# Subsets and balls

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 $$Y$$.

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$$y_n=\begin{cases}x_n&\text{ if }nThen $$(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 r2In 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.