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We have the metric defined: $d: X × X → [0,∞)$ by $$d\big((x_n)_n, (y_n)_n\big) = \sup\{|x_n − y_n|: n ∈ \Bbb N\}$$. Where $X$ is the set of all sequences $(x_n)_{n∈\Bbb N}$ of real numbers such that $\lim_{n→∞}x_n=0$. And $Y$ is the set of all sequences $(y_n)_{n\in\mathbb N}$ of real numbers such that $\sum_{n=1}^\infty |y_n|<∞$. And we know that $Y ⊆ X$.

Now we assume that $(x_n)_n ∈ X\setminus Y$ and $ϵ>0$. We have to show that the ball $B((x_n)_n; ϵ)$ contains elements from $Y$, and why this shows that $Y$ is not closed.

I know for the second part of the problem that if $X\setminus Y$ is closed then $Y$ is open. How can I show that the ball contains elements from $Y$ and that $X\setminus Y$ is closed.

Comment: So the limit is $\lim_{n→∞}x_n=0$ and it is in $X$, and the sum $\sum_{n=1}^\infty|x_n|=∞$ (will diverge) because it is not in $Y$. Basically I have to pick elements $(y_n)_n∈Y$ in the ball around $(x_n)_n$ with converging absolute values ($\sum_{n=1}^\infty |y_n|<∞$) in order to be in $Y$.

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If $(x_n)_{n\in\Bbb N}\in X$ and $r>0$, then there is some $N\in\Bbb N$ such that $n\geqslant N\implies|x_n|\leqslant\frac r2$. Now consider the sequence $(y_n)_{n\in\Bbb N}$ defined by$$y_n=\begin{cases}x_n&\text{ if }n<N\\0&\text{ otherwise.}\end{cases}$$Then $(y_n)_{n\in\Bbb N}\in Y$ and$$d\bigl((x_n)_{n\in\Bbb N},(y_n)_{n\in\Bbb N}\bigr)=\sup_{n\geqslant N}|x_n|\leqslant\frac r2<r.$$In other words, $(y_n)_{n\in\Bbb N}\in B_r\bigl((x_n)_{n\in\Bbb N}\bigr)$. So, $X\setminus Y$ is not open, and therefore $Y$ is not closed.

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