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.

  • $\begingroup$ What have you tried? $\endgroup$
    – azif00
    Jun 30 at 3:08
  • $\begingroup$ @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 $\ell_1$. $\endgroup$ Jun 30 at 3:31

1 Answer 1


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 $\partial P = P$.


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