# Proving that the sequence space is a metric space in $\mathbb{R}$

I need to show that $$(X, d)$$ is a metric space:

$$X$$ is the sequence space where its elements are sequences of real numbers, and $$d : X \times X \rightarrow \mathbb{R}$$ where d is defined as:

$$d(x,y) = \sum_{j=1}^\infty\frac{1}{2^j} \frac{|x_j - y_j|}{1 + |x_j - y_j|}$$

I think that I need to prove the right $$\frac{|x_j - y_j|}{1 + |x_j - y_j|}$$ is montone increasing but I have no clue how to start with proving the triangle inequality for this expression. Any help is appreciated.

• Show (e.g., by calculus) that $\varphi(t)=\frac{t}{1+t}$ is increasing. Commented Oct 9, 2018 at 12:16

Hint: First, the function $$f(x)=\frac{x}{1+x}$$ is monotone increasing when $$x \geq 0$$, which can be proved by methods like taking derivative.

Second, since $$f(x)=\frac{x}{1+x}$$ is monotone increasing, prove the triangle inequality holds for each component, which is $$\frac{|x_j - y_j|}{1 + |x_j - y_j|}+\frac{|y_j-z_j|}{1 + |y_j-z_j|} \geq \frac{|x_j-z_j|}{1 + |x_j-z_j|}$$.

Third, prove the metric satisfies the triangle inequality.

Let $$f(x) = \frac{x}{1 + x}$$. We wish to show that, for $$x, y \ge 0$$, we have $$f(x + y) \le f(x) + f(y)$$ (i.e. the function is subadditive over $$[0, \infty)$$). I hope you can see how this would prove triangle inequality for the proposed metric.

Suppose $$x, y \ge 0$$. Consider \begin{align*} f(x + y) \le f(x) + f(y) &\iff \frac{x + y}{1 + x + y} \le \frac{x}{1 + x} + \frac{y}{1 + y} \\ &\iff 1 - \frac{1}{1 + x + y} \le 2 - \frac{1}{1 + x} - \frac{1}{1 + y} \\ &\iff \frac{1}{1 + x} + \frac{1}{1 + y} \le 1 + \frac{1}{1 + x + y} \\ &\iff (1 + x + y)(1 + y) + (1 + x + y)(1 + x) \le (2 + x + y)(1 + x)(1 + y) \\ &\iff 1 + x + y \le (1 + x)(1 + y) \\ &\iff xy \ge 0, \end{align*} which is true.

You can also use the representation $$f(x) = 1 - \frac{1}{1 + x}$$ to quickly conclude that $$f$$ increases over $$[0, \infty)$$.

The most important idea is here, which shows that

If $$d: X \times X \to \mathbb{R}$$ is a metric, then so is $$d'(x,y) = \frac{d(x,y)}{1+d(x,y)}$$.

So in our case we apply it to $$d(x,y) = |x-y|$$ on the reals. The quotient metric has the pleasant property that it's bounded by $$1$$.

Then our formula just becomes a weighted sum of metrics, all with weights $$>0$$ and the weights $$\frac{1}{2^j}$$ serve mostly to ensure that the series is absolutely convergent (as the $$n$$-th term is bounded above by $$\frac{1}{2^n}$$, so the series sum never grows above $$1$$ too).

Symmetry is clear and the triangle inequality just becomes an infinite sum of component triangle inequalities (because we know that the quotients form a metric by themselves). The sum can only be $$0$$ iff all terms are etc. So it's then easy to check the metric properties.

We know the function $$f:R\rightarrow R$$ given by $$f(x)=\frac {x}{1+x}$$ is increasing on $$[0,\infty)$$ since $$f'>0\ on\ [0,\infty)$$.Hence $$|x_n-y_n|≤|x_n-z_n|+|z_n-y_n|$$ so that $$\frac {|x_n-y_n|}{1+|x_n-y_n|}≤ \frac {|x_n-z_n|+|z_n-y_n|}{1+|x_n-z_n|+|z_n-y_n|}=\frac {|x_n-z_n|}{1+|x_n-z_n|+|z_n-y_n|} + \frac {|z_n-y_n|}{1+|x_n-z_n|+|z_n-y_n|}≤\frac{|x_n-z_n|}{1+|x_n-z_n|}+\frac {|z_n-y_n|}{1+|z_n-y_n|}$$

Now multiply by $$\frac{1}{2^n}$$ both first and last expression and then take sum over $$n$$.