# How to complete this proof to show that the metric $d'(x,y) = d(x,y) / (1 + d(x,y))$ gives the same topology as $d(x,y)$ gives?

This is an exercise problem from Munkres's Topology (Exercise 11 of Section 20 "The Metric Topology", 2nd edition).

Exercise 11: Show that if $d$ is a metric for $X$, then $$d'(x,y) = d(x,y) / (1 + d(x,y))$$ is a bounded metric that gives the topology of $X$.

It is not hard to show that $d'$ is indeed a bounded metric.

For the rest (i.e., $d'$ induces the topology $X$), I found at this site [dbfin] a proof using the following lemma (which is given as the exercise 3 of same section).

Lemma: Let $X$ be a metric space with metric $d$. Let $X'$ denote a space having the same underlying set as $X$. If $d: X' \times X' \to \mathbb{R}$ is continuous, then the topology of $X'$ is finer than the topology of $X$. (Note: its proof can be found here; more elaborations can be found at Metric space and continuous function.)

The proof I found is quite brief:

Solution from [dbfin]: Since the functions $f: f(x) = x / (1 + x)$ and $f^{-1}$ are continuous, $d'$ is continuous in $d$ and vice versa, which means the topologies are the same.

I am quite confused about the condition of the Lemma (if $d: X' \times X' \to \mathbb{R}$ is continuous) and therefore don't understand the details of the proof.

My Problem: In order to show that the two topological spaces $X$ (induced by $d$) and $X’$ (induced by $d’$) are the same. We can show that (1) $X’$ is finer than $X$ and (2) $X$ is finer than $X’$.
To prove the former one (the latter one goes similarly), using the lemma above, we shall first justify its premise (i.e., the “if ” part) $d : X’ \times X’ \to \mathbb{R}$ is continuous. I don’t know how this is achieved in the [dbfin] solution. Could someone offer me more details?

P.S.: The "more standard" approach without the lemma above is given by @Stromael. However, I have found a similar proof at this site after I posted this problem. To avoid duplication, in this post, I would prefer to the solution with the lemma.

• Since a topology is completely defined by its open sets, you need to show that a set is open in the new metric if and only if it's open in the old metric. In this way it becomes a basic delta-epsilon proof. – user45878 Feb 6 '14 at 8:43
• Lemma and its proof are interesting. Thank you for your introducing. – HK Lee Feb 8 '14 at 2:51

• Is this a hint at the solution using the lemma mentioned in the post? If so, following your hint: Both the metric $d: X \times X \to \mathbb{R}$ and the function $f(x) = x / (1 + x), x \ge 0$ are continuous, therefore the function $d: X' \times X' \to \mathbb{R}$ is continuous, satisfying the premise of the lemma. Hence, $X'$ is finer than $X$. Is it right? – hengxin Feb 6 '14 at 14:49
Without doing some book-work, I can't cite any deep or powerful theorems that will solve this for you automatically, so my (at least initial) approach would be more elementary. You need to show that the open sets of $(X,d')$ are the same as those of $(X,d)$. It suffices to show that the open balls in the $d'$-topology are open sets in the $d$-topology, and vice versa.
Let $x\in X$, $r>0$ be fixed, and consider the set $B':=\{y\in X~|~d'(x,y)<r\}$. Can you see how, for each $y\in B'$, one can easily find an open ball in the $d$-topology that is centred on $y$? Then repeat the argument in the converse direction.