Lemma 14.1.8 (Riesz).label Let $E$ be an infinite-dimensional normed vector space over $K \in \RC$ and $\seq{\alpha_n}\subset (0, 1)$, then there exists $\seq{x_n}\subset \partial B_{E}(0, 1)$ such that
Proof. Let $N \in \natp$ and suppose inductively that $\bracsn{x_n|1 \le n < N}$ have been constructed to satisfy the lemma for all $1 \le n < N$. Let $F_{N} = \text{span}\bracsn{x_n|1 \le n < N}$, then $F_{N}$ is finite-dimensional, and closed by Proposition 13.9.8. As $E$ is infinite-dimensional, $E/F_{N}$ is non-trivial, and there exists $y_{N} \in E$ such that $\norm{y_N + F_N}_{E/F_N}> \alpha_{N}$. By Lemma 13.6.2, there exists $z_{N} \in B_{E}(0, 1)$ such that $z_{N} + F_{N} = y_{N} + F_{N}$. As $z_{N} + F_{N} \ne 0$, $z_{N} \ne 0$ as well. Let $x_{N} = z_{N}/\norm{z_N}_{E}\in \partial B_{E}(0, 1)$, then
Thus $d(x_{N}, \text{span}\bracsn{x_n|1 \le n < N}) = \norm{x_N + F_N}_{E/F}> \alpha_{N}$.$\square$
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