Suppose $X = \{\{x_n\}_{n \in \mathbb{N}} : x_n \in \mathbb{R}$ $\forall n\geq 1\}$. Prove that $d: X \times X \rightarrow \mathbb{R}$ such that $d(\{x_n\}_{n \in \mathbb{N}}, \{y_n\}_{n \in \mathbb{N}}) = \sum_{n = 1}^{\infty} 2^{-n}\frac{|x_n - y_n|}{1 + |x_n - y_n|}$ is a metric on $X$.

My attempt:

I was given the hint that I should check each of the following: (1) a function $f(x) = \frac{x}{x+1}$ is montone increasing, (2) the triangle inequality applies to $\frac{|x_n - y_n|}{(1 + |x_n - y_n|)}$ and (3) the metric itself satisfies the inequality.

(1) $f$ is monotone increasing:

If $x \geq 0$, $f'(x) = \frac{1}{(1 + x)^2} > 0$.

So $f$ is increasing.

(2) Triangle inequality applies:

We have $|x_n-y_n|≤|x_n-z_n|+|z_n-y_n|$ and so

$\frac {|x_n-y_n|}{1+|x_n-y_n|}≤ \frac {|x_n-z_n|+|z_n-y_n|}{1+|x_n-z_n|+|z_n-y_n|}=\frac {|x_n-z_n|}{1+|x_n-z_n|+|z_n-y_n|} + \frac {|z_n-y_n|}{1+|x_n-z_n|+|z_n-y_n|}≤\frac{|x_n-z_n|}{1+|x_n-z_n|}+\frac {|z_n-y_n|}{1+|z_n-y_n|}$

Therefore $\frac{|x_n - z_n|}{1 + |x_n - z_n|}+\frac{|z_n-y_n|}{1 + |z_n-y_n|} \geq \frac{|x_n-y_n|}{1 + |x_n-y_n|}$.

(3) Metric satisfies triangle inequality:

We know $\frac{|x_n - z_n|}{1 + |x_n - z_n|}+\frac{|z_n-y_n|}{1 + |z_n-y_n|} \geq \frac{|x_n-y_n|}{1 + |x_n-y_n|}$. So

$$2^{-n}(\frac{|x_n - z_n|}{1 + |x_n - z_n|}+\frac{|z_n-y_n|}{1 + |z_n-y_n|}) \geq 2^{-n}\frac{|x_n-y_n|}{1 + |x_n-y_n|}$$


$$\sum_{n = 1}^{\infty} 2^{-n}(\frac{|x_n - z_n|}{1 + |x_n - z_n|}+\frac{|z_n-y_n|}{1 + |z_n-y_n|}) \geq \sum_{n = 1}^{\infty} 2^{-n}\frac{|x_n-y_n|}{1 + |x_n-y_n|}$$


First of all, is this correct? If so, is that it? Is that sufficient to show that $d$ is a metric on $X$? Although I understood how to prove the hints, I still am not sure how these hints actually show that $d$ is a metric on $X$, presuming they are sufficient to show this. Any insight is much appreciated.


The definition of a metric $d$ has 3 requirements for all points $x, y \in X$.

  1. $d(x, y)=0 \iff x=y$
  2. $d(x, y)=d(y, x)$
  3. triangle inequality.

You've proved the last requirement successfully, now all you need to do is prove the first two (this is quite straight forward).

  • 1
    $\begingroup$ Actually nevermind that last comment. Thank you! $\endgroup$ Mar 12 at 2:58
  • 1
    $\begingroup$ Sorry, don't you also need to show that $d(x, y) \geq 0$? How would one go about doing that in this case? @jMdA $\endgroup$ Mar 12 at 3:02
  • 1
    $\begingroup$ No, that's a consequence of the triangle inequality and symmetry $2d(x,y) = d(x, y)+d(y,x)\geq d(x, x)=0$. $\endgroup$
    – jMdA
    Mar 12 at 3:04
  • $\begingroup$ One last thing. When showing the $\implies$ direction in the first one, is this correct: $d(\{x_n\}, \{y_n \}) = 0 \implies \sum_{n = 1}^{\infty} 2^{-n}\frac{|x_n - y_n|}{1 + |x_n - y_n|} = 0 \implies \lim_{n \rightarrow \infty} 2^{-n} \frac{|x_n - y_n|}{1 + |x_n - y_n|} = 0 \implies \frac{|x_n - y_n|}{1 + |x_n - y_n|} = 0 \implies x_n = y_n$ $\endgroup$ Mar 12 at 3:16
  • 1
    $\begingroup$ Not exactly, you have to use that it is a sum of non-negative numbers that sums to zero, therefore each of the terms has to be zero. The limit isn't enough for each of the terms to be zero. $\endgroup$
    – jMdA
    Mar 12 at 3:35

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