Any help on this would be appreciated.
I'm trying to prove that the subspace $(E,\rho)$ is compact.
$$E = \{\{x_n\}_n \in X: |x_n|\leq1/(3^n)\text{ for every }n\}$$
$$X=\{\{x_n\}_n \in X: \sum \limits^\infty |x_n| < \infty\}$$
$\rho(x,y$) = $\sum \limits^\infty$ |$x_n - y_n$|
So as I understand it, the metric space is the space of sequences of numbers that converge in absolute value. We're restricting that space to a subset E, and E is given as sequences where the first item in the sequence is less than 1/3, the second is less than 1/9, etc.
So I know that if $(X,\rho)$ is compact and $E$ is a subset of $X$, $(E, \rho)$ is compact iff $E$ is closed in $X$. So I need to prove that $E$ is closed.
I'm just having trouble showing that the set is closed. I wanted to prove by contradiction, and say that if $E$ isn't closed, there must be some sequence in $E$ where the sequence converges outside of $E$. I'm confused because the limit of a sequence is a number (don't all these sequences converge to $0$??), but how can the number $0$ be in $E$ if $E$ is a set of sequences? Yeah, I'm confused.
The space $\mathcal{l}^{1}(\mathbb{N})$ consisting of absolutely summable sequences $x=\{ x_{n} \}_{n=1}^{\infty}$ is an infinite dimensional Banach space under the norm $$ \|x\|_{1} = \sum_{n}|x_{n}|. $$ And your metric $\rho$ is generated by this norm: $\rho(x,y)=\|x-y\|_1$. The closed unit ball of $X$ is not compact because $X$ is inifinite dimensional. So you cannot just show that $E$ is closed subset of $X$ or of its closed unit ball in order to prove compactness. That's not enough.
To show that $E= \{ x \in X : |x_n| \le 1/3^{n} \}$ is compact, suppose $\{ x_{k}\}_{k=1}^{\infty} \subset E$. Each $x_k$ is itself a sequence with $n$ element $(x_k)_n$. That is, $$ x_k = \{ (x_k)_1, (x_k)_2, (x_k)_3,\cdots \}. $$ For a fixed $j$, the sequence $\{ (x_{k})_j \}_{k=1}^{\infty}$ is a bounded sequence with uniform bound $1/3^{j}$. So you can choose a subsequence $\{ x_{k_{l}}\}_{l=1}^{\infty}\subset E$ such that $\{(x_{k_{l}})_1\}_{l=1}^{\infty}$ converges. Then you can choose a subsequence $\{ x_{k_{l_{m}}}\}_{m=1}^{\infty}\subset E$ so that $\{ (x_{k_{l_{m}}})_1\}_{m=1}^{\infty}$ and $\{ (x_{k_{l_{m}}})_2\}_{m=1}^{\infty}$ both converge. By using the Cantor diagonalization process, you can find a subsequence $\{ x_{k_{n}}\}_{n=1}^{\infty}\subset E$ of the original sequence $\{ x_k \}_{k=1}^{\infty} \subset E$ such that the following limits exists $$ y_j = \lim_{k} \;(x_{k_{n}})_j ,\;\;\; j=1,2,3,\cdots. $$ Because $x_{k_{n}} \in E$, then $|(x_{k_{n}})_{j}| \le 1/3^{j}$ and, therefore, $|y_j| \le 1/3^{j}$ follows from the above limit, which puts $y \in E$. To show that $\{ x_{k_{n}} \}_{n=1}^{\infty}$ converges in $X$ to $y \in E$, let $\epsilon > 0$ be given. Choose $J$ large enough that $$ \sum_{j=J+1}^{\infty}\frac{2}{3^{j}} < \epsilon/2. $$ Then choose $N$ large enough that $$ \sum_{j=1}^{J}|(x_{k_{n}})_j-y_j| < \epsilon/2,\;\;\; n \ge N. $$ Then $\rho(x_{k_{n}},y) < \epsilon$ for $n \ge N$. Because $\epsilon > 0$ was arbitrary, then $\lim_{n} x_{k_{n}}=y$ in the metric space. Hence $E$ is sequentially compact and, therefore, compact.