# Prove that $(M,d)$ is a metric space

Hello I have problems with exercise

Let $$M$$ be the space of all real sequences. Given $$x = (x_i)_i \; , y = (y_i)_i \in M$$, define

$$d(x,y) = \displaystyle\sum_{i=1}^\infty{\displaystyle\frac{1}{i^2} \; min \{ |x_i-y_i| , 1 \}}$$

Prove that $$(M, d)$$ is a metric space. Let $$\{ x^{(k)} \}_k$$ be a sequence in $$M$$. Prove that $$x^{(k)} \rightarrow x$$ for $$d$$ if and only if $${x_i}^{(k)} \rightarrow x_i$$ for all $$i \in \mathbb{N}.$$

My attempt

$$(i) \; d(x,x)=0$$

$$d(x,x)= \displaystyle\sum_{i=1}^\infty{\displaystyle\frac{1}{i^2} \; min \{ |x_i-x_i| , 1 \}} = \displaystyle\sum_{i=1}^\infty{\displaystyle\frac{1}{i^2} \; min \{ |0| , 1 \}} = \displaystyle\sum_{i=1}^\infty{\displaystyle\frac{1}{i^2}} \cdot 0 = 0$$

$$(ii) \; If \; x \neq y$$ then $$d(x,y) > 0$$

$$d(x,y) = \displaystyle\sum_{i=1}^\infty{\displaystyle\frac{1}{i^2} \; min \{ |x_i-y_i| , 1 \}}$$ (This expression is always greater than zero)

$$(iii) \; d(x,y)=d(y,x)$$

$$d(x,y) = \displaystyle\sum_{i=1}^\infty{\displaystyle\frac{1}{i^2} \; min \{ |x_i-y_i| , 1 \}} = \displaystyle\sum_{i=1}^\infty{\displaystyle\frac{1}{i^2} \; min \{ |y_i-x_i| , 1 \}} = d(y,x)$$

$$(iv) \; d(x,z) \leq{} d(x,y) + d(y,z)$$ (I have problems with triangular inequality)

I don't know how to prove this part:

Prove that $$x^{(k)} \rightarrow x$$ for $$d$$ if and only if $${x_i}^{(k)} \rightarrow x_i$$ for all $$i \in \mathbb{N}.$$

Thanks

Concerning the triangular inequality, note that, for each $$i\in\Bbb N$$,\begin{align}\min\{|x_i-z_i|,1\}&\leqslant\min\{|x_i-y_i|+|y_i-z_i|,1\}\\&\leqslant\min\{|x_i-y_i|,1\}+\min\{|y_i-z_i|,1\};\end{align}see this question.
If $$\lim_{k\to\infty}x^{(k)}=x$$ and if $$i\in\Bbb N$$, take $$\varepsilon>0$$. Then, if $$k$$ is large enough, $$d\left(x^{(k)},x\right)<\frac\varepsilon{i^2}$$. In particular, $$\frac1{i^2}\min\left\{\left|x_i^{(k)}-x_i\right|,1\right\}<\frac\varepsilon{i^2}$$, and therefore $$\min\left\{\left|x_i^{(k)}-x_i\right|,1\right\}<\varepsilon$$, which implies that $$\left|x_i^{(k)}-x_i\right|<\varepsilon$$.
Finally, if $$\lim_{i\to\infty}x_i^{(k)}=x_i$$ for each $$i\in\Bbb N$$ and if $$\varepsilon>0$$, take $$M\in\Bbb N$$ such that $$\sum_{i=M+1}^\infty\frac1{i^2}<\frac\varepsilon2$$. For each $$i\leqslant M$$, take $$N_i\in\Bbb N$$ such that $$k\geqslant N_i\implies\frac1{i^2}\left|x_i^{(k)}-x_i\right|<\frac\varepsilon{2N}$$. Then$$k\geqslant\max\{N_1,N_2,\ldots,N_M\}\implies d\left(x^{(k)},x\right)<\frac\varepsilon2+\frac\varepsilon2=\varepsilon.$$