Proof, [Lemma III.9.1, SW99]. (1): Assume without loss of generality that $A \ne \emptyset$. Let $\bracsn{\lambda_n}_{0}^{\infty} \subset (0, \infty)$ such that $\sum_{n \in \natz}\lambda_{n} = 1$. For each $n \in \natz$, let $r_{n} = \lambda_{n+1}^{2}$ and $A_{n} \subset A$ be finite such that $A \subset \bigcup_{x \in A_n}B_{E}(x, r_{n})$.

Define $B_{0} = \lambda_{0}^{-1}A_{0}$. For each $n \in \natp$, write $A_{n} = \bracsn{x_j}_{1}^{k}$. By definition of $A_{n-1}$, there exists $\bracsn{y_j}_{1}^{k} \subset A_{n-1}$ such that $d(x_{j}, y_{j}) < r_{n-1}$ for all $1 \le j \le k$. For every $1 \le j \le k$, let $z_{j} = (x_{j} - y_{j})/\lambda_{n}$, and define $B_{n} = \bracsn{z_j}_{1}^{k}$.

By the above construction, $A_{n} \subset \sum_{j = 0}^{n} \lambda_{j} B_{j}$ for all $n \in \natz$. As $\sum_{n \in \natz}\lambda_{n} = 1$,

Finally, for each $n \in \natp$, $B_{n}$ is finite with $\norm{z}_{E} \le r_{n-1}/\lambda_{n}= \lambda_{n}$ for all $z \in B_{n}$. As $B_{0}$ is finite as well, any enumeration of $\bracs{0}\cup \bigcup_{n \in \natz}B_{n}$ yields a null sequence.

(2): Using (1) and Mazur’s Theorem, assume without loss of generality that there exists a null sequence $\seq{x_n}\subset E$ such that $A$ is the closed convex hull of $\seq{x_n}$.

Since $\seq{x_n}\subset E$ is a null sequence, there exists $\seq{\lambda_n}\subset [1, \infty)$ such that:

1. (i)

   $\lambda_{n} \to \infty$ as $n \to \infty$.

2. (ii)

   $\lambda_{n} x_{n} \to 0$ as $n \to \infty$.

Let $B = \ol{\aconv}(\seq{\lambda_n x_n})$, then by Mazur’s Theorem, $B$ is a compact, convex, and circled subset of $E$ with $\seq{x_n}\subset B$. In addition, $\seq{x_n}\subset E_{B}$ with $\norm{x_n}_{E_B}\le \lambda_{n}^{-1}$ for all $n \in \natp$. Thus $\seq{x_n}$ is a null sequence in $E_{B}$ as well.

Now, let $A'$ be the closed convex hull of $\seq{x_n}$ with respect to $E_{B}$. Since the inclusion $E_{B} \to E$ is continuous, $A'$ is a compact convex set in $E$ by Proposition 5.16.3. As such, $A' = A$ by Proposition 5.5.3 and Proposition 5.16.4. Therefore $A$ is a compact subset of $E_{B}$.$\square$