# Show that, for each nonempty subset $\mathcal{N}$ of $\mathbb{N}$, the function...

Show that, for each nonempty subset $$\mathcal{N}$$ of $$\mathbb{N}$$, the function $$d(x,y) = \sum_{n \in \mathcal{N}} n^{-1}|x_{n}-y_{n}|$$ is not a metric on the set $$c_{0}= \{x= (x_{n})_{n=1}^{\infty}: x_{n} \rightarrow 0$$ as $$n \rightarrow \infty \}$$

I have used the usual definition of the metric space to solve this problem.

1) $$d(x,y)=0$$ iff $$x=y$$.

This is clear, when we let $$x=y$$, we will get $$0$$.

2) $$d(x,y) = d(y,x)$$

This is also clear due to the property of absolute value.

3) I am not sure about how to use the triangle inequality. We know that $$x_{n}$$ converges to zero. Does that imply $$y_{n}$$ is convergent, too? I don't what exactly I should show here.

Any help is greatly appreciated.

The question is slightly tricky! $$d$$ is not a metric when the subset is the whole of $$\mathbb N$$ because it is not even finite: consider the sequence $$x_n=\frac 1 {\ln (n+1)}$$. If the subset is not the whole of $$\mathbb N$$ pick $$k$$ which is not in this set. If $$x_n=y_n=0$$ for all $$n \neq k$$, $$x_k=0$$ and $$y_k=1$$ the we get $$d((x_n), (y_n))=0$$ but $$(x_n) \neq (y_n)$$ so $$d$$ is not a metric.
Hint: you are not done with 1). You have done one direction of the if and only if statement, namely: if $$x=y$$ then $$d(x,y)=0$$. You must still consider the other direction, namely: if $$d(x,y)=0$$ then $$x=y$$.