2

For each $y \in \ell^1$, the following set is a compact subset of $\ell^1$ with the norm topology: $$ A_y = \{x \in \ell^1 \mid \forall n : |x_n| \leq |y_n|\} \subseteq \ell^1 $$

Indeed, let's begin by showing that each $A_y \subseteq \ell^1$, for $y \in \ell^1$, is a compact subset of $\ell^1$.

By A relative compactness criterion in $\ell^p$ , the compactness of $A_y$ is equivalent to all these three conditions:

  1. $A_y$ is bounded

  2. $A_y$ is closed

  3. $$\underset{N \to \infty}{\operatorname{lim}} \underset{x \in A_y}{\operatorname{sup}} \sum_{n = N}^\infty |x_n| = 0$$

Condition (1.) is clear since for $x \in A_y$ we have $\|x\|_1 \leq \|y\|_1$.

Condition (2.) can be seen as follows: For each $n \in \mathbb{N}$ define $$ A_{y, n} = \{x \in \ell^1 \mid |x_n| \leq |y_n|\} \subseteq \ell^1 $$ Then we can write $A_{y, n} = \pi_n^{-1}([-|y_n|, |y_n|])$, so $A_{y, n}$ is the preimage of a closed set under a continuous map, hence closed. Next, we have $$ A_y = \bigcap_{n \in \mathbb{N}} A_{y, n} $$ So $A_y$ is an intersection of closed sets, hence itself closed.

Finally, condition (3.) easily follows from $$ \underset{N \to \infty}{\operatorname{lim}} \underset{x \in A_y}{\operatorname{sup}} \sum_{n = N}^\infty |x_n| \leq \underset{N \to \infty}{\operatorname{lim}} \sum_{n = N}^\infty |y_n| = 0 $$

Now, let's say that a family of compact sets $\mathcal{A}$ is cofinal in the set of all compact subsets of $\ell^1$ if for each compact $K \subseteq \ell^1$, there is some $K' \in \mathcal{A}$ with $K \subseteq K'$.

Is the family $\{A_y\}_{y \in \ell^1}$ cofinal in the set of all compact subsets of $\ell^1$?

Equivalently, is the following statement true? $$ \forall K \subseteq \ell^1 \text{ compact } \exists y \in \ell^1 \ \forall x \in K \ \forall n \in \mathbb{N} : |x_n| \leq |y_n| $$

Smiley1000
  • 4,219

1 Answers1

3

What about $$ K = \{ \frac 1n e_n: \ n\in \mathbb N\} \cup \{0\} $$ which is compact in $l^1$ but not contained in any $A_y$.

Note that there is a related result: if $(x_k)$ converges in $l^1$ then there is a subsequence $(x_{k_n})$ and $y\in l^1$ such that $x_{k_n} \in A_y$ for all $n$. This is a consequence of the proof of completeness of $L^p$ spaces (and part of the proof that converging sequence have pointwise a.e. converging subsequences).

daw
  • 54,637
  • 2
  • 44
  • 85