Showing that $d_1(x,y) := |x - y|$ and $d_2(x,y) := \left| \frac{1}{x} - \frac{1}{y}\right|$ induce same topology 
Let $M := [1,\infty)$ and define $$d_1(x,y) := |x - y| \qquad
\text{and} \qquad d_2(x,y) := \left| \frac{1}{x} -
\frac{1}{y}\right|.$$ Show that $d_1$ and $d_2$ induce the same
  topology on $M$.

As far as I can see, there are two possibilities of showing this: We show either, that the two metrics are strongly equivalent, i.e. that $$c_1 d_1(x,y) \leq d_2(x,y) \leq c_2d_1(x,y)$$ holds for all $x,y \in M$ and $c_1,c_2 > 0$ or we use an exercise in the book by Lee (Topological manifolds, p.22):

$d_1$ and $d_2$ induce same topology $\Leftrightarrow$ For all $x \in
M$, $r > 0$ there exist $r_1,r_2 > 0$ such that $$B_{r_1}^{(d_1)}(x)
\subseteq B_{r}^{(d_2)}(x) \qquad \text{and} \qquad B_{r_2}^{(d_2)}(x)
\subseteq B_{r}^{(d_1)}(x).$$

I think that the two metrics are not strongly equivalent, so I go with the second approach. Let $r_1 = r$ and $y \in B_{r_1}^{(d_1)}(x)$. Then we have $$\left| \frac{1}{x} -
\frac{1}{y}\right| = \left| \frac{y - x}{xy}\right| \leq |x - y| < r.$$ Hence $y \in B_{r}^{(d_2)}(x)$. However, I am a bit clueless about how to show the second inclusion. So my questions:


*

*How do I show the second inclusion?

*Is there an easier way?

 A: It is true that the two metrics are not strongly equivalent, and this tells you that $r_2$ should depend on $x$ and $r$. We can argue as follows to show that such $r_2$ exists.
Since $$\left|\frac1x-\frac1y\right|<r_2\iff \frac1x-r_2<\frac1y<\frac1x+r_2,$$ we choose $r_2<\frac1x$, and the condition above is equivalent to
$$\frac{x}{1+xr_2}<y<\frac{x}{1-xr_2}.$$
Thus,
$$-\frac{x^2r_2}{1+xr_2}<y-x<\frac{x^2r_2}{1-xr_2}.$$
Note that the absolute values of both sides tend to $0$ as $r_2$ tends to $0$. We can thus find some $r_2$ such that the absolute values are less than the given $r$. 
(For example, choose $r_2$ such that $r_2<\frac{1}{2x}$ and $r_2<\frac{r}{2x^2}$.)
A: Choose $r_2$ such that:
$$r_2 < \frac{r}{x(x+r)}$$
note then that $r_2 < \frac1{x}$.
We have:
$$d_2(x,y) < r_2 \implies -r_2 < \frac1{y} - \frac1{x} < r_2 \implies \frac{x}{1+r_2x} < y < \frac{x}{1-r_2x}$$
From the first inequality and choice of $r_2$ we get: 
$$y-x > \frac{x}{x+2r} (-r) > -r$$
From the second inequality we directly get $y-x < r$. Hence $d(x,y) < r$.
A: $d_2(x,y) = d_1(f(x), f(y))$ for $f(x) = \frac{1}{x}$ which is continuous. Use this pointwise continuity to find the $r_1$ and $r_2$. Also $d_1(x,y) = d_2(f(x), f(y))$ as well,so it's symmetric.
