x)=x/(1+x)$. Then $d(x,y)=f(\rho(x,y))$. Notice that $f$ is an increasing and continuous function (you can verify this by examining the first derivative of $f$). Also notice that $f$ is a subadditive function. In fact, if $a,b\geq0$ then $$f(a+b)\leq f(ab+a+b)=\frac{ab+a+b}{1+ab+a+b}\leq\frac{2ab+a+b}{1+ab+a+b}=f(a)+f(b),$$ where the first inequality is due to the fact that $f$ is increasing.

Claim 1: $d$ is a metric on $X$.

It is enough to show that $d$ satisfies the triangle inequality. Since $f$ is increasing and subadditive we have that: $$d(x,y)=f(\rho(x,y))\leq f(\rho(x,z)+\rho(z,y))\leq f(\rho(x,z))+f(\rho(z,y))=d(x,z)+d(z,y).$$

Claim 2: For every $x\in X$ and every $\varepsilon>0$ there exists $\delta>0$ such that $B_\rho(x,\delta)\subset B_d(x,\varepsilon)$.

Since $f$ is continuos at $0$, for any $\varepsilon>0$ given, there exist $\delta>0$ such that: $$0\leq a<\delta\Longrightarrow f(a)<\varepsilon.$$ Thus, for any $x,y\in X$: $$\rho(x,y)<\delta\Longrightarrow d(x,y)<\varepsilon.$$

Claim 3: For any $a,b\geq0$: $f(b)<\frac{f(a)}{2}\Longrightarrow b<\frac{a}{2}$.

It is enough to prove its (equivalent) contrapositive: for any $a,b\geq0$: $a\leq2b\Longrightarrow f(a)\leq2 f(b)$.

Let $a,b\geq0$ such that $a\leq2b$. Since $f$ is increasing and subadditive, we have that: $$f(a)\leq f(b+b)\leq f(b)+f(b).$$

Claim 4: For every $x\in X$ and every $\varepsilon>0$ there exists $\delta>0$ such that $B_d(x,\delta)\subset B_\rho(x,\varepsilon)$.

By claim 3 we have that: $$d(x,y)=f(\rho(x,y))<\frac{f(\varepsilon)}{2}\Longrightarrow \rho(x,y)<\frac{\varepsilon}{2}<\varepsilon.$$

Then for any given $x\in X$ and $\varepsilon>0$, choose $\delta=\frac{f(\varepsilon)}{2}$ to satisfy the claim.

Thus we have proved that $d$ and $\rho$ induce the same topology on $X$. $\blacksquare$

Bonus Track: Actually the same proof works to prove this more general result: Let $(X,\rho)$ be a metric space and consider the function $d:X\times X\to \mathbb{R}$, defined by $d(x,y):=f(\rho(x,y))$, where $f:[0,\infty)\to\mathbb{R}$ is an increasing, continuous and subadditive function such that $f^{-1}(\{0\})=\{0\}$. Then $d$ is a metric and induces the same topology than $\rho$.

Remark: Notice that $\rho/(1+\rho)$ is a bounded metric. If $\rho$ is not a bounded metric then it cannot be strongly equivalent to $\rho/(1+\rho)$.

• oo. I love bonus tracks!
– user100463
May 16 '16 at 12:09