Showing that a subset of $\ell^1$ is totally bounded Let $A=\{ f \in \ell^1\, :\,\text{ for each natural number}\, n\,\text{we have}\, |f(n)|<1/2^n \}$.
Show that it is totally bounded in $\ell^1$ and find


*

*the interior of $A$ in $\ell^1$

*the closure of $A$ in $\ell^1$
To prove that $A$ is totally bounded it suffices to show A is compact in $\ell^1$, right? But I do not know how to prove it. And also I would like to know how to find interior and closure of $A$.
 A: First, in a complete metric space, being totally bounded is the same as being precompact.
There is a simple characterization of precompact (totally bounded) sets of $\ell^1$.

A subset $A\subset \ell^1$ is precompact if and only if
  
  
*
  
*For all $n$ we have $\sup_{f\in A} |f(n)| < \infty$, and
  
*$\sup_{f\in A} \sum\limits_{n=m}^\infty |f(n)| \to 0$ as $m\to\infty$.
  

This is a special case of the Kolmogorov-Riesz theorem. 
Let us look at your set $A$. The first condition is trivially satisfied: at every point the value of $|f(n)|$ can be no greater than $\frac1{2^n }<\infty$. The second condition is also satisfied: 
$$\sum_{n=m}^\infty |f(n)|  < \sum_{n=m}^\infty \frac1{2^n}=\frac1{2^{m-1}}\overset{m\to\infty}{\longrightarrow} 0$$
Thus, $A$ is totally bounded.
Moreover, it is not hard to see that $A$ cannot contain any interior points. That is, for every point in $A$ we can find points not in $A$ that are arbitrarily close to it. Let $f\in A$ and $\varepsilon>0$. Let $n_0$ be so large that $\frac2{2^{n_0}}<\varepsilon$. Then set
$$\tilde{f}(n)=\left\{\begin{array}{ll}f(n), & \text{if}\,n\not=n_0,\\
\frac1{2^{n_0}}, & \text{if}\, n=n_0.\end{array}\right.$$
Then $\tilde{f}\not\in A$, because it is too large at $n_0$, but it is $\varepsilon$-close to $f$:
$$\|f-\tilde{f}\|_1 = |f(n_0)-\frac1{2^{n_0}}| < \frac2{2^{n_0}} < \varepsilon.$$
So the interior of $A$ is the empty set.
The closure of $A$ is given by
$$\overline{A}=\{f\in\ell^1\,:\, |f(n)|\le \frac1{2^n}\,\forall\,n\}.$$
This is also not hard to see (try it!).
