Showing that a subset of $\ell^1$ is totally bounded

Question

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

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

2. 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 $A$.

Answer 1

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

1. For all $n$ we have $\sup_{f\in A} |f(n)| < \infty$, and
2. $\sup_{f\in A} \sum\limits_{n=m}^\infty |f(n)| \to 0$ 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!).