Spring 2011 - Problem 5

Solution.

1. Consider

   $$A = \set{x \mid x\p{n} \in \set{0, 1}} \setminus \set{0}$$

   i.e., the set of binary strings excluding the zero string. This is uncountable: if that were not the case, then there exists a bijection $A \to \Z$, which we denote $A = \set{x_k}_{k\in\Z}$. But consider $x\p{n} = 1 - x_n\p{n}$ for $n \in \Z$, i.e., at each $n \in \Z$, $x\p{n} \neq x_n\p{n}$ and $x \in A$. By construction, $x \in A$ and by assumption, $A = \set{x_k}_{k\in\Z}$, so there exists $k \in \Z$ such that $x = x_k$. But $x\p{k} \neq x_k\p{k}$, so $x \neq x_k$, a contradiction. Hence, $A$ must be uncountable.

   By construction, for any $x \in A$, there exists $n \in \Z$ such that $x\p{n} = 1$. Since $\abs{x\p{n}} \leq 1$ for all $n \in \Z$, this means $\norm{x}_{\ell^\infty}$. Similarly, for $x_\alpha, x_\beta \in A$ with $\alpha \neq \beta$, then $x_\alpha \neq x_\beta$, so there exists $n \in \Z$ with $x_\alpha\p{n} \neq x_\beta\p{n}$. By construction, $\abs{x_\alpha\p{n} - x_\beta\p{n}} = 1$ and so $\abs{x_\alpha - x_\beta}$ attains its maximum, i.e., $\norm{x_\alpha - x_\beta}_{\ell^\infty} = 1$.

2. Let $\set{x_\alpha}_{\alpha \in A}$ be the uncountable subset we constructed in (1) and suppose $\set{y_n}_n$ is a countable dense subset. Then for any $\alpha \in A$, there exists $n_\alpha \in \N$ such that $\norm{x_\alpha - y_{n_\alpha}}_{\ell^\infty} \leq \frac{1}{4}$, by the axiom of choice,

   Notice that if $\alpha \neq \beta$, then

   $$\begin{aligned} 1 &\leq \norm{x_\alpha - x_\beta}_{\ell^\infty} \\ &\leq \norm{x_\alpha - y_{n_\alpha}}_{\ell^\infty} + \norm{y_{n_\alpha} - y_{n_\beta}}_{\ell^\infty} + \norm{y_{n_\beta} - x_\beta}_{\ell^\infty} \\ &\leq \frac{1}{2} + \norm{y_{n_\alpha} - y_{n_\beta}}_{\ell^\infty} \\ \implies \frac{1}{2} &\leq \norm{y_{n_\alpha} - y_{n_\beta}}_{\ell^\infty}. \end{aligned}$$

   Thus, $y_{n_\alpha} \neq y_{n_\beta}$, so the mapping $A \ni \alpha \mapsto n_\alpha$ is injective. But this is impossible, since this implies that $A$ is at most countable, so no dense subset could have existed to begin with.

3. We know that in a Banach space, if $X^*$ is separable, then so is $X$ (see Spring 2011 - Problem 5). Thus, if $\ell^1\p{\Z}$ were reflexive, we would have

   $$\p{\ell^\infty\p{\Z}}^* \simeq \ell^1\p{\Z},$$

   and $\ell^1\p{\Z}$ is separable, e.g., via

   $$\set{\sum_{k=1}^n q_ne_{k_n} \st n \in \N,\ q_n \in \Q,\ k_n \in \Z}$$

   where $e_n\p{k} = 1$ if $k = n$ and $0$ otherwise. But this implies that $\ell^\infty\p{\Z}$ is also separable, which we established is impossible in (2). Thus, $\ell^1\p{\Z}$ is not reflexive, which completes the proof.