Show $d_{1}(x_{n},x)\rightarrow{0}$ if and only if $d_{2}(x_{n},x)\rightarrow{0}$.

Question

Let $X=(0,\infty)$. Define two metrics on $X$ by

$d_{1}(x,y)=|x-y|$ and

$d_{2}(x,y)=|x-y|+|\frac{1}{x}-\frac{1}{y}|$

for all $x, y \in{X}$.

Let $(x_{n})$ be a sequence in $X$ and $x\in{X}$.

Show $d_{1}(x_{n},x)\rightarrow{0}$ if and only if $d_{2}(x_{n},x)\rightarrow{0}$.

Note: $d_{1}$ and $d_{2}$ are then said to be equivalent metrics on $X$.

Since $d_{1}$ and $d_{2}$ are equivalent metrics, will it be sufficient to show $c*d_{2}(x,y)\leq{d_{1}(x,y)}\leq{C*d_{2}(x,y)}$?

So far I have the following: $d_{1}(x,y)=|x-y|\leq{|x-y|+|\frac{1}{x}-\frac{1}{y}|}=d_{2}(x,y)$. Therefore, $d_{1}(x,y)\leq{C*d_{2}(x,y)}$ where $C=1$. Is there a $c$ that would satisfy the left-hand side of the inequality?

Any help would be appreciated. (this is not homework)

Answer 1

The first condition (both metrics induce the same notion of convergent sequence) is called topological equivalence of metrics. The second condition (existence of $c>0,C>0$ such that ...) is called strong equivalence of metrics. Strong imlpies topological but not vice versa.

Your example is a case where where we have topological but not strong equivalence: $d_1(\frac1n,\frac2n)=\frac1n$, but $d_2(\frac1n,\frac2n)=n+\frac1n$, so any $c>0$ with $cd_2(x,y)\le d_1(x,y)$ for all $x,y$ would have to be $<\frac1{n^2}$ for all $n$, contradiction. Note also that $(\frac1n)_{n\in\mathbb N}$ is a Cauchy sequence (non-convergent in $X$ though) with respect to $d_1$, but not with respect to $d_2$.

Answer 2

Let's prove this using $\epsilon - \delta$ argument. Fix $x$, and consider $d_2(x_n,x) = d_1(x_n,x) + \dfrac{d_1(x_n,x)}{|x_n|\cdot |x|}$. We now prove the forward direction. That is if $d_1(x_n,x) \to 0$ as $n \to \infty$, then $d_2(x_n,x) \to 0$. let $\epsilon = \dfrac{x}{2} > 0$, then there is $N_1$ such that if $n > N_1$ then $x - |x_n| = |x| - |x_n| \leq d_1(x_n,x) < \epsilon = \dfrac{x}{2}$. So $|x_n| > \dfrac{x}{2}$. Thus: $d_2(x_n,x) < d_1(x_n,x)\cdot \left(1 + \dfrac{2}{x^2}\right)$. Now there is an $N_2$ such that if $n > N_2$ then $d_1(x_n,x) < \dfrac{\epsilon x^2}{x^2 + 2}$. So choose $N = max\{N_1, N_2\}$, then if $n > N$ then $d_2(x_n,x) < \dfrac{\epsilon x^2}{x^2 + 2}\cdot \left(1 + \dfrac{2}{x^2}\right) = \epsilon$. So we've shown that: $d_2(x_n,x) \to 0$ as $n \to \infty$.

For the reverse direction, observe that as you've done it: $d_1(x_n,x) < d_2(x_n,x)$. This means if for a given $\epsilon > 0$, if $d_2(x_n,x) < \epsilon$, then $d_1(x_n,x) < d_2(x_n,x) < \epsilon$. So $d_1(x_n,x) < \epsilon$. This means if $d_2(x_n,x) \to 0$, then $d_1(x_n,x) \to 0$ also.