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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.

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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.

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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 $x=y$.

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