I have to prove that this is a metric: $$d(x,y):=\sum_{i=0}^\infty \frac{|x_i -y_i|}{2^i (1+|x_i-y_i|)}$$ The only thing that I can't prove is the triangle inequality, it's really hard. I know that $|x_i -y_i|\le |x_i -z_i|+|z_i -y_i|$, and I want to get: $$\frac{1}{2^i (1+|x_i-y_i|)} \le \frac{1}{2^i (1+|x_i-z_i|)}+\frac{1}{2^i (1+|z_i-y_i|)}$$ (I think...) The closest thing I've got is: $2^i (1+|x_i-y_i|)\le 2^i (1+|x_i-z_i|)+2^i (1+|z_i-y_i|)$ but... that's not quite what i want. I don't know how to get this...

  • $\begingroup$ It might help to fix your numerators. Your desired inequality should have $|x-y|$ and similar expressions in the numerators, not $1$. $\endgroup$
    – Potato
    Commented Aug 21, 2013 at 4:12

1 Answer 1


Do you know that

$$d(a,b) = \frac{|a-b|}{1 + |a-b|}$$

is a metric on $\mathbb{C}$? First prove that this is a metric. The triangle inequality which you prove here will immediately imply what you want.

To prove triangle inequality in this case, you may use that the function $f(t) = \frac{t}{1+t}$ is monotonic.

EDIT: As requested in the comments, I will complete the proof as follows:

$$ |a-b| + |b-c| \geq |a-c| \Rightarrow \frac{|a-b|+ |b-c|}{1 + |a-b| + |b-c|} \geq \frac{|a-c|}{1 + |a-c|} = d(a,c)$$

(using the fact that the function $f$ is increasing)


$$d(a,b) + d(b,c) = \frac{|a-b|}{1 + |a-b|} + \frac{|b-c|}{1+|b-c|} \geq \frac{|a-b|+ |b-c|}{1 + |a-b| + |b-c|}$$

Combining above two, we get the triangle inequality.

Generalities: In fact there is nothing special about the metric on $|\cdot|$ on $\mathbb{C}$. If $(X,d)$ is any metric space, then $d_{1}$ is an equivalent metric on $X$ where

$$d_{1}(x,y) = \frac{d(x,y)}{1 + d(x,y)}$$

Note that $d_{1}(x,y) < 1 \;\forall x,y \in X$.

And then $d_{2}$ is a metric on $X^{\mathbb{N}}$ (the space of all $X-$ valued sequences) where $d_{2}$ is given as

$$ d_{2}\left((x_{i}),(y_{i})\right) = \sum\limits_{i=1}^{\infty}\frac{1}{2^{i}}d_{1}(x_{i},y_{i}) $$

  • $\begingroup$ You can use the triangle inequality for $|\cdot|$ and the fact that the function I gave is monotonic (infact increasing) to conclude the triangle inequality for $d$. $\endgroup$ Commented Aug 21, 2013 at 4:38
  • $\begingroup$ The inequality in your last display does not seem obvious to me. (My point is that simply noting the function is monotonic is not enough. You really do have to do some work.) $\endgroup$
    – Potato
    Commented Aug 21, 2013 at 7:30
  • $\begingroup$ Thanks @Vishal, how ever i feel that the inequalitys are wrong, shouldn't they be facing the other way? Also, I don't understand whay the first implication is true, and when did you used f(t)? Thanks $\endgroup$
    – Ana Galois
    Commented Aug 21, 2013 at 11:47
  • $\begingroup$ @AnaGalois Which inequality are you referring to? $\endgroup$ Commented Aug 21, 2013 at 12:19
  • $\begingroup$ @Vishal. Maybe I'm just tired of this problem, but can you explain why you say this: |a-b|+|b-c|=<|a-c|? Isn't it |a-b|+|b-c|>=|a-c|? Thanks $\endgroup$
    – Ana Galois
    Commented Aug 21, 2013 at 12:28

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