# Using epsilon-balls on metric spaces

Need help with a metric space epsilon ball question..

If $$\ell_1$$ is the set of all sequences of real numbers $$(x_n)_{n=1}^\infty$$ such that $$\sum_{i=1}^{\infty} |x_i| < \infty$$ with metric $$d_1((x_n)_{n=1}^\infty, (y_n)_{n=1}^\infty) = \sum_{i=1}^{\infty} |x_i - y_i|$$

Show an example of completeness.

• What have you tried? Jun 30 at 3:08
• @CraigLutic : Did you mean "$P\subseteq\ell_1$" instead of "$P\in\ell_1$"? Because the latter means $P$ is just a single sequence in $\ell_1$. Jun 30 at 3:31

Consider the complement, $$\ell_{1} - P$$, which is the set of all $$\textbf{x}\in \ell_{1}$$ such that there is some $$n \in \mathbb{N}$$ for which $$|x_{n+1}| > |x_{n}|$$. I claim $$\overline{\ell_{1} - P} = \ell_{1}$$.
Let $$\textbf{x} \in \ell_{1}$$ and let $$B_{d_{1}}(\textbf{x},r)$$ be a neighbourhood of $$\textbf{x}$$. We note that $$x_{n} \to 0$$ as $$\sum_{n=1}^{\infty}|x_{n}|$$ converges. This implies that $$x_{n}$$ is Cauchy, hence, for all $$\varepsilon > 0$$, there exists some $$N \in \mathbb{N}$$ such that for all $$n,m > N$$, $$|x_{n} - x_{m}| < \varepsilon$$. There then exists some $$N \in \mathbb{N}$$ such that $$|x_{N} - x_{N+1}| < r/2$$. If $$\textbf{x}$$ satisfies the property that $$|x_{n+1}| > |x_{n}|$$ for some $$n \in \mathbb{N}$$, then we need not to work anymore. If not, define a new sequence $$\textbf{y} = (y_{n})_{n=1}^{\infty}$$ such that $$y_{n} = x_{n}$$ for each $$n$$ except at $$N$$. We replace $$x_{N+1}$$ with $$x_{N+1}+r/2$$. This gives us the property we want. We have that $$\sum_{n=1}^{\infty}|x_{n} - y_{n}| = |y_{N+1} - x_{N+1}| = |r/2| < r$$. Hence, $$\textbf{y} \in \ell_{1} - P$$ and $$\textbf{y} \in B_{d_{1}}(\textbf{x},r)$$. Therefore, $$\textbf{y} \in \overline{\ell_{1} - P}$$. The claim follows.
The above gives us that $$\text{Int}(P) = \emptyset$$. If we find $$\overline{P}$$, we also get $$\partial P$$. Now consider a sequence of elements in $$P$$, $$(\textbf{x}_{n})_{n=1}^{\infty}$$ that converges to a point $$\textbf{x} \in \ell_{1}$$. Non-strict monotonicity is preserved under limits, so $$\textbf{x} \in P$$. This shows that $$P$$ is closed and so $$\overline{P} = P$$ and also gives us that $$\partial P = P$$.